New Developments in Geometric Function Theory II

Theorem 1. Let f(z) be defined by (1) and p(z) by (19). Then, $f \in \mathcal{G}S^*_{\mathcal{F}\psi}(m,\beta)$ where $m \in [-1,1]$ and $\beta \geq 1$ , if $$\left|\frac{a_1}{s}\right| \le \left|\frac{3(4c_1^{\beta}(m) - p_1)}{4}\right| \tag{23}$$ and $$\left| \frac{a_2}{s} \right| \le \left| \frac{5 \left[ 8p_2 - 7p_1^2 + 16c_1^{\beta}(m)p_1 + 32\left( (c_1^{\beta}(m))^2 - c_2^{\beta}(m) \right) \right]}{32} \right|. \tag{24}$$ Proof. Let $f \in \mathcal{G}S^*_{\mathcal{F}\psi}(m,\beta)$ where $m \in [-1,1]$ and $\beta \geq 1$ , then $$\frac{\mathcal{G}\left(\mathcal{F}_{\psi}'\right)(z)}{\mathcal{F}_{\psi}(z)} = \sqrt{1 + \omega(z)}.$$ (25) The left-hand side of (25) gives $$\frac{\mathcal{G}\left(\mathcal{F}_{\psi}'\right)(z)}{\mathcal{F}_{\psi}(z)} = 1 + \left(c_1^{\beta}(m) - \frac{a_1}{3s}\right)z + \left(c_2^{\beta}(m) - \frac{c_1^{\beta}(m)a_1}{3s} + \frac{a_2}{5s} - \frac{a_1^2}{9s^2}\right)z^2 + \dots, \tag{26}$$ while the right-hand side gives $$\sqrt{1 + \frac{p(z) - 1}{p(z) + 1}} = 1 + \frac{p_1}{4}z + \left(\frac{p_2}{4} - \frac{5p_1^2}{32}\right)z^2 + \dots$$ (27) Comparing the coefficients of z and $z^2$ in (26) and (27), we obtain $$\frac{a_1}{s} = \frac{3(4c_1^{\beta}(m) - p_1)}{4}$$ and $$\frac{a_2}{s} = \frac{5\left[8p_2 - 7p_1^2 + 16c_1^{\beta}(m)p_1 + 32\left((c_1^{\beta}(m))^2 - c_2^{\beta}(m)\right)\right]}{32},$$ which completes the proof. $\Box$ The next two theorems are concerned with the investigation of certain coefficient problems for the class of functions $\mathcal{G}S^*_{\mathcal{F}\psi}(m,\beta)$ involving the sigmoid function given by (20).

Theorem 2. Let f(z) be defined by (1) and $\gamma(z)$ by (20). Then, $f \in \mathcal{G}S^*_{\mathcal{F}\psi}(m,\beta)$ where $m \in [-1,1]$ and $\beta \geq 1$ if the following condition holds true $$\left| \frac{a_2}{s} - \mu \frac{a_1^2}{s^2} \right| \le \left| \frac{9(4c_1^{\beta}(m) - p_1)^2}{16} \left[ \frac{5[8p_2 - 7p_1^2 + 16c_1^{\beta}(m)p_1 + 32((c_1^{\beta}(m))^2 - c_2^{\beta}(m))]}{18(4c_1^{\beta}(m) - p_1)^2} - \mu \right] \right|. \tag{28}$$ Proof. Let $f \in \mathcal{GS}^*_{\mathcal{F}\psi}(m,\beta)$ where $m \in [-1,1]$ $\beta \geq 1$ and $\mu \in \mathbb{R}$ . Then $$\begin{split} \frac{a_2}{2} - \mu \frac{a_1^2}{s^2} &= \left[ \frac{5[8p_2 - 7p_1^2 + 16c_1^\beta(m)p_1 + 32((c_1^\beta(m))^2 - c_2^\beta(m))]}{32} - \mu \left( \frac{3(4c_1^\beta(m) - p_1)}{4} \right)^2 \right], \\ \frac{a_2}{2} - \mu \frac{a_1^2}{s^2} &= \frac{9(4c_1^\beta(m) - p_1)^2}{16} \left[ \frac{5[8p_2 - 7p_1^2 + 16c_1^\beta(m)p_1 + 32((c_1^\beta(m))^2 - c_2^\beta(m))]}{18(4c_1^\beta(m) - p_1)^2} - \mu \right], \end{split}$$ which finally gives $$\left| \frac{a_2}{2} - \mu \frac{a_1^2}{s^2} \right| \le \left| \frac{9(4c_1^{\beta}(m) - p_1)^2}{16} \left[ \frac{5[8p_2 - 7p_1^2 + 16c_1^{\beta}(m)p_1 + 32((c_1^{\beta}(m))^2 - c_2^{\beta}(m))]}{18(4c_1^{\beta}(m) - p_1)^2} - \mu \right] \right|.$$

Theorem 3. Let f(z) be defined by (1) and $\gamma(z)$ by (20). Then, $f \in \mathcal{GS}^*_{\mathcal{F}\psi}(m,\beta)$ where $m \in [-1,1]$ and $\beta \geq 1$ , if $$\left| \frac{a_1}{s} \right| \le \left| \frac{3(8c_1^{\beta}(m) - 1)}{8} \right| \tag{29}$$ and $$\left| \frac{a_2}{s} \right| \le \left| \frac{5 \left[ 256 (c_1^{\beta}(m))^2 - 128 c_2^{\beta}(m) - 48 c_1^{\beta}(m) - 3 \right]}{128} \right|. \tag{30}$$ Proof. Let $f \in \mathcal{GS}^*_{\mathcal{F}\psi}(m,\beta)$ , where $m \in [-1,1]$ and $\beta \geq 1$ . Then $$\frac{\mathcal{G}(\mathcal{F}'_{\psi})(z)}{\mathcal{F}_{\psi}(z)} = \sqrt{1 + \omega(z)}.$$ The left-hand side of (25) gives $$\frac{\mathcal{G}\left(\mathcal{F}_{\psi}'\right)(z)}{\mathcal{F}_{\psi}(z)} = 1 + \left(c_1^{\beta}(m) - \frac{a_1}{3s}\right)z + \left(c_2^{\beta}(m) - \frac{c_1^{\beta}(m)a_1}{3s} + \frac{a_2}{5s} - \frac{a_1^2}{9s^2}\right)z^2 + \dots, \tag{31}$$ while the right-hand side gives $$\sqrt{1 + \frac{\gamma(z) - 1}{\gamma(z) + 1}} = 1 + \frac{1}{8}z - \frac{5}{128}z^2 + \dots$$ (32) Comparing the coefficients of z and $z^2$ in (26) and (32), we obtain $$\frac{a_1}{s} = \frac{3(8c_1^{\beta}(m) - 1)}{8}$$ and $$\frac{a_2}{c} = \frac{5\left[256(c_1^{\beta}(m))^2 - 128c_2^{\beta}(m) - 48c_1^{\beta}(m) - 3\right]}{128},$$ which completes the proof. $\Box$ Theorems 4–6 involve the investigation of certain coefficient problems of the class $\mathcal{GS}^*_{\mathcal{F}\psi}(m,\beta)$ with respect to the Bell numbers (21).

Theorem 4. Let f(z) be defined by (1) and Q(z) by (21). Then, $f \in \mathcal{GS}^*_{\mathcal{F}\psi}(m,\beta)$ where $m \in [-1,1]$ and $\beta \geq 1$ if the following condition holds true $$\left| \frac{a_2}{s} - \mu \frac{a_1^2}{s^2} \right| \le \left| \frac{5 \left[ 256(c_1^{\beta}(m))^2 - 128c_2^{\beta}(m) - 48c_1^{\beta}(m) - 3 \right]}{36(8c_1^{\beta}(m) - 1)^2} - \mu \right|. \tag{33}$$ Proof. Let $f \in \mathcal{GS}^*_{\mathcal{F}\psi}(m,\beta)$ where $m \in [-1,1]$ and $\beta \geq 1$ and $\mu \in \mathbb{R}$ . Then $$\frac{a_2}{s} - \mu \frac{a_1^2}{s^2} = \frac{5 \left[ 256(c_1^{\beta}(m))^2 - 128c_2^{\beta}(m) - 48c_1^{\beta}(m) - 3 \right]}{128} - \mu \left( \frac{3(8c_1^{\beta}(m) - 1)}{8} \right)^2,$$ $$\frac{a_2}{2} - \mu \frac{a_1^2}{s^2} = \frac{9(8c_1^{\beta}(m) - 1)^2}{64} \left( \frac{5 \Big[ 256(c_1^{\beta}(m))^2 - 128c_2^{\beta}(m) - 48c_1^{\beta}(m) - 3 \Big]}{36(8c_1^{\beta}(m) - 1)^2} - \mu \right),$$ which finally gives: $$\left|\frac{a_2}{s} - \mu \frac{a_1^2}{s^2}\right| \leq \left|\frac{5\Big[256(c_1^\beta(m))^2 - 128c_2^\beta(m) - 48c_1^\beta(m) - 3\Big]}{36(8c_1^\beta(m) - 1)^2} - \mu\right|.$$

Theorem 5. Let f(z) be defined by (1) and Q(z) by (21). Then, $f \in \mathcal{G}S^*_{\mathcal{F}\psi}(m,\beta)$ where $m \in [-1,1]$ and $\beta \geq 1$ , if $$\left| \frac{a_1}{s} \right| \le \left| \frac{3(4c_1^{\beta}(m) - 1)}{4} \right| \tag{34}$$ and $$\left| \frac{a_2}{s} \right| \le \left| \frac{5\left[ 5 + 16c_1^{\beta}(m) + 32\left( 2(c_1^{\beta}(m))^2 - c_2^{\beta}(m) \right) \right]}{32} \right|. \tag{35}$$ Proof. Let $f \in \mathcal{G}S^*_{\mathcal{F}\psi}(m,\beta)$ where $m \in [-1,1]$ and $\beta \geq 1$ . Then $$\frac{\mathcal{G}\Big(\mathcal{F}'_{\psi}\Big)(z)}{\mathcal{F}_{\psi}(z)} = \sqrt{1 + \omega(z)}.$$ The left-hand side of (25) gives $$\frac{\mathcal{G}\left(\mathcal{F}'_{\psi}\right)(z)}{\mathcal{F}_{\psi}(z)} = 1 + \left(c_{1}^{\beta}(m) - \frac{a_{1}}{3s}\right)z + \left(c_{2}^{\beta}(m) - \frac{c_{1}^{\beta}(m)a_{1}}{3s} + \frac{a_{2}}{5s} - \frac{a_{1}^{2}}{9s^{2}}\right)z^{2} + \dots,$$ while the right-hand side gives $$\sqrt{1 + \frac{\mathcal{Q}(z) - 1}{\mathcal{Q}(z) + 1}} = 1 + \frac{1}{4}z + \frac{3}{32}z^2 + \dots$$ (36) When the coefficients of z and $z^2$ in (26) and (36), are compared, the following values are obtained: $$\frac{a_1}{s} = \frac{3(4c_1^{\beta}(m) - 1)}{4}$$ and $$\frac{a_2}{s} = \frac{5\left[5 + 16c_1^{\beta}(m) + 32\left(2(c_1^{\beta}(m))^2 - c_2^{\beta}(m)\right)\right]}{32}.$$ Hence, the proof is completed. $\Box$ In Theorem 2, the coefficient estimates were established using the Carathéodory function p(z) given by (19). We now consider in the next theorem the coefficient estimates for the class of functions $\mathcal{G}S^*_{\mathcal{F}\psi}(m,\beta)$ using the Bell numbers $\mathcal{Q}(z)$ as given by (21).

Theorem 6. Let f(z) be defined by (1) and Q(z) by (21). Then, $f \in \mathcal{GS}^*_{\mathcal{F}\psi}(m,\beta)$ where $m \in [-1,1]$ and $\beta \geq 1$ if the following condition holds true $$\left| \frac{a_2}{s} - \mu \frac{a_1^2}{s^2} \right| \le \left| \frac{9(4c_1^{\beta}(m) - 1)^2}{16} \left\lceil \frac{5\left[5 + 16c_1^{\beta}(m) + 32\left(2(c_1^{\beta}(m))^2 - c_2^{\beta}(m)\right)\right]}{18(4c_1^{\beta}(m) - 1)^2} - \mu \right| \right|. \tag{37}$$ Proof. Let $f \in \mathcal{GS}^*_{\mathcal{F}\psi}(m,\beta)$ where $m \in [-1,1]$ , $\beta \geq 1$ and $\mu \in \mathbb{R}$ . Then $$\frac{a_2}{2} - \mu \frac{a_1^2}{s^2} = \left[ \frac{5\left[5 + 16c_1^{\beta}(m) + 32\left(2(c_1^{\beta}(m))^2 - c_2^{\beta}(m)\right)\right]}{32} - \mu \left(\frac{3(4c_1^{\beta}(m) - 1)}{4}\right)^2 \right],$$ $$\frac{a_2}{2} - \mu \frac{a_1^2}{s^2} = \frac{9(4c_1^{\beta}(m) - 1)^2}{16} \left[ \frac{5\left[5 + 16c_1^{\beta}(m) + 32\left(2(c_1^{\beta}(m))^2 - c_2^{\beta}(m)\right)\right]}{18(4c_1^{\beta}(m) - p_1)^2} - \mu \right],$$ which finally gives $$\left|\frac{a_2}{2} - \mu \frac{a_1^2}{s^2}\right| \leq \left|\frac{9(4c_1^\beta(m) - 1)^2}{16} \left[\frac{5\left[5 + 16c_1^\beta(m) + 32\left(2(c_1^\beta(m))^2 - c_2^\beta(m)\right)\right]}{18(4c_1^\beta(m) - p_1)^2} - \mu\right]\right|.$$

Theorem 1. Let $f \in \Sigma$ given by (5) belongs to the class $\mathfrak{G}^t_{\Sigma}(x, p, q, \lambda, \beta, \gamma)$ . Then $$|a_2| \leq$$ $$\frac{tx\sqrt{2tx}}{\lambda e^{e^{\frac{1}{2}(1-\lambda^2)}}\sqrt{\left|\left[5(1+2\gamma+6\beta)(tx)^2-8e^{e^{(1-\lambda^2)}}(1+\gamma+2\beta)^2(ptx^2+aq)\right]\right|}}$$ and $$|a_3| \le \frac{t^2 x^2}{4\lambda^2 (1 + \gamma + 2\beta)^2 e^{2e^{(1 - \lambda^2)}}} + \frac{2tx}{5\lambda^2 (1 + 2\gamma + 6\beta) e^{e^{(1 - \lambda^2)}}}.$$ Proof. Let $f \in \mathfrak{G}^t_{\Sigma}(x, p, q, \lambda, \beta, \gamma)$ . From Definition 1, we can write $$(1 - \gamma) \frac{\mathbb{P}_{\lambda} f(\xi)}{\xi} + \gamma (\mathbb{P}_{\lambda} f(\xi))' + \beta \xi (\mathbb{P}_{\lambda} f(\xi))'' = \Omega(x, \varkappa(\xi)) + 1 - a \tag{21}$$ and $$(1-\gamma)\frac{\mathbb{P}_{\lambda}f(w)}{w} + \gamma(\mathbb{P}_{\lambda}f(w))' + \beta w(\mathbb{P}_{\lambda}f(w))'' = \Omega(x,\tau(w)) + 1 - a, \tag{22}$$ where the analytical functions $\varkappa$ and $\tau$ have the form $$\varkappa(\xi) = c_1 \xi + c_2 \xi^2 + c_3 \xi^3 + \cdots, \ (\xi \in \mathfrak{U})$$ and $$\tau(w) = d_1 w + d_2 w^2 + d_3 w^3 + \cdots, (w \in \mathfrak{U}),$$ such that $\varkappa(0) = \tau(0) = 0$ and $|\varkappa(\xi)| < 1$ , $|\tau(w)| < 1$ for all $\xi, w \in \mathfrak{U}$ . From the equalities (21) and (22), we get $$(1 - \gamma) \frac{\mathbb{P}_{\lambda} f(\xi)}{\xi} + \gamma (\mathbb{P}_{\lambda} f(\xi))' + \beta \xi (\mathbb{P}_{\lambda} f(\xi))'' = 1 + h_2(x) c_1 \xi + \left[ h_2(x) c_2 + h_3(x) c_1^2 \right] \xi^2 + \cdots$$ (23) and $$(1-\gamma)\frac{\mathbb{P}_{\lambda}f(w)}{w} + \gamma(\mathbb{P}_{\lambda}f(w))' + \beta w(\mathbb{P}_{\lambda}f(w))'' = 1 + h_2(x)d_1w + \left[h_2(x)d_2 + h_3(x)d_1^2\right]w^2 + \cdots.$$ (24) It is common knowledge that if $$|\varkappa(\xi)| = \left| c_1 \xi + c_2 \xi^2 + c_3 \xi^3 + \dots \right| < 1, \ (\xi \in \mathfrak{U})$$ and $$|\tau(w)| = \left| d_1 w + d_2 w^2 + d_3 w^3 + \cdots \right| < 1, \ (w \in \mathfrak{U}),$$ then $$|c_i| \le 1 \text{ and } |d_i| \le 1 \text{ for all } j \in \mathbb{N}.$$ (25) Equating the coefficients of both sides in (23) and (24), we get $$2\lambda(1+\gamma+2\beta)e^{e^{(1-\lambda^2)}}a_2 = h_2(x)c_1,$$ (26) $$\frac{5}{2}\lambda^2(1+2\gamma+6\beta)e^{e^{\left(1-\lambda^2\right)}}a_3 = h_2(x)c_2 + h_3(x)c_1^2,\tag{27}$$ $$-2\lambda(1+\gamma+2\beta)e^{e^{(1-\lambda^2)}}a_2 = h_2(x)d_1,$$ (28) and $$\frac{5}{2}\lambda^2(1+2\gamma+6\beta)e^{e^{\left(1-\lambda^2\right)}}\left[2a_2^2-a_3\right] = h_2(x)d_2 + h_3(x)d_1^2.$$ (29) It follows from (26) and (28) that $$c_1 = -d_1 \tag{30}$$ and $$8\lambda^{2}(1+\gamma+2\beta)^{2}e^{2e^{\left(1-\lambda^{2}\right)}}a_{2}^{2}=\left[h_{2}(x)\right]^{2}\left(c_{1}^{2}+d_{1}^{2}\right). \tag{31}$$ If we add (27) and (29), we get $$5\lambda^{2}(1+2\gamma+6\beta)e^{e^{\left(1-\lambda^{2}\right)}}a_{2}^{2} = h_{2}(x)(c_{2}+d_{2}) + h_{3}(x)\left(c_{1}^{2}+d_{1}^{2}\right). \tag{32}$$ Replacing the value of $(c_1^2 + d_1^2)$ from (31) in the right hand side of (32), we have $$\left[5(1+2\gamma+6\beta)-8(1+\gamma+2\beta)^{2}e^{e^{(1-\lambda^{2})}}\frac{h_{3}(x)}{[h_{2}(x)]^{2}}\right]\lambda^{2}e^{e^{(1-\lambda^{2})}}a_{2}^{2}$$ $$=h_{2}(x)(c_{2}+d_{2}). \tag{33}$$ Using (3), (25) and (33), we find that $|a_2| \leq$ $$\frac{tx\sqrt{2tx}}{\lambda e^{e^{\frac{1}{2}(1-\lambda^2)}}\sqrt{\left|\left[5(1+2\gamma+6\beta)[tx]^2-8e^{e^{(1-\lambda^2)}}(1+\gamma+2\beta)^2(ptx^2+aq)\right]\right|}}.$$ Moreover, if we subtract (29) from (27), we obtain $$5\lambda^{2}(1+2\gamma+6\beta)e^{e^{\left(1-\lambda^{2}\right)}}\left(a_{3}-a_{2}^{2}\right)=h_{2}(x)(c_{2}-d_{2})+h_{3}(x)\left(c_{1}^{2}-d_{1}^{2}\right). \tag{34}$$ Then, in view of (30) and (31), Equation (34) becomes $$a_3 = \frac{[h_2(x)]^2}{8\lambda^2 (1 + \gamma + 2\beta)^2 e^{2e^{(1-\lambda^2)}}} \left(c_1^2 + d_1^2\right) + \frac{h_2(x)}{5\lambda^2 (1 + 2\gamma + 6\beta) e^{e^{(1-\lambda^2)}}} (c_2 - d_2).$$ By applying (3), we conclude that $$|a_3| \le \frac{t^2 x^2}{4\lambda^2 (1 + \gamma + 2\beta)^2 e^{2e^{(1-\lambda^2)}}} + \frac{2tx}{5\lambda^2 (1 + 2\gamma + 6\beta) e^{e^{(1-\lambda^2)}}}.$$

Theorem 2. Let $f \in \Sigma$ given by (5) belongs to the class $\mathfrak{G}^t_{\Sigma}(x, p, q, \lambda, \beta, \gamma)$ . Then $$|a_{3} - \eta a_{2}^{2}| \leq \begin{cases} \frac{2|tx|}{5\lambda^{2}(1+2\gamma+6\beta)e^{e^{(1-\lambda^{2})}}}, & |\eta - 1| \leq \delta \\ \frac{2(tx)^{3}|1-\eta|}{\lambda^{2}e^{e^{(1-\lambda^{2})}}\left|\left[5(1+2\gamma+6\beta)t^{2}x^{2}-8(1+\gamma+2\beta)^{2}e^{e^{(1-\lambda^{2})}}(ptx^{2}+aq)\right]\right|}, & |\eta - 1| \geq \delta, \end{cases}$$ where $$\delta = \left| 1 - \frac{8(1+\gamma+2\beta)^2 e^{e^{\left(1-\lambda^2\right)}} \left(ptx^2 + aq\right)}{5(1+2\gamma+6\beta)t^2x^2} \right|.$$ Proof. From (33) and (34) $$\begin{split} &a_{3} - \eta a_{2}^{2} \\ &= (1 - \eta) \frac{\left[h_{2}(x)\right]^{3} (c_{2} + d_{2})}{\lambda^{2} e^{e^{(1 - \lambda^{2})}} \left[5(1 + 2\gamma + 6\beta)\left[h_{2}(x)\right]^{2} - 8(1 + \gamma + 2\beta)^{2} e^{e^{(1 - \lambda^{2})}} h_{3}(x)\right]} \\ &+ \frac{h_{2}(x)}{5\lambda^{2} (1 + 2\gamma + 6\beta) e^{e^{(1 - \lambda^{2})}}} (c_{2} - d_{2}) \\ &= h_{2}(x) \left[h(\eta) + \frac{1}{5\lambda^{2} (1 + 2\gamma + 6\beta) e^{e^{(1 - \lambda^{2})}}}\right] c_{2} \\ &+ h_{2}(x) \left[h(\eta) - \frac{1}{5\lambda^{2} (1 + 2\gamma + 6\beta) e^{e^{(1 - \lambda^{2})}}}\right] d_{2}, \end{split}$$ where $$Y(\eta) = \frac{[h_2(x)]^2 (1 - \eta)}{\lambda^2 e^{e^{(1 - \lambda^2)}} \left[ 5(1 + 2\gamma + 6\beta) [h_2(x)]^2 - 8(1 + \gamma + 2\beta)^2 e^{e^{(1 - \lambda^2)}} h_3(x) \right]}'$$ Then, in view of (3), we conclude that $$\left| a_3 - \eta a_2^2 \right| \le \begin{cases} \frac{2|h_2(x)|}{5\lambda^2(1 + 2\gamma + 6\beta)e^{e^{\left(1 - \lambda^2\right)}}} & 0 \le |Y(\eta)| \le \frac{1}{5\lambda^2(1 + 2\gamma + 6\beta)e^{e^{\left(1 - \lambda^2\right)}}}, \\ 2|h_2(x)||Y(\eta)| & |Y(\eta)| \ge \frac{1}{5\lambda^2(1 + 2\gamma + 6\beta)e^{e^{\left(1 - \lambda^2\right)}}}. \end{cases}$$

Corollary 1. Let $f \in \Sigma$ given by (5) belongs to the class $\mathfrak{G}^t_{\Sigma}(x, p, q, \lambda, \gamma)$ . Then $$\frac{tx\sqrt{2tx}}{\lambda e^{e^{\frac{1}{2}(1-\lambda^{2})}}\sqrt{\left|\left[5(1+2\gamma)(tx)^{2}-8e^{e^{(1-\lambda^{2})}}(1+\gamma)^{2}(ptx^{2}+aq)\right]\right|}},$$ $$|a_{3}| \leq \frac{t^{2}x^{2}}{4\lambda^{2}(1+\gamma)^{2}e^{2e^{(1-\lambda^{2})}}} + \frac{2tx}{5\lambda^{2}(1+2\gamma)e^{e^{(1-\lambda^{2})}}},$$ and $$|a_{3}-\eta a_{2}^{2}| \leq \begin{cases} \frac{2|tx|}{5\lambda^{2}(1+2\gamma)e^{e^{(1-\lambda^{2})}}}, & |\eta-1| \leq \Phi\\ \frac{2(tx)^{3}|1-\eta|}{\lambda^{2}e^{e^{(1-\lambda^{2})}}\left|\left[5(1+2\gamma)t^{2}x^{2}-8(1+\gamma)^{2}e^{e^{(1-\lambda^{2})}}(ptx^{2}+aq)\right]\right|}, & |\eta-1| \geq \Phi, \end{cases}$$ where $$\Phi = \left|1 - \frac{8(1+\gamma)^{2}e^{e^{(1-\lambda^{2})}}(ptx^{2}+aq)}{5(1+2\gamma)t^{2}x^{2}}\right|.$$

Corollary 2. Let $f \in \Sigma$ given by (5) belongs to the class $\mathfrak{G}^t_{\Sigma}(x, p, q, \lambda)$ . Then $$|a_{2}| \leq \frac{tx\sqrt{2tx}}{\lambda e^{e^{\frac{1}{2}(1-\lambda^{2})}}\sqrt{\left|\left[15(tx)^{2} - 32e^{e^{(1-\lambda^{2})}}(ptx^{2} + aq)\right]\right|}},$$ $$|a_{3}| \leq \frac{t^{2}x^{2}}{16\lambda^{2}e^{2e^{(1-\lambda^{2})}}} + \frac{2tx}{15\lambda^{2}e^{e^{(1-\lambda^{2})}}},$$ $$and$$ $$|a_{3} - \eta a_{2}^{2}| \leq \begin{cases} \frac{2|tx|}{15\lambda^{2}e^{e^{(1-\lambda^{2})}}}, & |\eta - 1| \leq \left|1 - \frac{32e^{e^{(1-\lambda^{2})}}(ptx^{2} + aq)}{15t^{2}x^{2}}\right| \\ \frac{2(tx)^{3}|1 - \eta|}{\lambda^{2}e^{e^{(1-\lambda^{2})}}\left|\left[15t^{2}x^{2} - 32e^{e^{(1-\lambda^{2})}}(ptx^{2} + aq)\right]\right|}, & |\eta - 1| \geq \left|1 - \frac{32e^{e^{(1-\lambda^{2})}}(ptx^{2} + aq)}{15t^{2}x^{2}}\right|.$$

Corollary 3. Let $f \in \Sigma$ given by (5) belongs to the class $\mathfrak{G}^t_{\Sigma}(x, p, q, \lambda, 0)$ . Then $$\begin{aligned} |a_{2}| &\leq \\ \frac{tx\sqrt{2tx}}{\lambda e^{e^{\frac{1}{2}(1-\lambda^{2})}}\sqrt{\left|\left[5(tx)^{2}-8e^{e^{(1-\lambda^{2})}}(ptx^{2}+aq)\right]\right|}},\\ |a_{3}| &\leq \frac{t^{2}x^{2}}{4\lambda^{2}e^{2e^{(1-\lambda^{2})}}} + \frac{2tx}{5\lambda^{2}e^{e^{(1-\lambda^{2})}}},\\ and\\ |a_{3}-\eta a_{2}^{2}| &\leq \begin{cases} \frac{2|tx|}{5\lambda^{2}e^{e^{(1-\lambda^{2})}}}, & |\eta-1| \leq \Phi\\ \frac{2(tx)^{3}|1-\eta|}{\lambda^{2}e^{e^{(1-\lambda^{2})}}\left|\left[5t^{2}x^{2}-8e^{e^{(1-\lambda^{2})}}(ptx^{2}+aq)\right]\right|}, & |\eta-1| \geq \Phi, \end{cases} \end{aligned}$$ where $$\Phi = \left| 1 - \frac{8e^{e^{\left(1 - \lambda^2\right)}} \left(ptx^2 + aq\right)}{5t^2x^2} \right|.$$

Corollary 4. Let $f \in \Sigma$ given by (5) belongs to the class $\mathfrak{G}^t_{\Sigma}(x, p, q, \beta, \gamma)$ . Then $$|a_{2}| \leq \frac{tx\sqrt{2tx}}{e\sqrt{\left|\left[5(1+2\gamma+6\beta)(tx)^{2}-8e(1+\gamma+2\beta)^{2}(ptx^{2}+aq)\right]\right|}}}$$ $$|a_{3}| \leq \frac{t^{2}x^{2}}{4e^{2}(1+\gamma+2\beta)^{2}} + \frac{2tx}{5e(1+2\gamma+6\beta)}.$$ and $$|a_{3}-\eta a_{2}^{2}| \leq \begin{cases} \frac{2|tx|}{5e(1+2\gamma+6\beta)'} & |\eta-1| \leq \delta \\ \frac{2(tx)^{3}|1-\eta|}{e|\left[5(1+2\gamma+6\beta)t^{2}x^{2}-8e(1+\gamma+2\beta)^{2}(ptx^{2}+aq)\right]\right|}, & |\eta-1| \geq \delta, \end{cases}$$ where $$\delta = \left|1 - \frac{8e(1+\gamma+2\beta)^{2}(ptx^{2}+aq)}{5(1+2\gamma+6\beta)t^{2}x^{2}}\right|.$$

Lemma 1. (See [2].) Let $f(z) = z|z|^{2\beta}h(z)\overline{g(z)}$ be a log-harmonic mapping on E, $0 \notin hg(E)$ . Then, $f \in ST_{LH}(\alpha)$ if and only if $\varphi(z) = \frac{zh(z)}{g(z)} \in S^*(\alpha)$ . In [2], the authors studied the class of $\alpha - spirallike$ functions and proved that, if $f(z) = z|z|^{2\beta}h(z)\overline{g(z)}$ is a log-harmonic mapping on E, $0 \notin hg(E)$ , then f is $\alpha - spirallike$ if $$\operatorname{Re}\left(e^{-i\alpha}\frac{zf_z-\overline{z}f_{\overline{z}}}{f}\right)>0, \qquad 0\leq \alpha<1$$ for all $z \in E$ . We remark that a simply connected domain $\Omega$ in $\mathbb C$ containing the origin is said to be $\alpha-spirallike$ , $-\frac{\pi}{2}<\alpha<\frac{\pi}{2}$ if $w\exp(-te^{i\alpha})\in\Omega$ for all $t\geq 0$ whenever $w\in\Omega$ and that f is an $\alpha-spirallike$ function, if f(E) is an $\alpha-spirallike$ domain. Motivated by this, we define the class of $\alpha-spirallike$ log-harmonic mappings of order $\rho$ as follows:

Lemma 2. ([2]) If $f(z) = z|z|^{2\beta}h(z)\overline{g(z)}$ is log-harmonic on E and $0 \notin hg(E)$ , with $\operatorname{Re}\beta > -\frac{1}{2}$ , then $f \in S_{LH}^{\alpha}(\rho)$ if and only if $\psi(z) = \frac{zh(z)}{g(z)e^{2i\alpha}} \in S^{\alpha}(\rho)$ . In the celebrated paper [11], the authors introduce a new way of studying harmonic functions in Geometric Function Theory. Additionally, many authors investigated the linear combinations of harmonic functions in a plane; see, for example, [12–14]. In Section 2 of this paper, taking the convex-exponent product combination of two elements, a specified class of new log-harmonic functions is constructed. Indeed, we show that, if $f(z)=zh(z)\bar{g}(z)$ is spirallike log-harmonic of order $\rho$ , then by choosing suitable parameters of $\alpha$ and $\gamma$ , the function $F(z)=f(z)|f(z|^{2\gamma})$ is log-harmonic spirallike of order $\alpha$ . Additionally, in Section 3, we provide some examples that are constructed from Section 2.

Theorem 1. Let $f(z) = zh(z)\overline{g(z)} \in ST_{LH}(\rho)$ , $(0 \le \rho < 1)$ with respect to $a \in B_0$ , $\phi \in S^*(\gamma)$ , $(0 \le \gamma < 1)$ and $\alpha$ , $\beta$ be real numbers with $\alpha + \beta = 1$ . Then, $F(z) = f(z)^{\alpha}K(z)^{\beta}$ is starlike log-harmonic mapping of order $\alpha\rho + \beta\gamma$ with respect to a, where $$K(z) = \phi(z) \exp\left\{2\operatorname{Re} \int_0^z \frac{a(s)}{1 - a(s)} \frac{\phi'(s)}{\phi(s)} ds\right\}.$$ Proof. By definition of F, we have $$\frac{F_z}{F} = \alpha \frac{f_z}{f} + \beta \frac{K_z}{K} \quad and \quad \frac{F_{\overline{z}}}{F} = \alpha \frac{f_{\overline{z}}}{f} + \beta \frac{K_{\overline{z}}}{K}. \tag{6}$$ Additionally direct computations show that $$\frac{K_z}{K} = \frac{1}{1 - a(z)} \frac{\phi'(z)}{\phi(z)}, \quad and \quad \frac{\overline{K_{\overline{z}}}}{\overline{K}} = \frac{a(z)}{1 - a(z)} \frac{\phi'(z)}{\phi(z)}. \tag{7}$$ Now, in view of Equations (6) and (7), $$\hat{a}(z) = \frac{\frac{\overline{F_z}}{\overline{F}}}{\frac{F_z}{F}} = \frac{\alpha \frac{\overline{f_z}}{\overline{f}} + \beta \frac{\overline{K_z}}{\overline{K}}}{\alpha \frac{f_z}{f} + \beta \frac{K_z}{K}} = a(z) \frac{\alpha \frac{f_z}{f} + \beta \frac{K_z}{K}}{\alpha \frac{f_z}{f} + \beta \frac{K_z}{K}} = a(z).$$ On the other hand, $$\operatorname{Re} \frac{zF_{z} - \overline{z}F_{\overline{z}}}{F} = \operatorname{Re} \left( \alpha \frac{zf_{z}}{f} + \beta \frac{zK_{z}}{K} \right) - \operatorname{Re} \left( \alpha \frac{z\overline{f_{\overline{z}}}}{\overline{f}} + \beta \frac{z\overline{K_{\overline{z}}}}{\overline{K}} \right)$$ $$= \alpha \operatorname{Re} \left( \frac{zf_{z}}{f} - \frac{z\overline{f_{\overline{z}}}}{\overline{f}} \right) + \beta \operatorname{Re} \left( \frac{zK_{z}}{K} - \frac{z\overline{K_{\overline{z}}}}{\overline{K}} \right)$$ $$> \alpha \rho + \beta \gamma.$$ The above relation shows that F is a log-harmonic starlike function of order $\alpha \rho + \beta \gamma$ , and the proof is complete. $\Box$

Theorem 2. Let $f(z)=zh(z)\overline{g(z)}\in S_{LH}^{\beta}(\rho)$ with respect to $a\in B_0$ and $\gamma$ be a constant with $\text{Re}\gamma>-\frac{1}{2}$ . Then, $F(z)=f(z)|f(z)|^{2\gamma}$ is an $\alpha-$ spirallike log-harmonic mapping of order $\rho$ with respect to $$\hat{a}(z) = \frac{(1+\bar{\gamma})a(z) + \bar{\gamma}}{1+\gamma+\gamma a(z)},$$ where $|\beta| < \frac{\pi}{2}$ and $\alpha = \tan^{-1} \left( \frac{\tan \beta + 2 Im \gamma}{1 + 2 Re \gamma} \right)$ . Proof. By definition of F, we have $$F(z) = f(z)|f(z)|^{2\gamma} = z^{1+\gamma}\overline{z}^{\gamma}H(z)\overline{G(z)},$$ where $$H(z) = h^{1+\gamma}(z)g^{\gamma}(z)$$ and $G(z) = h^{\bar{\gamma}}(z)g^{1+\bar{\gamma}}(z)$ . With a straightforward calculation and using Equation (5), $$\frac{zF_z}{F} = (1+\gamma)\left(1+\frac{zh'(z)}{h(z)}\right) + \gamma\frac{zg'(z)}{g(z)} = \left(1+\frac{zh'(z)}{h(z)}\right)((1+\gamma) + \gamma a(z)),$$ and $$\frac{\bar{z}F_{\bar{z}}}{F} = \gamma \left(1 + \frac{\overline{zh'(z)}}{\overline{h(z)}}\right) + (1 + \gamma) \frac{\overline{zg'(z)}}{\overline{g(z)}} = \left(1 + \frac{\overline{zh'(z)}}{\overline{h(z)}}\right) (\gamma + (1 + \gamma)\overline{a(z)}).$$ If we consider $$\hat{a}(z) = \frac{\overline{\left(\frac{\bar{z}F_{\bar{z}}(z)}{F(z)}\right)}}{\frac{\bar{z}F_{\bar{z}}(z)}{F(z)}},$$ then $$\hat{a}(z) = \frac{\bar{\gamma} + (1 + \bar{\gamma})a(z)}{(1 + \gamma) + \gamma a(z)}.$$ Now, in view of |a(z)| < 1, it easy to see that $|\hat{a}(z)| < 1$ provided that $\left|\frac{\overline{\gamma}}{1+\overline{\gamma}}\right| < 1$ , which evidently holds $|\gamma|^2 < |1+\overline{\gamma}|^2$ since $\text{Re}\gamma > -\frac{1}{2}$ , and this means that F is a log-harmonic function. Additionally, by putting $$\psi(z) = \frac{zH(z)}{G(z)^{e^{2i\alpha}}},$$ we have $$\psi(z) = \frac{zH(z)}{G(z)e^{2i\alpha}} = \frac{zh(z)^{1+\gamma}g(z)^{\gamma}}{(h\bar{\gamma}(z)\sigma^{1+\bar{\gamma}}(z))e^{2i\alpha}}.$$ Then, we obtain $$\begin{split} e^{-i\alpha} \frac{z\psi'(z)}{\psi(z)} &= e^{-i\alpha} + \left[ (1+\gamma)e^{-i\alpha} - \overline{\gamma}e^{i\alpha} \right] \frac{zh'(z)}{h(z)} - \left[ (1+\overline{\gamma})e^{i\alpha} - \gamma e^{-i\alpha} \right] \frac{zg'(z)}{g(z)} \\ &= (-\gamma e^{-i\alpha} + \overline{\gamma}e^{i\alpha}) + \left[ (1+\gamma)e^{-i\alpha} - \overline{\gamma}e^{i\alpha} \right] \left( 1 + \frac{zh'(z)}{h(z)} \right) \\ &- \left[ (1+\overline{\gamma})e^{i\alpha} - \gamma e^{-i\alpha} \right] \frac{zg'(z)}{g(z)}. \end{split}$$ The condition on $\alpha$ ensures that $$(1+\gamma)e^{-i\alpha} - \overline{\gamma}e^{i\alpha} = \frac{\cos\alpha}{\cos\beta}e^{-i\beta}$$ and $(1+\overline{\gamma})e^{i\alpha} - \gamma e^{-i\alpha} = \frac{\cos\alpha}{\cos\beta}e^{i\beta}$ , because by letting $\gamma = \gamma_1 + i\gamma_2$ , the first equality holds true if and only if $\cos \beta \cos \alpha - i(1+2\gamma_1) \sin \alpha \cos \beta + i2\gamma_2 \cos \beta \cos \alpha = \cos \alpha \cos \beta - i \cos \alpha \sin \beta$ or, equivalently, after simplification $$2\gamma_2 \cot \beta - (1 + 2\gamma_1) \tan \alpha \cot \beta = -1$$ or $$\alpha = \tan^{-1} \left( \frac{\tan \beta + 2 \text{Im} \gamma}{1 + 2 \text{Re} \gamma} \right).$$ Thus, by hypothesis, $$\operatorname{Re}\left\{e^{-i\alpha}\frac{z\psi'(z)}{\psi(z)}\right\} = \frac{\cos\alpha}{\cos\beta}\operatorname{Re}\left(e^{-i\beta}\left(1 + \frac{zh'(z)}{h(z)}\right) - e^{i\beta}\frac{zg'(z)}{g(z)}\right) > \rho\cos\alpha$$ and it follows that F is an $\alpha$ -spirallike log-harmonic mapping of order $\rho$ in which the dilation is $\hat{a}(z)$ . $\square$

Theorem 3. Let $f_k(z) = zh_k(z)\overline{g_k}(z) \in S_{LH}^{\beta}(\rho)$ with k = 1, 2 and with respect to the same $a \in B_0$ and $\gamma$ be a constant with $\text{Re}\gamma > -\frac{1}{2}$ . Moreover, let $$F_1(z) = f_1(z)|f_1(z)|^{2\gamma}$$ and $F_2(z) = f_2(z)|f_2(z)|^{2\gamma}$ . Then, $F(z)=F_1^\lambda(z)F_2^{1-\lambda}(z)$ is an $\alpha$ -spirallike log-harmonic mapping of order $\rho$ with respect to $$\hat{a}(z) = \frac{(1+\bar{\gamma})a(z) + \bar{\gamma}}{1+\gamma + \gamma a(z)},$$ where $|\beta| < \frac{\pi}{2}$ and $\alpha = tan^{-1} \left( \frac{tan \beta + 2Im \gamma}{1 + 2Re \gamma} \right)$ . Proof. According to the definitions of $F_1$ and $F_2$ , we have $$F_1^{\lambda}(z) = (f_1(z)|f_1(z)|^{2\gamma})^{\lambda}$$ = $(z|z|^{2\gamma}h_1^{1+\gamma}(z)g_1^{\gamma}(z)\overline{h_1^{\overline{\gamma}}(z)g_1^{1+\overline{\gamma}}(z)})^{\lambda}$ and $$\begin{split} F_2^{1-\lambda}(z) &= (f_2(z)|f_2(z)|^{2\gamma})^{1-\lambda} \\ &= (z|z|^{2\gamma} h_2^{1+\gamma}(z) g_2^{\gamma}(z) \overline{h_2^{\overline{\gamma}}(z) g_2^{1+\overline{\gamma}}(z)})^{1-\lambda}. \end{split}$$ Putting the values of $F_1^{\lambda}$ and $F_2^{1-\lambda}$ on F, we obtain $$\begin{split} F(z) &= (z|z|^{2\gamma} h_1^{1+\gamma}(z) g_1^{\gamma}(z) \overline{h_1^{\overline{\gamma}}(z) g_1^{1+\overline{\gamma}}(z)} \big)^{\lambda} (z|z|^{2\gamma} h_2^{1+\gamma}(z) g_2^{\gamma}(z) \overline{h_2^{\overline{\gamma}}(z) g_2^{1+\overline{\gamma}}(z)} \big)^{1-\lambda} \\ &= z|z|^{2\gamma} H(z) \overline{G(z)}, \end{split}$$ where $$H(z) = h_1(z)^{\lambda(1+\gamma)} g_1(z)^{\lambda\gamma} h_2(z)^{(1-\lambda)(1+\gamma)} g_2(z)^{(1-\lambda)\gamma}$$ (8) and $$G(z) = h_1(z)^{\lambda \overline{\gamma}} g_1(z)^{\lambda (1+\overline{\gamma})} h_2(z)^{(1-\lambda)\overline{\gamma}} g_2(z)^{(1-\lambda)(1+\overline{\gamma})}.$$ (9) Now, we show that the second dilation of F i.e., $\mu(z)$ satisfies the condition $|\mu(z)| < 1$ . For this, since $$\mu(z) = \frac{\frac{\overline{F}_{\overline{z}}(z)}{\overline{F}(z)}}{\frac{F_{\overline{z}}(z)}{F(z)}},$$ we have $$\mu(z) = \frac{\lambda \frac{\overline{F_{1z}(z)}}{\overline{F_{1}(z)}} + (1 - \lambda) \frac{\overline{F_{2z}(z)}}{\overline{F_{2}(z)}}}{\lambda \frac{\overline{F_{1z}(z)}}{\overline{F_{1}(z)}} + (1 - \lambda) \frac{\overline{F_{2z}(z)}}{\overline{F_{2}(z)}}}$$ $$= \frac{\lambda [\overline{\gamma}(1 + \frac{zh'_{1}}{h_{1}}) + (1 + \overline{\gamma}) \frac{zg'_{1}}{g_{1}}] + (1 - \lambda) [\overline{\gamma}(1 + \frac{zh'_{2}}{h_{2}}) + (1 + \overline{\gamma}) \frac{zg'_{2}}{g_{2}}]}{\lambda [(1 + \gamma)(1 + \frac{zh'_{1}}{h_{1}}) + \gamma \frac{zg'_{1}}{g_{1}}] + (1 - \lambda) [(1 + \gamma)(1 + \frac{zh'_{2}}{h_{2}}) + \gamma \frac{zg'_{2}}{g_{2}}]}$$ $$= \frac{\lambda (1 + \frac{zh'_{1}}{h_{1}}) [\overline{\gamma} + (1 + \overline{\gamma})a(z)] + (1 - \lambda)(1 + \frac{zh'_{2}}{h_{2}}) [\overline{\gamma} + (1 + \overline{\gamma})a(z)]}{\lambda (1 + \frac{zh'_{1}}{h_{1}}) [(1 + \gamma) + \gamma a(z)] + (1 - \lambda)(1 + \frac{zh'_{2}}{h_{2}}) [(1 + \gamma) + \gamma a(z)]}$$ $$= \frac{[\lambda (1 + \frac{zh'_{1}}{h_{1}}) + (1 - \lambda)(1 + \frac{zh'_{2}}{h_{2}}) [\overline{\gamma} + (1 + \overline{\gamma})a(z)]}{[\lambda (1 + \frac{zh'_{1}}{h_{1}}) + (1 - \lambda)(1 + \frac{zh'_{2}}{h_{2}}) ][(1 + \gamma) + \gamma a(z)]}$$ $$= \frac{[\overline{\gamma} + (1 + \overline{\gamma})a(z)]}{[(1 + \gamma) + \gamma a(z)]}$$ $$= \frac{(1 + \overline{\gamma})}{(1 + \gamma)} \frac{a(z) + \frac{\overline{\gamma}}{1 + \overline{\gamma}}}{1 + \frac{a(z)\gamma}{1 + z}},$$ (10) and the condition $\text{Re}\gamma > -\frac{1}{2}$ ensures that $|\mu(z)| < 1$ in E, which implies that F is a locally univalent log-harmonic mapping. Now, to prove $$F(z) = F_1^{\lambda}(z)F_2^{1-\lambda}(z) \in S_{LH}^{\alpha}(\rho),$$ we have to show that $\psi(z) = \frac{zH(z)}{G(z)^{e^{2i\alpha}}} \in S^{\alpha}(\rho)$ . However, a direct calculation shows that $$\psi(z) = \frac{zH(z)}{G(z)^{e^{2i\alpha}}} = \frac{[zh_1^{\lambda(1+\gamma)}(z)g_1^{\lambda\gamma}(z)h_2^{(1-\lambda)(1+\gamma)}(z)g_2^{(1-\lambda)\gamma}(z)]}{[h_1^{\lambda\overline{\gamma}}(z)g_1^{\lambda(1+\overline{\gamma})}(z)h_2^{(1-\lambda)\overline{\gamma}}(z)g_2^{(1-\lambda)(1+\overline{\gamma})}(z)]^{e^{2i\alpha}}}.$$ Now, $$\begin{split} &e^{-i\alpha}\frac{z\psi'(z)}{\psi(z)} \\ &= e^{-i\alpha}\left[1 + \lambda(((1+\gamma) - e^{2i\alpha}\overline{\gamma})\frac{zh_1'(z)}{h_1(z)} - ((1+\overline{\gamma})e^{2i\alpha} - \gamma)\frac{zg_1'(z)}{g_1(z)})\right] \\ &+ e^{-i\alpha}\left[(1-\lambda)(((1+\gamma) - e^{2i\alpha}\overline{\gamma})\frac{zh_2'(z)}{h_2(z)} - ((1+\overline{\gamma})e^{2i\alpha} - \gamma)\frac{zg_2'(z)}{g_2(z)})\right] \\ &= -\gamma e^{-i\alpha} + e^{i\alpha}\overline{\gamma} \\ &+ \lambda\left[((1+\gamma)e^{-i\alpha} - e^{i\alpha}\overline{\gamma})(1+\frac{zh_1'(z)}{h_1(z)}) - ((1+\overline{\gamma})e^{i\alpha} - \gamma e^{-i\alpha})\frac{zg_1'(z)}{g_1(z)}\right] \\ &+ (1-\lambda)\left[((1+\gamma)e^{-i\alpha} - e^{i\alpha}\overline{\gamma})(1+\frac{zh_2'(z)}{h_2(z)}) - ((1+\overline{\gamma})e^{i\alpha} - \gamma e^{-i\alpha})\frac{zg_2'(z)}{g_2(z)}\right]. \end{split}$$ By hypothesis, we know that $$(1+\gamma)e^{-i\alpha} - \overline{\gamma}e^{i\alpha} = \frac{\cos\alpha}{\cos\beta}e^{-i\beta}$$ and $(1+\overline{\gamma})e^{i\alpha} - \gamma e^{-i\alpha} = \frac{\cos\alpha}{\cos\beta}e^{i\beta}$ , Re $$\left\{e^{-i\alpha} \frac{z\psi'(z)}{\psi(z)}\right\}$$ $$= \lambda \frac{\cos \alpha}{\cos \beta} \operatorname{Re}\left(e^{-i\beta} \left(1 + \frac{zh_1'(z)}{h_1(z)}\right) - e^{i\beta} \frac{zg_1'(z)}{g_1(z)}\right)$$ $$+ (1 - \lambda) \frac{\cos \alpha}{\cos \beta} \operatorname{Re}\left(e^{-i\beta} \left(1 + \frac{zh_2'(z)}{h_2(z)}\right) - e^{i\beta} \frac{zg_1'(z)}{g_1(z)}\right)$$ $$> \rho \cos \alpha$$ and the proof is completed. $\Box$

Theorem 4. Let $f_k(z) = zh_k(z)\overline{g}_k(z) \in S_{LH}^{\beta}(\rho)$ with respect to $a_k \in B_0(k=1,2)$ . Moreover, suppose that $\text{Re}\gamma > -\frac{1}{2}$ , $$F_1(z) = f_1(z)|f_1(z)|^{2\gamma}$$ and $F_2(z) = f_2(z)|f_2(z)|^{2\gamma}$ . If $$\operatorname{Re}\left[(1-a_1(z)\overline{a}_2(z))\left(1+\frac{zh_1'(z)}{h_1(z)}\right)\overline{\left(1+\frac{zh_2'(z)}{h_2(z)}\right)}\right]\geq 0 \qquad (\textit{for any } z\in E),$$ then $$F(z) = F_1^{\lambda}(z)F_2^{1-\lambda}(z) \in S_{LH}^{\alpha}(\rho),$$ where $$|\beta| < \frac{\pi}{2}$$ , $0 \le \lambda \le 1$ and $\alpha = \tan^{-1}\left(\frac{\tan \beta + 2Im\gamma}{1 + 2Re\gamma}\right)$ . Proof. Using the same argument as in Theorem 3, we have $$F(z) = z|z|^{2\gamma}H(z)\overline{G(z)},$$ where H(z) and G(z) are defined by Equations (8) and (9). Now, we show that the second dilation of F, i.e., $\mu(z)$ , satisfies the condition $|\mu(z)| < 1$ . For this, since $$\mu(z) = \frac{\frac{\overline{F}_{\overline{z}}(z)}{\overline{F}(z)}}{\frac{F_{z}(z)}{\overline{F}(z)}},$$ using a similar argument to the relation Equation (10) of Theorem 3, we have $$|\mu(z)| = \left| \frac{\lambda(1 + \frac{zh_1'}{h_1})[\overline{\gamma} + (1 + \overline{\gamma})a_1(z)] + (1 - \lambda)(1 + \frac{zh_2'}{h_2})[\overline{\gamma} + (1 + \overline{\gamma})a_2(z)]}{\lambda(1 + \frac{zh_1'}{h_1})[(1 + \gamma) + \gamma a_1(z)] + (1 - \lambda)(1 + \frac{zh_2'}{h_2})[(1 + \gamma) + \gamma a_2(z)]} \right|.$$ However, by hypothesis, we obtain $$\left| \lambda (1 + \frac{zh'_1}{h_1})[(1+\gamma) + \gamma a_1(z)] + (1-\lambda)(1 + \frac{zh'_2}{h_2})[(1+\gamma) + \gamma a_2(z)] \right|^2$$ $$- \left| \lambda (1 + \frac{zh'_1}{h_1})[\overline{\gamma} + (1+\overline{\gamma})a_1(z)] + (1-\lambda)(1 + \frac{zh'_2}{h_2})[\overline{\gamma} + (1+\overline{\gamma})a_2(z)] \right|^2$$ $$= (2\operatorname{Re}\gamma + 1) \left( \lambda^2 \left| 1 + \frac{zh'_1}{h_1} \right|^2 (1 - |a_1|^2) + (1-\lambda)^2 \left| 1 + \frac{zh'_2}{h_2} \right|^2 (1 - |a_2|^2) \right)$$ $$+ (2\operatorname{Re}\gamma + 1) \left( 2\lambda(1-\lambda)\operatorname{Re}[(1-a_1\overline{a}_2)(1 + \frac{zh'_1}{h_1})(1 + \frac{zh'_2}{h_2})] \right) > 0.$$ Therefore, $|\mu(z)| < 1$ in E, which implies that F is a locally univalent mapping. Moreover, by following a similar proof to that in Theorem 3, we observe that $$F(z) = F_1^{\lambda}(z)F_2^{1-\lambda}(z) \in S_{LH}^{\alpha}(\rho),$$ and the proof is completed. $\Box$

Theorem 5. Let $f_k(z) = zh_k(z)\overline{g}_k(z)$ be univalent log-harmonic functions with respect to $a_k \in B_0(k=1,2)$ and $\text{Re}\gamma > -\frac{1}{2}$ . Moreover, suppose that $zh_kg_k = \phi_k(z)$ , where $$\phi_k(z) = zexp\left\{2\int_0^z \frac{a_k(t)}{t(1 - a_k(t))}dt\right\}$$ and $$F_1(z) = f_1(z)|f_1(z)|^{2\gamma}$$ and $F_2(z) = f_2(z)|f_2(z)|^{2\gamma}$ . Then, $$F(z) = F_1^{\lambda}(z)F_2^{1-\lambda}(z) \in S_{LH}^{\alpha}(1)$$ where $0 \le \lambda \le 1$ and $\alpha = \tan^{-1} \left( \frac{2 \text{Im} \gamma}{1 + 2 \text{Re} \gamma} \right)$ . Proof. Since $zh_kg_k=\phi_k(z)$ , by definition of $a_k(z)$ and $\phi_k(z)$ , we obtain $$1 + \frac{zh'_k(z)}{h_k(z)} = \frac{1}{1 - a_k(z)} \quad (k = 1, 2).$$ Let $$\mu(z) = \frac{\frac{\overline{F}_{\overline{z}}(z)}{\overline{F}(z)}}{\frac{F_z(z)}{\overline{F}(z)}}.$$ Using a similar argument to the relation in Equation (10) of Theorem 3, we obtain $$|\mu(z)| = \left| \frac{\lambda(1 - a_2(z))[\overline{\gamma} + (1 + \overline{\gamma})a_1(z)] + (1 - \lambda)((1 - a_1(z))[\overline{\gamma} + (1 + \overline{\gamma})a_2(z)]}{\lambda(1 - a_2(z))[(1 + \gamma) + \gamma a_1(z)] + (1 - \lambda)(1 - a_1(z))[(1 + \gamma) + \gamma a_2(z)]} \right|.$$ Now, |u(z)| < 1 is equivalent to $$\begin{split} \psi(\lambda) &:= |\lambda(1-a_2(z))[(1+\gamma)+\gamma a_1(z)] + (1-\lambda)(1-a_1(z))[(1+\gamma)+\gamma a_2(z)]|^2 \\ &- |\lambda(1-a_2(z))[\overline{\gamma}+(1+\overline{\gamma})a_1(z)] + (1-\lambda)((1-a_1(z))[\overline{\gamma}+(1+\overline{\gamma})a_2(z)]|^2 \\ &= (2\text{Re}\gamma+1)[\lambda^2|1-a_2(z)|^2(1-|a_1(z)|^2) \\ &+ 2\lambda(1-\lambda)\text{Re}[(1-a_2(z))(1-\overline{a_1(z)})(1-a_1(z)\overline{a_2(z)})] \\ &+ (1-\lambda)^2|1-a_1(z)|^2(1-|a_2(z)|^2)] > 0. \end{split}$$ However, by taking the derivative of $\psi(\lambda)$ , we have $$\psi'(\lambda) = 2(2\text{Re}\gamma + 1)$$ $$\left[\text{Re}[(1 - a_2(z))(1 - \overline{a_1(z)})(1 - a_1(z)\overline{a_2(z)})] - |1 - a_1(z)|^2(1 - |a_2(z)|^2)\right],$$ which shows that $\psi$ is a continuous monotonic function of $\lambda$ in the interval [0,1]. Since $$\psi(0) = (2\operatorname{Re}\gamma + 1)|1 - a_2(z)|^2(1 - |a_1(z)|^2) > 0$$ and $$\psi(1) = (2\operatorname{Re}\gamma + 1)|1 - a_1(z)|^2(1 - |a_2(z)|^2) > 0,$$ we deduce that $\psi(\lambda) > 0$ for all $\lambda \in [0,1]$ , which implies that F is a locally univalent mapping. Now, to prove $$F = F_1^{\lambda} F_2^{1-\lambda} \in S_{LH}^{\alpha} \tag{11}$$ we have to show that $\psi(z) = \frac{zH(z)}{G(z)e^{2i\alpha}} \in S^{\alpha}(1)$ , where H(z) and G(z) are defined by Equations (8) and (9). A direct computation such as that in Theorem 3 shows that $$\frac{(1+\gamma)e^{-i\alpha}-\overline{\gamma}e^{i\alpha}}{\cos\alpha}=\frac{(1+\overline{\gamma})e^{i\alpha}-\gamma e^{-i\alpha}}{\cos\alpha}=1.$$ Additionally, we note that $$1 + \frac{zh_1'}{h_1} - \frac{zg_1'}{g_1} = 1 + \frac{zh_2'}{h_2} - \frac{zg_2'}{g_2} = 1.$$ Using these relation and the same argument as that made in Theorem 3, we obtain $\psi(z)=\frac{zH(z)}{G(z)e^{2i\alpha}}\in S^{\alpha}(1)$ , and the proof is complete. $\qed$

Theorem 6. Let $f_k(z) = zh_k(z)\overline{g}_k(z)(k=1,2)$ be log-harmonic functions with respect to $a_k \in B_0$ . Moreover, suppose that $zh_kg_k = z$ and $$F_1(z) = f_1(z)|f_1(z)|^{2\gamma}$$ and $F_2(z) = f_2(z)|f_2(z)|^{2\gamma}$ . Then, $$F(z) = F_1^{\lambda}(z)F_2^{1-\lambda}(z) \in S_{LH}^{\alpha}(1),$$ where $0 \le \lambda \le 1$ and $\alpha = tan^{-1} \Big( \frac{2Im\gamma}{(1+2Re\gamma)} \Big)$ . Proof. Since $zh_kg_k = z$ , by definition of $a_k(z)$ , we obtain $$1 + \frac{zh'_k(z)}{h_k(z)} = \frac{1}{1 + a_k(z)} \quad (k = 1, 2).$$ Using the same argument as that in Theorem 5, we obtain our result, but we omit the details. $\Box$

Theorem 1. Assume that the function f(z) is an entire function of q −1 -exponential growth of order 1 and a finite type α less than ξ1, or it is an entire function of q −1 -exponential growth of an order of less than 1. Then, f(z) has a convergent q-Lidstone representation $$f(z) = \sum_{n=0}^{\infty} \left[ D_{q^{-1}}^{2n} f(1) A_n(z) - D_{q^{-1}}^{2n} f(0) B_n(z) \right], \tag{3}$$ where (An)ⁿ and (Bn)ⁿ are the q-Lidstone polynomials defined, respectively, by the generating functions $$\frac{E_q(zw) - E_q(-zw)}{E_q(w) - E_q(-w)} = \sum_{n=0}^{\infty} A_n(z)w^{2n},$$ (4) $$\frac{E_q(zw)E_q(-w) - E_q(-zw)E_q(w)}{E_q(w) - E_q(-w)} = \sum_{n=0}^{\infty} B_n(z) \frac{w^n}{[n]_q!}.$$ (5) Moreover, A0(z) = z, B0(z) = 1 − z, and for n ∈ N, An(z) and Bn(z) satisfy the q-difference equation $$D_{q-1}^2 y_n(z) = y_{n-1}(z) \quad \text{with} \quad y_n(0) = y_n(1) = 0.$$ (6) In [16], AL-Towailb and Mansour proved that the condition $$D_{q^{-1}}^n f(0) = o(\xi_1^n) \quad \text{as } n \to \infty$$ (7) is both sufficient and necessary for expanding an entire function f(z) in the q-Lidstone series $$f(1)A_0(z) - f(0)B_0(z) + D_{q^{-1}}^2 f(1)A_1(z) - D_{q^{-1}}^2 f(0)B_1(z) + \dots,$$ and we noted that Condition (7) is insufficient for the convergence of the following arrangement of the q-Lidstone series: $$\sum_{n=0}^{\infty} D_{q^{-1}}^{2n} f(1) A_n(z) - \sum_{n=0}^{\infty} D_{q^{-1}}^{2n} f(0) B_n(z),$$ and not necessary for the convergence of (3). This paper aimed to obtain a sufficient and necessary condition for a real-valued function to have an absolutely convergent q-Lidstone series expansion (3). To achieve this aim, we introduced generalizations for the class of completely convex functions (2) on a closed interval of form [0, a] (a > 0), and the class of minimal completely convex functions on the interval [0, 1]. This paper is organized as follows. The following section gives the essential notions and basic definitions of q-calculus. Section 3 contains some properties and basic results on q-Lidstone polynomials, which we need in our investigation. In Section 4, we define a q-analog of the class of completely convex functions for the difference operator Dq−¹ . Then, we study the relation of this class to a problem of the representation of functions by the q-Lidstone series. In Section 5, we provide a necessary and sufficient condition for a real function to have an absolutely convergent q-Lidstone series expansion.

Lemma 1. For any $x \in [0,1]$ , we have $$\operatorname{Sin}_{q}\xi_{1}x \leq \xi_{1}x. \tag{11}$$ Proof. Let $f(x) = \xi_1 x - \operatorname{Sin}_q \xi_1 x$ , $x \in [0,1]$ . Then, $D_{q^{-1}} f(x) = \xi_1 (1 - \operatorname{Cos}_q \xi_1 x) \geq 0$ . Therefore, by using (8), we obtain $$f(x) \le f(\frac{x}{q}) \quad (x \in [0,1]),$$ which implies $f(x) \ge \lim_{n \to \infty} f(q^n x) = 0$ . Then, Inequality (11) holds. $\square$

Proposition 1. ([16]). Let $\{\xi_k\}_{k\in\mathbb{N}}$ be the sequence of the positive zeros of $\operatorname{Sin}_q(x)$ and $m\in\mathbb{N}_0$ . Then, $$(-1)^{n-1}A_n(x) = \frac{2\operatorname{Sin}_q(\xi_1 x)}{\xi_1^{2n+1}\operatorname{Sin}_q'(\xi_1)} + \mathcal{O}(\xi_2^{-(2n+1)}); \tag{12}$$ $$(-1)^{n-1}B_n(x) = \frac{\operatorname{Sin}_q(\xi_1 x)\operatorname{Cos}_q(\xi_1)}{(1-q)(\xi_1)^{2n+1}\operatorname{Sin}_q'(\xi_1)} + \mathcal{O}(\xi_1^{-2n}(2n)^{-m}), \tag{13}$$ for a sufficiently large n.

Proposition 2. ([17]). If $f \in C_a^{2n}([0,1])$ , then $$f(x) = \sum_{m=0}^{n-1} \left[ D_{q-1}^{2m} f(1) A_m(x) - D_{q-1}^{2m} f(0) B_m(x) \right] + \int_0^1 G_n(x, qt) D_{q-1}^{2n} f(q^2t) d_q t, \quad (14)$$ where $$G(x,t) = G_1(x,t) = \begin{cases} -qt(1-x), & 0 \le t < x \le 1; \\ -qx(1-t), & 0 \le x < t \le 1, \end{cases}$$ (15) $$G_n(x,qt) = \int_0^1 G(x,qy) \, G_{n-1}(qy,qt) \, d_q y \quad (n \in \mathbb{N}).$$ (16) Moreover, $$\int_{0}^{1} G_{n}(x,qt) d_{q}t = A_{n}(x) - B_{n}(x) \quad (n \in \mathbb{N}).$$ (17) Remark 1 ([14]). For $x \in [0,1]$ and $n \in \mathbb{N}_0$ , we have $$(-1)^n A_n(x) \ge 0$$ and $(-1)^{n-1} B_n(x) \ge 0.$ (18)

Proposition 3. Let $\xi_1$ be the smallest positive zero of $Sin_q(x)$ . Then, there exist some constants $M_1$ and $M_2$ and a positive integer $n_0$ such that the following inequalities hold $$0 \le (-1)^n A_n(x) \le \frac{M_1}{\xi_1^{2n}};\tag{19}$$ $$0 \le (-1)^{n-1} B_n(x) \le \frac{M_2}{\xi_1^{2n}},\tag{20}$$ for all $x \in [0,1]$ and $n \ge n_0$ . Proof. From (12), there is a positive real number $C_1$ and $n_0 \in \mathbb{N}$ such that $$\left| (-1)^{n-1} A_n(x) - 2 \frac{\operatorname{Sin}_q(\xi_1 x)}{\xi_1^{2n+1} \operatorname{Sin}_q'(\xi_1)} \right| \le \frac{C_1}{\xi_2^{2n}}, \tag{21}$$ for all $x \in [0, 1]$ and $n \ge n_0$ . Consequently, $$0 \le (-1)^n A_n(x) \le \frac{C_1}{\xi_2^{2n}} - 2 \frac{\operatorname{Sin}_q(\xi_1 x)}{\xi_1^{2n+1} \operatorname{Sin}_q'(\xi_1)}.$$ (22) Note that $\xi_1 < \xi_2$ and $\operatorname{Sin}_q(\xi_1 x)$ is bounded on [0,1]. Then, from (22), we obtain $$0 \leq (-1)^{n} A_{n}(x) \leq \frac{C_{1}}{\xi_{1}^{2n}} + \frac{2}{\xi_{1}^{2n+1}} \left| \frac{\operatorname{Sin}_{q}(\xi_{1}x)}{\operatorname{Sin}'_{q}(\xi_{1})} \right|$$ $$\leq \frac{C_{1}}{\xi_{1}^{2n}} + \frac{C_{2}}{\xi_{1}^{2n}} = \frac{M_{1}}{\xi_{1}^{2n}}.$$ $$(23)$$ Similarly, we obtain (20) from (13). $\Box$

Proposition 4. There exists a constant M such that $$0 \le \int_0^1 (-1)^n G_n(x, qt) d_q t \le \frac{M}{\xi_1^{2n}}.$$ Proof. The proof follows immediately from Equation (17) and Proposition 3. $\Box$

Proposition 5. For any fixed point $x_0 \in (0,1)$ and sufficiently large n, there exist some constants $M_1$ and $M_2$ such that $$(-1)^n A_n(x_0) \ge \frac{M_1}{\xi_1^{2n}};\tag{24}$$ $$(-1)^{n-1}B_n(x_0) \ge \frac{M_2}{\xi_1^{2n}}. (25)$$ Proof. From (12), we obtain $$(-1)^n A_n(x) \xi_1^{2n+1} = L(x) + \mathcal{O}((\frac{\xi_1}{\xi_2})^{2n+1}) \quad (n \to \infty),$$ where $L(x) = \frac{-2\operatorname{Sin}_q(\xi_1 x)}{\operatorname{Sin}_q'(\xi_1)}$ . Notice, for any fixed $x_0 \in (0,1)$ , $L(x_0) > 0$ and $$\lim_{n\to\infty} (-1)^n A_n(x_0) \xi_1^{2n+1} = L(x_0).$$ This implies that the sequence $(-1)^n A_n(x_0) \xi_1^{2n+1}$ is bounded below by a positive number. I.e., (24) holds. Similarly, we obtain the Inequality (25) from (13). $\Box$ Now, using the previous results, we prove the following theorem.

Theorem 2. If the series $$S = a_0 A_0(x) + b_0 B_0(x) + a_1 A_1(x) + b_1 B_1(x) + \dots$$ (26) converges for a single value $x_0 \in (0,1)$ , then the series $\sum_{n=0}^{\infty} (-1)^n \left[ \frac{a_n + b_n}{\xi_1^{2n}} \right]$ is absolutely convergent. Proof. Since the series (26) converges for $x_0 \in (0,1)$ , we have $$\lim_{n \to \infty} a_n A_n(x_0) = 0, \quad \lim_{n \to \infty} b_n B_n(x_0) = 0.$$ Then, from the inequalities (24) and (25), we obtain $$a_n = \mathcal{O}(\xi_1^{2n})$$ and $b_n = \mathcal{O}(\xi_1^{2n})$ . (27) From (12), (13), and (27), we conclude that the series $$S_{1} = \sum_{n=0}^{\infty} \left\{ a_{n} \left[ A_{n}(x_{0}) + \frac{2(-1)^{n} \operatorname{Sin}_{q}(\xi_{1}x_{0})}{\xi_{1}^{2n+1} \operatorname{Sin}'_{q}(\xi_{1})} \right] + b_{n} \left[ B_{n}(x_{0}) + \frac{(-1)^{n} \operatorname{Cos}_{q} \xi_{1} \operatorname{Sin}_{q}(\xi_{1}x_{0})}{(1-q)\xi_{1}^{2n+1} \operatorname{Sin}'_{q}(\xi_{1})} \right] \right\}$$ converges absolutely. This implies that $S_1 - S$ is also convergent. Notice that $$S_{1} - S = \sum_{n=0}^{\infty} \left[ \frac{2\operatorname{Sin}_{q}(\xi_{1}x_{0})}{\xi_{1}\operatorname{Sin}_{q}'(\xi_{1})} \frac{(-1)^{n}}{\xi_{1}^{2n}} a_{n} + \frac{\operatorname{Cos}_{q}\xi_{1}\operatorname{Sin}_{q}(\xi_{1}x_{0})}{(1-q)\xi_{1}\operatorname{Sin}_{q}'(\xi_{1})} \frac{(-1)^{n}}{\xi_{1}^{2n}} b_{n} \right]$$ $$> \frac{2\operatorname{Sin}_{q}(\xi_{1}x_{0})}{\xi_{1}\operatorname{Sin}_{q}'(\xi_{1})} \sum_{n=0}^{\infty} \left[ \frac{(-1)^{n}}{\xi_{1}^{2n}} a_{n} + \frac{(-1)^{n}}{\xi_{1}^{2n}} b_{n} \right].$$ Therefore, we obtain the result. $\Box$

Proposition 6. If a function $f \in C_a^{\infty}[0,a]$ is q-completely convex, then $$(-1)^n D_{q^{-1}}^{2n} f(0) \ge 0 \quad (n \in \mathbb{N}_0). \tag{30}$$ Proof. The proof follows directly by taking the limit as $k \to \infty$ in (28) and using that $D_{q^{-1}}^{2n}f$ is continuous at zero for all $n \in \mathbb{N}_0$ . $\square$

Proposition 7. Let $f \in C_q^{\infty}(0,1)$ be a q-completely convex function on [0,1]. Then, for a sufficiently large n, we have $$D_{q^{-1}}^{2n}f(0) = \mathcal{O}(\xi_1^{2n}); \tag{31}$$ $$D_{q-1}^{2n}f(1) = \mathcal{O}(\xi_1^{2n}). \tag{32}$$ Proof. From Proposition 1 and Inequality (28), every term of (14) is non-negative. Therefore, $$0 \le A_n(x) D_{q-1}^{2n} f(0) \le f(x); \tag{33}$$ $$0 \le (-B_n(x))D_{q^{-1}}^{2n}f(1) \le f(x) \quad (x \in [0,1]; \ n \in \mathbb{N}_0).$$ (34) Thus, by using (24) and (33), we obtain $$0 \le (-1)^n D_{q^{-1}}^{2n} f(0) \le \frac{f(x_0)}{(-1)^n A_n(x_0)} \le K \xi_1^{2n} \quad (n \to \infty),$$ for some constant K > 0 and $x_0 \in (0,1)$ . Then, we have (31). Similarly, we obtain the asymptotic behavior in (32). $\square$

Proposition 8. Let f be a q-completely convex function on [0,1]. Then, there exists a positive constant C such that for all $x \in A_q$ $$0 \le (-1)^n D_{q-1}^{2n} f(x) \le C \left(\frac{\xi_1}{x}\right)^{2n},\tag{35}$$ where $\xi_1$ is the smallest positive zero of $Sin_a(x)$ . Proof. If f is q-completely convex on [0,1], then it is q-completely convex on [0,x] for all $x \in A_q$ . Consequently, the function $\widetilde{f}(t) := f(xt)$ is q-completely convex on [0,1]. Therefore, from Proposition (7), we have $$0 \leq (-1)^n D_{q^{-1}}^{2n} \widetilde{f}(1) = (-1)^n x^{2n} D_{q^{-1}}^{2n} f(x) = \mathcal{O}(\xi_1^{2n}),$$ which is nothing else but (35). $\Box$

Lemma 2. Let f(x) and $-D_{q^{-1}}^2 f(x)$ be non-negative on $A_q^*$ , and continuous at 0. Assume that there exists a number $x_0 \in A_q$ such that $f(x_0) \le \alpha$ $(\alpha \in \mathbb{R})$ . Then, $$f(x) \le \frac{(1+q)\alpha}{(1-q)x_0}$$ , for all $x \in A_q^*$ . Proof. First, let $x \in A_q^*$ and $x \ge x_0$ . Then, by using the assumption $D_{q^{-1}}^2 f(x) \le 0$ , we have $$\int_{x_0}^x D_q^2 f(\frac{t}{q^2}) d_q t \le 0.$$ Therefore, $D_q f(x) \leq D_q f(x_0)$ , and $$\int_{x_0}^x D_q f(t) \, d_q t \le (x - x_0) D_q f(x_0) \quad (x \in A_q^*, \, x_0 \le x). \tag{36}$$ Since $f(x) \ge 0$ on $A_q^*$ , from (8) and Inequality (36), we obtain $$f(x) \le f(x_0) + \frac{(x - x_0)}{(1 - q)x_0} f(x_0) = \frac{x - x_0 q}{(1 - q)x_0} f(x_0) < \frac{\alpha}{(1 - q)x_0}, \tag{37}$$ for all $x \in A_q$ and $x_0 \le x$ . Similarly, if $x \in A_q$ and $x < x_0$ , then $$f(x) \le \frac{x_0 - x}{(1 - q)x_0} f(qx_0) < \frac{f(qx_0)}{(1 - q)x_0}.$$ (38) On the other hand, since $D_{a^{-1}}^2 f(x) \le 0$ , we have $$(1+q)f(qx) \ge qf(x) + f(q^2x) \quad (x \in A_q^*).$$ Therefore, from the condition $f(x) \ge 0$ , we obtain $$(1+q)f(qx) \ge qf(\frac{x}{q}) + f(qx) > f(qx) \quad (x \in A_q^*).$$ (39) So, from the inequalities (38) and (39), we obtain $$f(x) < \frac{(1+q)\alpha}{(1-q)x_0} \quad (x \in A_q^*, \ x < x_0). \tag{40}$$ Hence, the relations (37) and (40) yield the required result. $\Box$

Corollary 1. If $f \in C_q^{\infty}[0,1]$ is a q-completely convex function, then there exists a positive constant M such that $$0 \le (-1)^n D_{q^{-1}}^{2n} f(x) \le M \xi_1^{2n} \quad (n \in \mathbb{N}_0, x \in A_q^*). \tag{41}$$ Proof. The proof follows from Proposition 8 and Lemma 2 by taking $x_0 = 1$ and $M = \frac{1+q}{1-q}C$ . $\square$

Lemma 3. If $f \in C_q^{\infty}[0,1]$ is a q-completely convex function on [0,1], then there exists a constant K > 0 such that $$|D_{q^{-1}}^n f(x)| \le K\xi_1^n \ (x \in A_q^*), \tag{42}$$ where $\xi_1$ is the smallest positive zero of $Sin_q(z)$ . Proof. From Corollary 1, it suffices to prove (42) when n is an odd integer. We set $g(x) = (-1)^n D_{q^{-1}}^{2n} f(x)$ . Since f(x) is a q-completely convex on $0 \le x \le 1$ , again from Corollary 1, there exists the constant M > 0 (independent of n) such that for all $x \in A_q^*$ $$0 \le g(x) \le M\xi_1^{2n},$$ $$0 \le -D_{q^{-1}}^2 g(x) \le M\xi_1^{2n+2}.$$ (43) Therefore, for every $x \in A_q^* - \{1\}$ , we have $$0 \le \int_{qx}^{q^2} -D_{q-1}^2 g(t) \, d_q t \le M q(q-x) \xi_1^{2n+2}.$$ So, by using the fundamental theorem of the q-calculus, we obtain $$0 \le (-1)^n D_{q^{-1}}^{2n+1} f(x) - (-1)^n D_{q^{-1}}^{2n+1} f(1) \le M \xi_1^{2n+2},$$ and hence, $$(-1)^n D_{q^{-1}}^{2n+1} f(1) \leq (-1)^n D_{q^{-1}}^{2n+1} f(x) \leq (-1)^n D_{q^{-1}}^{2n+1} f(1) + M \xi_1^{2n+2},$$ for all $x \in A_q^* - \{1\}$ . Consequently, $$|D_{q^{-1}}^{2n+1}f(x)| \le |D_{q^{-1}}^{2n+1}f(1)| + M\xi_1^{2n+2}. \tag{44}$$ On the other hand, since $D_{q^{-1}}^2g(x) < 0$ , one can verify that for all $x \in A_q^*$ $$(1+q)g(\frac{x}{q}) \ge g(x) + qg(\frac{x}{q^2}),$$ and then $$qg(\frac{x}{q}) \le (1+q)g(x) - g(qx) \quad (x \in A_q^*).$$ Thus, if x = 1, we obtain $$\left| (D_{q^{-1}}^{2n} f)(\frac{1}{q}) \right| = (-1)^n (D_{q^{-1}}^{2n} f)(\frac{1}{q}) \le \frac{(1+q)}{q} \left| D_{q^{-1}}^{2n} f(1) \right|. \tag{45}$$ Hence, from (8), (43) and (45), we have $$\left| D_{q^{-1}}^{2n+1} f(1) \right| = \left| D_{q^{-1}} g(1) \right| \le \frac{|g(1)| + |g(1/q)|}{1/q - 1} \le \frac{2q + 1}{q} M \xi_1^{2n}. \tag{46}$$ However, $\xi_1 > q^{-3/2}$ , this implies $$\left| D_{q-1}^{2n+1} f(1) \right| \le \sqrt{q} (2q+1) M \xi_1^{2n+1}. \tag{47}$$ By substituting (47) in (44), we obtain $$|D_{q^{-1}}^{2n+1}f(x)| \le \sqrt{q}(2q+1)M\xi_1^{2n+1} + M\xi_1^{2n+2} \le M_1M\xi_1^{2n+1}$$ for all $n \in \mathbb{N}$ and $x \in A_q^*$ , where $M_1 = \sqrt{q}(2q+1) + q^{-3/2}$ . Since $D_{q^{-1}}^{2n+1}f(x)$ is continuous at zero, then we obtain $D_{q^{-1}}^{2n+1}f(x)=\mathcal{O}(\xi_1^{2n+1})$ for a sufficiently large n. This completes the proof. $\square$

Theorem 3. Let $f \in C_q^{\infty}[0,1]$ be a q-completely convex on [0,1]. If f is analytic at zero, then the following q-Lidstone series expansion holds for all $x \in [0,1]$ . $$f(x) = \sum_{n=0}^{\infty} \left[ D_{q-1}^{2n} f(1) A_n(x) - D_{q-1}^{2n} f(0) B_n(x) \right]. \tag{48}$$ Moreover, f(x) is the restriction of an entire function of $q^{-1}$ -exponential growth of order 1 and a finite type less than $\xi_1$ and the expansion (48) holds for all x on the entire complex plane. Proof. Since f is analytic at 0, there exists 0 < c < 1 and the open interval $\Omega_c = (-c, c)$ such that f(x) has the Maclaurin series expansion $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = \sum_{n=0}^{\infty} q^{\frac{n(n-1)}{2}} \frac{D_{q-1}^n f(0)}{[n]_q!} x^n \quad (x \in \Omega_c).$$ (49) From Lemma 3, there exists a constant K such that $$\left| f(x) \right| \le \sum_{n=0}^{\infty} \left| q^{\frac{n(n-1)}{2}} \frac{D_{q^{-1}}^{n} f(0)}{[n]_{q}!} x^{n} \right| \le K \sum_{n=0}^{\infty} q^{\frac{n(n-1)}{2}} \frac{(\xi_{1} x)^{n}}{[n]_{q}!} = K E_{q}(\xi_{1} x), \tag{50}$$ where $E_q(.)$ is Jackson's q-exponential function defined in (10). Notice that, by the known properties of $E_q(.)$ (see [11]), $E_q(x)$ is an entire function that has a $q^{-1}$ -exponential growth of order 1, and it converges everywhere in the complex plane. Therefore, f(x) is the restriction of an entire function of $q^{-1}$ -exponential growth of order 1 and a finite type less than $\xi_1$ . So, according to Theorem 1, we obtain the result. $\square$

Theorem 4. Let $n \in \mathbb{N}_0$ , $(a_n)_n$ and $(b_n)_n$ be two sequences of non-negative integers. Assume that the series $$\sum_{n=0}^{\infty} \left[ (-1)^n a_n A_n(x) - (-1)^n b_n B_n(x) \right]$$ converges to a function f(x), $0 \le x \le 1$ . Then, f(x) is a minimal q-completely convex on the interval [0,1]. Proof. From the assumption, we have $$f(x) = \sum_{n=0}^{\infty} \left[ (-1)^n a_n A_n(x) - (-1)^n b_n B_n(x) \right], \ 0 \le x \le 1.$$ (51) Taking the $q^{-1}$ -derivative for (51) 2k times and using (6), we obtain $$(-1)^{k} D_{q^{-1}}^{2k} f(x) = \sum_{n=k}^{\infty} (-1)^{n-k} a_n A_{n-k}(x) - (-1)^{n-k} b_n B_{n-k}(x)$$ $$= \sum_{m=0}^{\infty} (-1)^{m} a_{m+k} A_m(x) - (-1)^{m} b_{m+k} B_m(x).$$ (52) From Proposition 5, since $(a_n)_n$ and $(b_n)_n$ are positive sequences, the right-hand side of Equation (52) is non-negative, and f(x) is q-completely convex in [0,1]. On the other hand, from Proposition 3 and Equation (52), there exists a constant M > 0 such that $$(-1)^k D_{q-1}^{2k} f(x) \le M \sum_{m=0}^{\infty} \left[ a_{m+k} + b_{m+k} \right] \xi_1^{-2m} = M \xi_1^{2k} \sum_{n=k}^{\infty} \frac{a_n + b_n}{\xi_1^{2n}}.$$ (53) According to Theorem 2, the power series $T_k = \sum_{n=k}^{\infty} \frac{a_n + b_n}{\xi_1^{2n}}$ converges to zero as $k \to \infty$ . Hence, for given $\epsilon > 0$ and $x_0 \in A_q$ , there exists an integer $k_0 \in \mathbb{N}$ such that $$MT_k - \epsilon \operatorname{Sin}_q(\xi_1 x_0) < 0 \ (k \ge k_0).$$ This implies from (53) that the function $$(-1)^k D_{q^{-1}}^{2k} \Big( f(x) - \epsilon \operatorname{Sin}_q(\xi_1 x) \Big) = (-1)^k D_{q^{-1}}^{2k} f(x) - \epsilon \, \xi_1^{2k} \operatorname{Sin}_q(\xi_1 x)$$ is negative at $x_0$ . Therefore, the function f is a minimal g-completely convex in [0,1]. $\square$

Theorem 5. If f(x) is a minimal q-completely convex function on [0,1], then it can be expanded into a convergent q-Lidstone series: $$f(x) = f(1)A_0(x) - f(0)B_0(x) + D_{q^{-1}}^2 f(1)A_1(x) - D_{q^{-1}}^2 f(0)B_1(x) + \dots$$ (54) Proof. We denote by $S_n(x)$ the nth partial sum of the series (54). Then, from the hypothesis on f(x) and Equation (14), we obtain $$S_n(x) \le f(x) \quad (0 \le x \le 1, \ n \in \mathbb{N}_0).$$ Moreover, for each x, $S_n(x)$ is a non-decreasing function of n. Thus, $\lim_{n\to\infty} S_n(x)$ exists and tends towards some function. To prove the result, we prove that $$\lim_{n\to\infty} S_n(x) = f(x) \quad (x\in[0,1]).$$ Suppose the contrary, and assume that for some $x_0 \in [0, 1]$ $$f(x_0) - \lim_{n \to \infty} S_n(x_0) = \triangle > 0.$$ Then, by using Equation (14), we have $$f(x_0) - S_{2n}(x_0) = \int_0^1 G_n(x_0, qt) D_{q-1}^{2n} f(q^2t) d_q t \ge \Delta \qquad (n \in \mathbb{N}).$$ (55) Since f(x) is a minimal q-completely convex function on [0,1], then $f(x) - \epsilon \operatorname{Sin}_q \xi_1 x$ is not q-completely convex in $0 \le x \le 1$ for any $\epsilon > 0$ . That is, there exists $n_0 \in \mathbb{N}$ and $t_0 \in A_q$ , $$(-1)^{n_0} D_{q^{-1}}^{2n_0} f(t_0) - \epsilon \, \xi_1^{2n_0} \operatorname{Sin}_q(\xi_1 t_0) < 0.$$ From Inequality (11), we have $$(-1)^{n_0} D_{q^{-1}}^{2n_0} f(t_0) < \epsilon \, \xi_1^{2n_0+1} \, t_0$$ By applying Lemma 2 on the function $g(x) = (-1)^{n_0} D_{q^{-1}}^{2n_0} f(x)$ , we obtain $$(-1)^{n_0} D_{q^{-1}}^{2n_0} f(t) \le \frac{1+q}{1-q} \epsilon \, \xi_1^{2n_0+1} \quad (t \in A_q).$$ Therefore, by choosing $\epsilon < \frac{1-q}{(1+q)\xi_1 M} \triangle$ , where M is the constant of Proposition 4, we obtain $$0 \le \int_0^1 G_{n_0}(x_0, qt) D_{q^{-1}}^{2n_0} f(q^2 t) d_q t < \triangle,$$ which contradicts Inequality (55), and then the result is proved. $\Box$ The following theorem is the main result of this section.

Theorem 6. A real function f(x) can be represented by an absolutely convergent q-Lidstone series if and only if it is the difference of two minimal q-completely convex functions on [0,1]. Proof. First, assume that f(x) = g(x) - h(x), where g(x) and h(x) are both minimal q-completely convex functions on [0,1]. According to Theorem 5, we have $$g(x) = \sum_{n=0}^{\infty} \left[ D_{q-1}^{2n} g(1) A_n(x) - D_{q-1}^{2n} g(0) B_n(x) \right], \tag{56}$$ $$h(x) = \sum_{n=0}^{\infty} \left[ D_{q^{-1}}^{2n} h(1) A_n(x) - D_{q^{-1}}^{2n} h(0) B_n(x) \right].$$ (57) Notice that each series only has positive terms. Thus, by subtracting (57) from (56), we obtain an absolutely convergent q-Lidstone series whose sum is f(x). Conversely, assume that f(x) can be represented by an absolutely convergent q-Lidstone series $$f(x) = \sum_{n=0}^{\infty} \left[ D_{q-1}^{2n} f(1) A_n(x) - D_{q-1}^{2n} f(0) B_n(x) \right].$$ (58) Set $$a_n = D_{q-1}^{2n} f(1)$$ , $b_n = D_{q-1}^{2n} f(0)$ , and $$g(x) = \sum_{n=0}^{\infty} \left[ (-1)^n \{ |a_n| - (-1)^n a_n \} A_n(x) + (-1)^{n+1} \{ |b_n| - (-1)^n b_n \} B_n(x) \right], (59)$$ $$h(x) = \sum_{n=0}^{\infty} \left[ (-1)^n |a_n| A_n(x) + (-1)^{n+1} |b_n| B_n(x) \right].$$ (60) Since series in (58) is absolutely convergent, then the two series in (59) and (60) both converge. Furthermore, note that every term of these series is positive. Hence, by using Theorem 4, g(x) and h(x) are minimal q-completely convex functions on [0, 1]. Since f(x) = h(x) − g(x), the proof is complete.

Theorem 1. Suppose that $0 \le \eta \le 1$ , $0 \le \lambda \le 1$ and $0 \le \delta \le 1$ . If $f \in \Sigma$ of the style (1) be an element of class $W_{\Sigma}(\eta, \delta, \lambda, \sigma, \theta, \alpha, \beta, p, q; h)$ , with $h(z) = 1 + e_1 z + e_2 z^2 + \cdots$ , then $$|a_2| \leq \frac{(\eta + \lambda(\delta+1))[\Psi_2(\sigma,\alpha,\beta)]^{\theta}_{p,q}|e_1|}{[\Psi_1(\sigma,\alpha,\beta)]^{\theta}_{p,q}} = \frac{|e_1|}{\Omega}$$ and $$|a_3| \leq \min \left\{ \max \left\{ \left| \frac{e_1}{\Delta} \right|, \left| \frac{e_2}{\Delta} - \frac{\varphi e_1^2}{\Omega^2 \Delta} \right| \right\}, \max \left\{ \left| \frac{e_1}{\Delta} \right|, \left| \frac{e_2}{\Delta} - \frac{(2\Delta + \varphi)e_1^2}{\Omega^2 \Delta} \right| \right\} \right\}, \tag{5}$$ where $$\Omega = \frac{(\eta + \lambda(\delta + 1))[\Psi_2(\sigma, \alpha, \beta)]^{\theta}_{p,q}}{[\Psi_1(\sigma, \alpha, \beta)]^{\theta}_{p,q}},$$ $$\Delta = \frac{2(\eta + \lambda(2\delta + 1))[\Psi_3(\sigma, \alpha, \beta)]^{\theta}_{p,q}}{[\Psi_1(\sigma, \alpha, \beta)]^{\theta}_{p,q}},\tag{6}$$ $$\varphi = \frac{[\eta(\eta-1) + \lambda(\delta+1)(2\eta + (\lambda-1)(\delta+1)) - 2(\eta + \lambda(3\delta+1))][\Psi_2(\sigma,\alpha,\beta)]_{p,q}^{2\theta}}{2[\Psi_1(\sigma,\alpha,\beta)]_{p,q}^{2\theta}}$$ Proof. Assume that $f \in \mathcal{W}_{\Sigma}(\eta, \delta, \lambda, \sigma, \theta, \alpha, \beta, p, q; e_1; e_2)$ . Then there consists two holomorphic functions $\phi, \psi : \mathbb{D} \longrightarrow \mathbb{D}$ showed by $$\phi(z) = r_1 z + r_2 z^2 + r_3 z^3 + \cdots \quad (z \in \mathbb{D})$$ (7) and $$\psi(w) = s_1 w + s_2 w^2 + s_3 w^3 + \dots \quad (w \in \mathbb{D}), \tag{8}$$ with $\phi(0) = \psi(0) = 0$ , $|\phi(z)| < 1$ , $|\psi(w)| < 1$ , $z, w \in \mathbb{D}$ so that $$\left(\frac{z\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)'}{W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)}\right)^{\eta}\left[(1-\delta)\frac{z\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)'}{W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)} + \delta\left(1 + \frac{z\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)''}{\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)'}\right)\right]^{\lambda}$$ $$= 1 + e_{1}\phi(z) + e_{2}\phi^{2}(z) + \cdots \tag{9}$$ and $$\left(\frac{w\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\right)'}{W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)}\right)^{\eta} \left[ (1-\delta)\frac{w\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\right)'}{W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)} + \delta\left(1 + \frac{w\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\right)''}{\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\right)'}\right)\right]^{\lambda}$$ $$= 1 + e_{1}\psi(w) + e_{2}\psi^{2}(w) + \cdots . \tag{10}$$ Unification of (7), (8), (9) and (10), yield $$\left(\frac{z\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)'}{W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)}\right)^{\eta} \left[ (1-\delta)\frac{z\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)'}{W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)} + \delta\left(1 + \frac{z\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)''}{\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)'}\right)\right]^{\lambda}$$ $$= 1 + e_{1}r_{1}z + \left[e_{1}r_{2} + e_{2}r_{1}^{2}\right]z^{2} + \cdots$$ (11) and $$\left(\frac{w\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\right)'}{W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)}\right)^{\eta}\left[\left(1-\delta\right)\frac{w\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\right)'}{W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)}+\delta\left(1+\frac{w\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\right)''}{\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\right)'}\right)\right]^{\lambda}$$ $$=1+e_{1}s_{1}w+\left[e_{1}s_{2}+e_{2}s_{1}^{2}\right]w^{2}+\cdots.$$ (12) It is clear that if $|\phi(z)| < 1$ and $|\psi(w)| < 1$ , $z, w \in \mathbb{D}$ , we obtain $$|r_j| \le 1$$ and $|s_j| \le 1 \ (j \in \mathbb{N})$ . Taking into account (11) and (12), after simplifying, we find that $$\frac{(\eta + \lambda(\delta + 1))[\Psi_2(\sigma, \alpha, \beta)]^{\theta}_{p,q}}{[\Psi_1(\sigma, \alpha, \beta)]^{\theta}_{p,q}} a_2 = e_1 r_1, \tag{13}$$ $$\begin{split} &\frac{2(\eta + \lambda(2\delta + 1))[\Psi_{3}(\sigma, \alpha, \beta)]^{\theta}_{p,q}}{[\Psi_{1}(\sigma, \alpha, \beta)]^{\theta}_{p,q}}a_{3} \\ &+ \frac{[\eta(\eta - 1) + \lambda(\delta + 1)(2\eta + (\lambda - 1)(\delta + 1)) - 2(\eta + \lambda(3\delta + 1))][\Psi_{2}(\sigma, \alpha, \beta)]^{2\theta}_{p,q}}{2[\Psi_{1}(\sigma, \alpha, \beta)]^{2\theta}_{p,q}}a_{2}^{2} \\ &= e_{1}r_{2} + e_{2}r_{1}^{2}, \end{split} \tag{14}$$ $$-\frac{(\eta + \lambda(\delta + 1))[\Psi_2(\sigma, \alpha, \beta)]^{\theta}_{p,q}}{[\Psi_1(\sigma, \alpha, \beta)]^{\theta}_{p,q}} a_2 = e_1 s_1$$ $$(15)$$ and $$\frac{2(\eta + \lambda(2\delta + 1))[\Psi_{3}(\sigma, \alpha, \beta)]^{\theta}_{p,q}}{[\Psi_{1}(\sigma, \alpha, \beta)]^{\theta}_{p,q}} \left(2a_{2}^{2} - a_{3}\right) \\ + \frac{[\eta(\eta - 1) + \lambda(\delta + 1)(2\eta + (\lambda - 1)(\delta + 1)) - 2(\eta + \lambda(3\delta + 1))][\Psi_{2}(\sigma, \alpha, \beta)]^{2\theta}_{p,q}}{2[\Psi_{1}(\sigma, \alpha, \beta)]^{2\theta}_{p,q}} a_{2}^{2} \\ = e_{1}s_{2} + e_{2}s_{1}^{2}.$$ (16) If we implement notation (6), then (13) and (14) becomes $$\Omega a_2 = e_1 r_1, \quad \Delta a_3 + \varphi a_2^2 = e_1 r_2 + e_2 r_1^2.$$ (17) This gives $$\frac{\Delta}{e_1} a_3 = r_2 + \left(\frac{e_2}{e_1} - \frac{\varphi e_1}{\Omega^2}\right) r_1^2,\tag{18}$$ and on using the given certain result ([48], p. 10): $$|r_2 - \mu r_1^2| \le \max\{1, |\mu|\} \tag{19}$$ for every $\mu \in \mathbb{C}$ , we get $$\left| \frac{\Delta}{e_1} \right| |a_3| \le \max \left\{ 1, \left| \frac{e_2}{e_1} - \frac{\varphi e_1}{\Omega^2} \right| \right\}. \tag{20}$$ In the same way, (15) and (16) becomes $$-\Omega a_2 = e_1 s_1, \quad \Delta (2a_2^2 - a_3) + \varphi a_2^2 = e_1 s_2 + e_2 s_1^2. \tag{21}$$ This gives $$-\frac{\Delta}{e_1}a_3 = s_2 + \left(\frac{e_2}{e_1} - \frac{(2\Delta + \varphi)e_1}{\Omega^2}\right)s_1^2.$$ (22) Applying (19), we obtain $$\left| \frac{\Delta}{e_1} \right| |a_3| \le \max \left\{ 1, \left| \frac{e_2}{e_1} - \frac{(2\Delta + \varphi)e_1}{\Omega^2} \right| \right\}. \tag{23}$$ Inequality (5) follows from (20) and (23). $\Box$ If we take the generating function $L_n^{\gamma}(\tau)$ given by (3) common generalized Laguerre polynomials as h(z), then from the equalities given(4), we get $e_1=1+\gamma-\tau$ and $e_2=\frac{\tau^2}{2}-(\gamma+2)\tau+\frac{(\gamma+1)(\gamma+2)}{2}$ . We obtain following corollary from Theorem 1.

Corollary 1. If $f \in \Sigma$ given by style (1) is in the family $W_{\Sigma}(\eta, \delta, \lambda, \sigma, \theta, \alpha, \beta, p, q; H_{\gamma}(\tau, z))$ , then $$|a_2| \leq \frac{(\eta + \lambda(\delta + 1))[\Psi_2(\sigma, \alpha, \beta)]^{\theta}_{p,q}|1 + \gamma - \tau|}{[\Psi_1(\sigma, \alpha, \beta)]^{\theta}_{p,q}} = \frac{|1 + \gamma - \tau|}{\Omega}$$ and $$\begin{aligned} |a_3| & \leq & \min \bigg\{ \max \bigg\{ \bigg| \frac{1+\gamma-\tau}{\Delta} \bigg|, \bigg| \frac{\frac{\tau^2}{2} - (\gamma+2)\tau + \frac{(\gamma+1)(\gamma+2)}{2}}{\Delta} - \frac{\varphi(1+\gamma-\tau)^2}{\Omega^2 \Delta} \bigg| \bigg\}, \\ & \max \bigg\{ \bigg| \frac{1+\gamma-\tau}{\Delta} \bigg|, \bigg| \frac{\frac{\tau^2}{2} - (\gamma+2)\tau + \frac{(\gamma+1)(\gamma+2)}{2}}{\Delta} - \frac{(2\Delta+\varphi)(1+\gamma-\tau)^2}{\Omega^2 \Delta} \bigg| \bigg\} \bigg\}, \end{aligned}$$ for all $\eta$ , $\lambda$ , $\delta$ so that $0 \le \eta \le 1$ , $0 \le \lambda \le 1$ and $0 \le \delta \le 1$ , where $\Omega$ , $\Delta$ , $\varphi$ are given by (6) and $H_{\gamma}(\tau, z)$ is given by (3).

Theorem 2. Suppose that $0 \le \xi \le 1$ and $0 \le \rho < 1$ . If $f \in \Sigma$ of the style (1) be an element of the class $\mathcal{K}_{\Sigma}(\xi, \rho, \sigma, \theta, \alpha, \beta, p, q; h)$ , with $h(z) = 1 + e_1 z + e_2 z^2 + \cdots$ , then $$|a_2| \leq \frac{(\xi+1)(1-\rho)[\Psi_2(\sigma,\alpha,\beta)]^{\theta}_{p,q}|e_1|}{[\Psi_1(\sigma,\alpha,\beta)]^{\theta}_{p,q}} = \frac{|e_1|}{Y}$$ and $$|a_3| \leq \min\left\{ \max\left\{ \left| \frac{e_1}{\Phi} \right|, \left| \frac{e_2}{\Phi} - \frac{\chi e_1^2}{Y^2 \Phi} \right| \right\}, \max\left\{ \left| \frac{e_1}{\Phi} \right|, \left| \frac{e_2}{\Phi} - \frac{(2\Phi + \chi)e_1^2}{Y^2 \Phi} \right| \right\} \right\}, \tag{24}$$ where $$Y = \frac{(\xi+1)(1-\rho)[\Psi_{2}(\sigma,\alpha,\beta)]^{\theta}_{p,q}}{[\Psi_{1}(\sigma,\alpha,\beta)]^{\theta}_{p,q}},$$ $$\Phi = \frac{2(2\xi+1)(1-\rho)[\Psi_{3}(\sigma,\alpha,\beta)]^{\theta}_{p,q}}{[\Psi_{1}(\sigma,\alpha,\beta)]^{\theta}_{p,q}},$$ $$\chi = \frac{(2\xi+1)(\rho^{2}-1)[\Psi_{2}(\sigma,\alpha,\beta)]^{2\theta}_{p,q}}{[\Psi_{1}(\sigma,\alpha,\beta)]^{2\theta}_{p,q}}.$$ (25) Proof. Assume that $f \in \mathcal{K}_{\Sigma}(\xi, \rho, \sigma, \theta, \alpha, \beta, p, q; e_1; e_2)$ . Then there consists two holomorphic functions $\phi, \psi : \mathbb{D} \longrightarrow \mathbb{D}$ such that $$(1-\xi)\frac{z\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)'}{(1-\rho)W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z) + \rho z\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)'} + \xi\left(\frac{\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)' + z\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)''}{\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)' + \rho z\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)''}\right)$$ $$= 1 + e_{1}\phi(z) + e_{2}\phi^{2}(z) + \cdots$$ (26) and $$(1 - \xi) \frac{w \left(W_{\alpha,\beta,p,q}^{\sigma,\theta} f^{-1}(w)\right)'}{(1 - \rho)W_{\alpha,\beta,p,q}^{\sigma,\theta} f^{-1}(w) + \rho w \left(W_{\alpha,\beta,p,q}^{\sigma,\theta} f^{-1}(w)\right)'} + \xi \left(\frac{\left(W_{\alpha,\beta,p,q}^{\sigma,\theta} f^{-1}(w)\right)' + w \left(W_{\alpha,\beta,p,q}^{\sigma,\theta} f^{-1}(w)\right)''}{\left(W_{\alpha,\beta,p,q}^{\sigma,\theta} f^{-1}(w)\right)' + \rho w \left(W_{\alpha,\beta,p,q}^{\sigma,\theta} f^{-1}(w)\right)''}\right)$$ $$= 1 + e_1 \psi(w) + e_2 \psi^2(w) + \cdots, \tag{27}$$ where $\phi$ and $\psi$ given by the style (7) and (8). Unification of (26) and (27), serve $$(1-\xi)\frac{z\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)'}{(1-\rho)W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z) + \rho z\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)'} + \xi\left(\frac{\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)' + z\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)''}{\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)' + \rho z\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\right)''}\right)$$ $$= 1 + e_{1}r_{1}z + \left[e_{1}r_{2} + e_{2}r_{1}^{2}\right]z^{2} + \cdots$$ (28) and $$(1 - \xi) \frac{w \left(W_{\alpha,\beta,p,q}^{\sigma,\theta} f^{-1}(w)\right)'}{(1 - \rho)W_{\alpha,\beta,p,q}^{\sigma,\theta} f^{-1}(w) + \rho w \left(W_{\alpha,\beta,p,q}^{\sigma,\theta} f^{-1}(w)\right)'} + \xi \left(\frac{\left(W_{\alpha,\beta,p,q}^{\sigma,\theta} f^{-1}(w)\right)' + w \left(W_{\alpha,\beta,p,q}^{\sigma,\theta} f^{-1}(w)\right)''}{\left(W_{\alpha,\beta,p,q}^{\sigma,\theta} f^{-1}(w)\right)' + \rho w \left(W_{\alpha,\beta,p,q}^{\sigma,\theta} f^{-1}(w)\right)''}\right)$$ $$= 1 + e_1 s_1 w + \left[e_1 s_2 + e_2 s_1^2\right] w^2 + \cdots$$ (29) It is clear that if $|\phi(z)| < 1$ and $|\psi(w)| < 1$ , $z, w \in \mathbb{D}$ , we obtain $$|r_i| \leq 1$$ and $|s_i| \leq 1 \ (j \in \mathbb{N}).$ Taking into account (28) and (29), after simplifying, we find that $$\frac{(\xi+1)(1-\rho)[\Psi_{2}(\sigma,\alpha,\beta)]^{\theta}_{p,q}}{[\Psi_{1}(\sigma,\alpha,\beta)]^{\theta}_{p,q}}a_{2}=e_{1}r_{1},$$ (30) $$\frac{2(2\xi+1)(1-\rho)[\Psi_{3}(\sigma,\alpha,\beta)]^{\theta}_{p,q}}{[\Psi_{1}(\sigma,\alpha,\beta)]^{\theta}_{p,q}}a_{3}+\frac{(2\xi+1)(\rho^{2}-1)[\Psi_{2}(\sigma,\alpha,\beta)]^{2\theta}_{p,q}}{[\Psi_{1}(\sigma,\alpha,\beta)]^{2\theta}_{p,q}}a_{2}^{2}=e_{1}r_{2}+e_{2}r_{1}^{2}, (31)$$ $$-\frac{(\xi+1)(1-\rho)[\Psi_{2}(\sigma,\alpha,\beta)]^{\theta}_{p,q}}{[\Psi_{1}(\sigma,\alpha,\beta)]^{\theta}_{p,q}}a_{2} = e_{1}s_{1}$$ (32) and $$\frac{2(2\xi+1)(1-\rho)[\Psi_{3}(\sigma,\alpha,\beta)]^{\theta}_{p,q}}{[\Psi_{1}(\sigma,\alpha,\beta)]^{\theta}_{p,q}} \left(2a_{2}^{2}-a_{3}\right) + \frac{(2\xi+1)(\rho^{2}-1)[\Psi_{2}(\sigma,\alpha,\beta)]^{2\theta}_{p,q}}{[\Psi_{1}(\sigma,\alpha,\beta)]^{2\theta}_{p,q}} a_{2}^{2}$$ $$= e_{1}s_{2} + e_{2}s_{1}^{2}.$$ (33) If we implement notation (25), then (30) and (31) becomes $$Ya_2 = e_1 r_1, \quad \Phi a_3 + \chi a_2^2 = e_1 r_2 + e_2 r_1^2.$$ (34) This gives $$\frac{\Phi}{e_1}a_3 = r_2 + \left(\frac{e_2}{e_1} - \frac{\chi e_1}{Y^2}\right)r_1^2,\tag{35}$$ and on using the given certain result ([48], p. 10): $$|r_2 - \mu r_1^2| \le \max\{1, |\mu|\} \tag{36}$$ for every $\mu \in \mathbb{C}$ , we get $$\left|\frac{\Phi}{e_1}\right||a_3| \le \max\left\{1, \left|\frac{e_2}{e_1} - \frac{\chi e_1}{Y^2}\right|\right\}. \tag{37}$$ In the same way, (32) and (33) becomes $$-Ya_2 = e_1s_1, \quad \Phi(2a_2^2 - a_3) + \chi a_2^2 = e_1s_2 + e_2s_1^2. \tag{38}$$ This gives $$-\frac{\Phi}{e_1}a_3 = s_2 + \left(\frac{e_2}{e_1} - \frac{(2\Phi + \chi)e_1}{Y^2}\right)s_1^2. \tag{39}$$ Applying (36), we obtain $$\left| \frac{\Phi}{e_1} \right| |a_3| \le \max \left\{ 1, \left| \frac{e_2}{e_1} - \frac{(2\Phi + \chi)e_1}{Y^2} \right| \right\}.$$ (40) Inequality (24) follows from (37) and (40). $\Box$ If we take the generating function $L_n^{\eta}(\tau)$ given by (3) common generalized Laguerre polynomials as h(z), then from the equalities given(4), we get $e_1 = 1 + \gamma - \tau$ and $e_2 = \frac{\tau^2}{2} - (\gamma + 2)\tau + \frac{(\gamma+1)(\gamma+2)}{2}$ . We obtain following corollary from Theorem 2.

Corollary 2. If $f \in \Sigma$ of the style (1) be an element of the class $\mathcal{K}_{\Sigma}(\xi, \rho, \sigma, \theta, \alpha, \beta, p, q; H_{\gamma}(\tau, z))$ , then $$|a_2| \leq \frac{(\xi+1)(1-\rho)[\Psi_2(\sigma,\alpha,\beta)]^{\theta}_{p,q}|1+\gamma-\tau|}{[\Psi_1(\sigma,\alpha,\beta)]^{\theta}_{p,q}} = \frac{|1+\gamma-\tau|}{Y}$$ and $$\begin{aligned} |a_3| & \leq & \min \left\{ \max \left\{ \left| \frac{1+\gamma-\tau}{\Phi} \right|, \left| \frac{\frac{\tau^2}{2} - (\gamma+2)\tau + \frac{(\gamma+1)(\gamma+2)}{2}}{\Phi} - \frac{\chi(1+\gamma-\tau)^2}{Y^2\Phi} \right| \right\}, \\ & \max \left\{ \left| \frac{1+\gamma-\tau}{\Phi} \right|, \left| \frac{\frac{\tau^2}{2} - (\gamma+2)\tau + \frac{(\gamma+1)(\gamma+2)}{2}}{\Phi} - \frac{(2\Phi+\chi)(1+\gamma-\tau)^2}{Y^2\Phi} \right| \right\} \right\}, \end{aligned}$$ for all $\xi$ , $\rho$ so that $0 \le \xi \le 1$ and $0 \le \rho < 1$ , where Y, $\Phi$ , $\chi$ are introduced by (25) and $H_{\gamma}(\tau, z)$ is given by (3). We investigate the "Fekete-Szegö Inequalities" for the families $\mathcal{W}_{\Sigma}(\eta, \delta, \lambda, \sigma, \theta, \alpha, \beta, p, q; h)$ and $\mathcal{K}_{\Sigma}(\xi, \rho, \sigma, \theta, \alpha, \beta, p, q; h)$ in next theorems.

Theorem 3. If $f \in \Sigma$ of the style (1) be an element of family $\mathcal{W}_{\Sigma}(\eta, \delta, \lambda, \sigma, \theta, \alpha, \beta, p, q; h)$ , then $$\left|a_3 - \zeta a_2^2\right| \leq \frac{|e_1|}{\Delta} \min\left\{\max\left\{1, \left|\frac{e_2}{e_1} + \frac{(\zeta \Delta - \varphi)e_1}{\Omega^2}\right|\right\}, \max\left\{1, \left|\frac{e_2}{e_1} - \frac{(2\Delta + \varphi - \zeta \Delta)e_1}{\Omega^2}\right|\right\}\right\},$$ for all $\zeta, \eta, \lambda, \delta$ such that $\zeta \in \mathbb{R}$ , $0 \le \eta \le 1$ , $0 \le \lambda \le 1$ and $0 \le \delta \le 1$ , where $\Omega, \Delta, \varphi$ are given by (6) and $e_1, e_2, a_2$ and $a_3$ as defined in Theorem 1. Proof. We implement the impressions from the Theorem 1's proof. From (17) and from (18), we get $$a_3 - \zeta a_2^2 = \frac{e_1}{\Delta} \left( r_2 + \left( \frac{e_2}{e_1} + \frac{(\zeta \Delta - \varphi)e_1}{\Omega^2} \right) r_1^2 \right)$$ by using the certain result $|r_2 - \mu r_1^2| \le \max\{1, |\mu|\}$ , we get $$|a_3 - \zeta a_2^2| \le \frac{|e_1|}{\Delta} \max \left\{ 1, \left| \frac{e_2}{e_1} + \frac{(\zeta \Delta - \varphi)e_1}{\Omega^2} \right| \right\}.$$ In the same way, from (21) and from (22), we get $$a_3 - \zeta a_2^2 = -\frac{e_1}{\Delta} \left( s_2 + \left( \frac{e_2}{e_1} - \frac{(2\Delta + \varphi - \zeta \Delta)e_1}{\Omega^2} \right) s_1^2 \right)$$ and on using $|s_2 - \mu s_1^2| \le \max\{1, |\mu|\}$ , we get $$|a_3 - \zeta a_2^2| \le \frac{|e_1|}{\Delta} \max \left\{ 1, \left| \frac{e_2}{e_1} - \frac{(2\Delta + \varphi - \zeta \Delta)e_1}{\Omega^2} \right| \right\}.$$

Corollary 3. If $f \in \Sigma$ of the style (1) be an element of $W_{\Sigma}(\eta, \delta, \lambda, \sigma, \theta, \alpha, \beta, p, q; H_{\gamma}(\tau, z))$ , then $$\begin{vmatrix} a_3 - \zeta a_2^2 \\ \leq \frac{|1 + \gamma - \tau|}{\Delta} \min \left\{ \max \left\{ 1, \left| \frac{\frac{\tau^2}{2} - (\gamma + 2)\tau + \frac{(\gamma + 1)(\gamma + 2)}{2}}{1 + \gamma - \tau} + \frac{(\zeta \Delta - \varphi)(1 + \gamma - \tau)}{\Omega^2} \right| \right\}, \\ \max \left\{ 1, \left| \frac{\frac{\tau^2}{2} - (\gamma + 2)\tau + \frac{(\gamma + 1)(\gamma + 2)}{2}}{1 + \gamma - \tau} - \frac{(2\Delta + \varphi - \zeta \Delta)(1 + \gamma - \tau)}{\Omega^2} \right| \right\} \right\},$$ for each $\zeta$ , $\eta$ , $\lambda$ , $\delta$ such that $\zeta \in \mathbb{R}$ , $0 \le \eta \le 1$ , $0 \le \lambda \le 1$ and $0 \le \delta \le 1$ , where $\Omega$ , $\Delta$ , $\varphi$ are given by (6) and $H_{\gamma}(\tau, z)$ is presented by (3).

Theorem 4. If $f \in \Sigma$ of the style (1) is in the family $\mathcal{K}_{\Sigma}(\xi, \rho, \sigma, \theta, \alpha, \beta, p, q; h)$ , then $$\left|a_3 - \zeta a_2^2\right| \leq \frac{|e_1|}{\Phi} \min\left\{\max\left\{1, \left|\frac{e_2}{e_1} + \frac{(\zeta\Phi - \chi)e_1}{Y^2}\right|\right\}, \max\left\{1, \left|\frac{e_2}{e_1} - \frac{(2\Phi + \chi - \zeta\Phi)e_1}{Y^2}\right|\right\}\right\},$$ for all $\zeta$ , $\xi$ , $\rho$ such that $\zeta \in \mathbb{R}$ , $0 \le \xi \le 1$ and $0 \le \rho < 1$ , where Y, $\Phi$ , $\chi$ are given by (25) and $e_1$ , $e_2$ , $a_2$ and $a_3$ as defined in Theorem 2. Proof. We implement the impressions from the Theorem 2's proof. From (34) and from (35), we get $$a_3 - \zeta a_2^2 = \frac{e_1}{\Phi} \left( r_2 + \left( \frac{e_2}{e_1} + \frac{(\zeta \Phi - \chi)e_1}{Y^2} \right) r_1^2 \right)$$ by using the certain result $|r_2 - \mu r_1^2| \le \max\{1, |\mu|\}$ , we get $$|a_3 - \zeta a_2^2| \le \frac{|e_1|}{\Phi} \max \left\{ 1, \left| \frac{e_2}{e_1} + \frac{(\zeta \Phi - \chi)e_1}{Y^2} \right| \right\}.$$ In the same way, from (38) and from (39), we get $$a_3 - \zeta a_2^2 = -\frac{e_1}{\Phi} \left( s_2 + \left( \frac{e_2}{e_1} - \frac{(2\Phi + \chi - \zeta \Phi)e_1}{Y^2} \right) s_1^2 \right)$$ and on using $|s_2 - \mu s_1^2| \le \max\{1, |\mu|\}$ , we get $$|a_3 - \zeta a_2^2| \le \frac{|e_1|}{\Phi} \max \left\{ 1, \left| \frac{e_2}{e_1} - \frac{(2\Phi + \chi - \zeta \Phi)e_1}{Y^2} \right| \right\}.$$

Corollary 4. If $f \in \Sigma$ of the style (1) be an element of $\mathcal{K}_{\Sigma}(\xi, \rho, \sigma, \theta, \alpha, \beta, p, q; H_{\gamma}(\tau, z))$ , then $$\left. \begin{array}{l} \left| a_3 - \zeta a_2^2 \right| \\ \leq & \frac{\left| 1 + \gamma - \tau \right|}{\Phi} \min \left\{ \max \left\{ 1, \left| \frac{\frac{\tau^2}{2} - (\gamma + 2)\tau + \frac{(\gamma + 1)(\gamma + 2)}{2}}{1 + \gamma - \tau} + \frac{(\zeta \Phi - \chi)(1 + \gamma - \tau)}{Y^2} \right| \right\}, \\ \max \left\{ 1, \left| \frac{\frac{\tau^2}{2} - (\gamma + 2)\tau + \frac{(\gamma + 1)(\gamma + 2)}{2}}{1 + \gamma - \tau} - \frac{(2\Phi + \chi - \zeta \Phi)(1 + \gamma - \tau)}{Y^2} \right| \right\} \right\}, \end{aligned}$$ for each $\zeta$ , $\xi$ , $\rho$ such that $\zeta \in \mathbb{R}$ , $0 \le \xi \le 1$ and $0 \le \rho < 1$ , where Y, $\Phi$ , $\chi$ are given by (25) and $H_{\gamma}(\tau, z)$ is presented by (3).

Lemma 1. ([52]). Let the Schwarz function $\omega(\eta)$ be given by $$\omega(\eta) = \omega_1 \eta + \omega_2 \eta^2 + \omega_3 \eta^3 + \ldots + \omega_k \eta^k + \ldots \qquad (\eta \in \nabla);$$ then $$|\omega_1| \le 1,$$ $$|\omega_2| \le 1 - |\omega_1|^2,$$ (10) $|\omega_2 - t\omega_1^2| < 1 + (|t| - 1)|\omega_1|^2$ .

Theorem 1. Let $\Phi$ be given by (1). For $\chi \geq 1$ , $0 \leq \alpha < 1$ , $(\mu, \beta, \gamma, \delta \geq 0)$ , $\gamma > \delta, \beta > \mu$ and $k \in \mathbb{N}_0$ . If $\Phi \in D^{\ell}_{\Sigma_a}(\chi, \delta, \gamma, \mu; \varphi)$ and $\varrho_m = 0$ ; $m = 2, ..., \mathbb{k} - 1$ , then $$|\varrho_{\mathbb{k}}| \le \frac{2(1-q)}{\left|1 + (q-q^{\mathbb{k}})\chi\right| \left|\Omega_{\mathbb{k}}^{\ell}\right|}; \qquad (\mathbb{k} = 4, 5, 6, \ldots). \tag{13}$$ Proof. Since $\Phi \in D_{\Sigma_a}^{\ell}(\chi, \delta, \gamma, \mu; \varphi)$ of form (1), we have: $$(1-\chi)\frac{Y_{q,\mu,\beta,\gamma,\delta}^{(\ell)}\Phi(\eta)}{\eta} + \chi D_q Y_{q,\mu,\beta,\gamma,\delta}^{(\ell)}\Phi(\eta) = 1 + \sum_{\mathbb{k}=1}^{\infty} \left(1-\chi(1-[\mathbb{k}]_q)\right) \Omega_{\mathbb{k}}^{\ell} \varrho_{\mathbb{k}} \eta^{\mathbb{k}-1}$$ (14) and for $h = \Phi^{-1}$ , we have $$(1-\chi)\frac{Y_{q,\mu,\beta,\gamma,\delta}^{(\ell)}h(\rho)}{\rho} + \chi D_{q}Y_{q,\mu,\beta,\gamma,\delta}^{(\ell)}h(\rho) = 1 + \sum_{k=1}^{\infty} \left(1-\chi(1-[k]_{q})\right)\Omega_{k}^{\ell}b_{k}\rho^{k-1}$$ $$= 1 + \sum_{k=1}^{\infty} \left(1-\chi(1-[k]_{q})\right)\Omega_{k}^{\ell}\left(\frac{1}{k}K_{k-1}^{-k}(\varrho_{2},\varrho_{3},\ldots,\varrho_{k})\right)\rho^{k-1},$$ (15) where $\Omega_{\Bbbk}^{\ell}$ and $K_{\Bbbk-1}^{-\Bbbk}$ are given by (4) and (9), respectively. Since $\Phi$ , $\Phi^{-1} \in D_{\Sigma_q}^{\ell}(\chi, \delta, \gamma, \mu; \varphi)$ . Then, by using the definition of subordination, two $$u(\eta) = \sum_{k=1}^{\infty} \exists_k \eta^k$$ and $v(\rho) = \sum_{k=1}^{\infty} \exists_k \rho^k$ , which are analytic in $\nabla$ , such that $$\varphi(u(\eta)) = (1 - \chi) \frac{Y_{q,\mu,\beta,\gamma,\delta}^{(\ell)} \Phi(\eta)}{\eta} + \chi D_q Y_{q,\mu,\beta,\gamma,\delta}^{(\ell)} \Phi(\eta), \quad (\eta \in \nabla)$$ (16) $$\varphi(v(\rho)) = (1 - \chi) \frac{Y_{q,\mu,\beta,\gamma,\delta}^{(\ell)} h(\rho)}{\rho} + \chi D_q Y_{q,\mu,\beta,\gamma,\delta}^{(\ell)} h(\rho), \quad (\rho \in \nabla), \tag{17}$$ where $$\varphi(u(\eta)) = 1 + \sum_{k=1}^{\infty} \sum_{\ell=1}^{k} \varphi_k E_k^{\ell}(\mathbf{J}_1, \mathbf{J}_2, \dots, \mathbf{J}_k) \eta^k.$$ (18) and $$\varphi(v(\rho)) = 1 + \sum_{k=1}^{\infty} \sum_{\ell=1}^{k} \varphi_k E_k^{\ell}(\exists_1, \exists_2, \dots, \exists_k) \rho^k.$$ (19) From (14), (16), and (18), we have $$\left(1 - \chi(1 - [\mathbb{k}]_q)\right)\Omega_{\mathbb{k}}^{\ell}\varrho_{\mathbb{k}} = \sum_{\ell=1}^{\mathbb{k}-1} \varphi_k E_{n-1}^{\ell}(\mathbb{I}_1, \mathbb{I}_2, \dots, \mathbb{I}_{n-1}) \qquad (n \ge 2), \tag{20}$$ Similarly, from (15), (17), and (19), we have $$\left(1 - \chi(1 - [\mathbb{k}]_q)\right) \Omega_{\mathbb{k}}^{\ell} b_{\mathbb{k}} = \sum_{\ell=1}^{\mathbb{k}-1} \varphi_k E_{\mathbb{k}-1}^{\ell} (\mathbb{k}_{1}, \mathbb{k}_{2}, \dots, \mathbb{k}_{\ell-1}) \quad (\mathbb{k} \ge 2),$$ (21) by the given assumption $$\rho_m = 0$$ , $(2 \le m \le k - 1)$ , which is equivalent to $$\exists_m = \exists_m = 0; \quad (1 \le m \le \mathbb{k} - 2),$$ and from Equations (20) and (21), we have $b_{\mathbb{k}} = -\varrho_{\mathbb{k}}$ and so $$\begin{cases} \left(1 - \chi(1 - [\mathbb{k}]_q)\right) \Omega_{\mathbb{k}}^{\ell} \varrho_{\mathbb{k}} = \varphi_1 \mathbb{I}_{\mathbb{k} - 1}, \\ \left(1 - \chi(1 - [\mathbb{k}]_q)\right) \Omega_{\mathbb{k}}^{\ell} \varrho_{\mathbb{k}} = -\varphi_1 \mathbb{I}_{\mathbb{k} - 1}. \end{cases}$$ (22) Taking the absolute value for Equation (22), we obtain $$\begin{aligned} |\varrho_{\mathbb{k}}| &\leq \frac{|\varphi_{1}||\mathbf{I}_{\mathbb{k}-1}|}{\left(1 - \chi(1 - [\mathbb{k}]_{q})\right)\Omega_{\mathbb{k}}^{\ell}} \\ &= \frac{|\varphi_{1}||\mathbf{I}_{\mathbb{k}-1}|}{\left(1 - \chi(1 - [\mathbb{k}]_{q})\right)\Omega_{\mathbb{k}}^{\ell}}, \qquad (n \geq 4). \end{aligned}$$ Using Caratheodory's Lemma, we obtain $$|\varrho_{\Bbbk}| \leq \frac{2(1-q)}{\left|1-q+(q-q^{\Bbbk})\chi\right|\left|\Omega_{\Bbbk}^{\ell}\right|}.$$ This completes the proof of the theorem. $\Box$ In the next theorem, we estimate the initial coefficients of the functions from the indicated class $D_{\Sigma_g}^\ell(\chi,\delta,\gamma,\mu;\varphi)$ .

Theorem 2. For $\chi \geq 1$ , $0 \leq \alpha < 1$ , $(\mu, \beta, \gamma, \delta \geq 0)$ , $\gamma > \delta, \beta > \mu$ and $k \in \mathbb{k}_0$ , if $\Phi \in D_{\Sigma_q}^{\ell}(\chi, \delta, \gamma, \mu; \varphi)$ where $\Phi(\eta)$ is given by (1), then we have the following consequence $$\begin{split} |\varrho_2| & \leq \min \left\{ \frac{2}{(1+\chi q)\Omega_2^{\ell}}, \frac{2}{\sqrt{(1+\chi(q^2+q))\Omega_3^{\ell}}} \right\}, \\ |\varrho_3| & \leq \min \left\{ \frac{4}{\left(1+\chi(2+q)\Omega_2^{\ell}\right)^2} + \frac{2}{(1+\chi(q^2+q))\Omega_3^{\ell}}, \frac{6}{(1+\chi(q^2+q))\Omega_3^{\ell}} \right\}, \end{split}$$ and $$|2\varrho_2^2-\varrho_3|\leq \frac{4}{\left|\left(1+\chi([3]_q-1)\right)\Omega_3^\ell\right|}.$$ Proof. Replacing k by 2 and 3 in (20) and (21), respectively, we obtain: $$(1 - \chi(1 - [2]_q))\Omega_2^{\ell}\varrho_2 = \varphi_1 J_1, \tag{24}$$ $$(1 - \chi(1 - [3]_q))\Omega_3^{\ell} \varrho_3 = \varphi_1 \mathbb{1}_2 + \varphi_2 c_1^2, \tag{25}$$ $$(1 - \chi(1 - [2]_a))\Omega_2^{\ell} \rho_2 = -\varphi_1 \overline{1}_1, \tag{26}$$ and $$\left(1 - \chi(1 - [3]_q)\right)\Omega_3^{\ell}(2\varrho_2^2 - \varrho_3) = \varphi_1 \mathcal{I}_2 + \varphi_2 d_1^2. \tag{27}$$ From (24) and (26), we have $\exists_1 = -\exists_1$ and $$|\varrho_{2}| = \frac{|\varphi_{1} \Im_{1}|}{\left|1 - \chi(1 - [2]_{q})\right| \Omega_{2}^{\ell}} = \frac{|\varphi_{1} \Im_{1}|}{\left|1 - \chi(1 - [2]_{q})\right| \Omega_{2}^{\ell}} \le \frac{2}{1 + \chi(1 + [2]_{q})\Omega_{2}^{\ell}}.$$ (28) Now, by adding (25) and (27) $$2\Big(1-\chi(1-[3]_q)\Big)\Omega_3^{\ell}\varrho_2^2 = \varphi_1(\gimel_2+\daleth_1)+\varphi_2(c_1^2+d_1^2),$$ or, equivalently, $$|\varrho_2| \le \frac{2}{\sqrt{\left(1 + \chi(q^2 + q)\right)\Omega_3^{\ell}}}.$$ (29) Next, in order to find the bounds of $|\varrho_3|$ , subtract (25) from (27), we have $$2(1+\chi([3]_q-1))\Omega_3^{\ell}(\varrho_3-\varrho_2^2) = \varphi_1(\Im_2-\Im_2) + \varphi_2(c_1^2-d_1^2), \tag{30}$$ or $$2\left(1+\chi([3]_{q}-1)\right)\Omega_{3}^{\ell}(\varrho_{3}-\varrho_{2}^{2}) \leq \varphi_{2}(\mathbb{I}_{2}-\mathbb{I}_{2}),$$ $$|\varrho_{3}| \leq \varrho_{2}^{2} + \frac{|\varphi_{2}(\mathbb{I}_{2}-\mathbb{I}_{2})|}{2\left|\left(1+\chi([3]_{q}-1)\right)\Omega_{3}^{\ell}\right|}.$$ (31) Equivalent to $$|\varrho_3| \le \varrho_2^2 + \frac{|\varphi_2(\gimel_2 - \urcorner_2)|}{2\Big|\Big(1 + \chi(q^2 + q)\Big)\Omega_3^\ell\Big|},$$ Substituting the value $\varrho_2$ from (29) and (30) into (31), one obtains $$|\varrho_3| \leq \frac{4}{\big(1+\chi(2+q)\Omega_2^\ell\big)^2} + \frac{2}{\big(1+\chi(q^2+q)\big)\Omega_3^\ell},$$ and $$|\varrho_3| \leq \frac{6}{\left(1 + \chi(q^2 + q)\right)\Omega_3^\ell}.$$ Finally, from (30), by applying the Caratheodory Lemma, we obtain $$|2\varrho_2^2 - \varrho_3| = \frac{|\varphi_1 \mathcal{I}_2 + \varphi_2 d_1^2|}{\left| (1 - \chi(1 - [3]_q))\Omega_3^\ell \right|} \le \frac{4}{\left| (1 + \chi([3]_q - 1))\Omega_3^\ell \right|}.$$ (32) This completes the proof of Theorem 2. $\Box$

Corollary 1. ([37]). Let $\chi \geq 1$ . A bi-univalent function $\Phi$ given by (1) belongs to the class $D_{\Sigma}(q,\chi;\varphi)$ ( $\chi \geq 1$ ). If $\varrho_m = 0$ ; $m = 2,\ldots, k-1$ . Then $$|\varrho_{\Bbbk}| \le \frac{2(1-q)}{1-q+(q-q^{\Bbbk})\chi} \qquad (n \ge 4).$$ Applying the limit $q \to 1^-$ in Theorem 1 and considering the case when $\ell = 0$ , we obtain the following corollary.

Corollary 2. ([37]). Let $\chi \geq 1$ . A bi-univalent function $\Phi$ given by (1) belongs to the class $R_{\sigma}(\chi, \varphi)(\chi \geq 1)$ . If $\varrho_m = 0$ ; m = 2, ..., k-1. Then $$|\varrho_{\Bbbk}| \le \frac{2}{1 + \chi(\Bbbk - 1)}$$ $(n \ge 4).$ For k = 0 in Theorem 2, we obtain the following corollary.

Corollary 3. ([37]). Let $\chi \geq 1$ . A bi-univalent function $\Phi$ given by (1) belongs to the class $D_{\Sigma}(q;\chi,\varphi)$ . Then $$(1) \quad |\varrho_2| \leq \frac{2}{1+q\chi},$$ (2) $$|\varrho_3| \le \frac{4}{(1+q\chi)^2} + \frac{2}{1+(q^2+q)\chi}$$ (3) $$|2\varrho_2^2 - \varrho_3| \le \frac{4}{1 + (g^2 + g)\chi}$$ . For k = 0 and $q \to 1^-$ in Theorem 2, we obtain the following corollary.

Corollary 4. A bi-univalent function $\Phi$ given by (1) belongs to the class $R_{\sigma}(\chi, \varphi)(\chi \geq 1)$ . Then $$(1) \quad |\varrho_2| \leq \frac{2}{1+\chi},$$ (2) $$|\varrho_3| \leq \frac{4}{(1+3\chi)^2} + \frac{2}{1+2\chi}$$

Lemma 1. (see Pescar [33]). Let $\alpha$ and $\beta$ be complex number such that $$\Re(\alpha) > 0$$ and $|\beta| \leq 1(\beta \neq -1)$ . If the function $h \in A$ satisfies the following inequality: $$\left|\beta|\rho|^{2\alpha} + \left(1 - |\rho|^{2\alpha}\right) \frac{\rho h''(\rho)}{\alpha h'(\rho)}\right| \le 1$$ for all $\rho \in \mathbb{D}$ , then the function $F_{\alpha} \in \mathcal{A}$ defined by $$F_{\alpha}(\rho) = \left(\alpha \int_{0}^{\rho} t^{\alpha - 1} h'(t) dt\right)^{1/\alpha} \tag{8}$$ is in the class S.

Lemma 2. (see Pascu [34]). Let $\omega \in \mathbb{C}$ such that $\Re(\omega) > 0$ . If $h \in \mathcal{A}$ satisfies the following inequality: $$\left(\frac{1-|\rho|^{2\Re(\varpi)}}{\Re(\varpi)}\right)\left|\frac{\rho h''(\rho)}{h'(\rho)}\right| \leq 1$$ for all $\rho \in \mathbb{D}$ . Then, for all $\alpha \in \mathbb{C}$ such that $$\Re(\alpha) \geq \Re(\omega)$$ , the function $F_{\alpha}$ defined by (8) is in the class S.

Lemma 3. Let $\xi > -1$ and $\varrho > 0$ . Then, for $\forall \varrho \in \mathbb{D}$ , the function $\mathbb{E}_{\xi,\varrho}$ defined by (4) provides the following inequalities: $$\left| \mathbb{E}'_{\xi,\varrho}(\rho) - \frac{\mathbb{E}_{\xi,\varrho}(\rho)}{\rho} \right| \le \frac{\varrho(\xi+1)}{(\xi-\varrho+1)^2} (\varrho - 1 < \xi), \tag{9}$$ $$\left| \frac{\rho \mathbb{E}'_{\xi,\varrho}(\rho)}{\mathbb{E}_{\xi,\varrho}(\rho)} - 1 \right| \le \frac{\varrho(\xi+1)}{(\xi-\varrho+1)(\xi-2\varrho+1)} (2\varrho - 1 < \xi), \tag{10}$$ $$\left| \rho \mathbb{E}'_{\xi,\varrho}(\rho) \right| \le \left( \frac{\xi + 1}{\xi - \varrho + 1} \right)^2 (\varrho - 1 < \xi), \tag{11}$$ $$\left| \frac{\rho \mathbb{E}'_{\xi,\varrho}(\rho)}{\mathbb{E}_{\xi,\varrho}(\rho)} \right| \le \frac{(\xi+1)^2}{(\xi-\varrho+1)(\xi-2\varrho+1)} (2\varrho-1<\xi) \tag{12}$$ and $$\left| \frac{\rho \mathbb{E}_{\xi,\varrho}^{"}(\rho)}{\mathbb{E}_{\xi,\varrho}^{"}(\rho)} \right| \leq \frac{(\xi - \varrho + 1)^3 + 2\varrho(\xi + 1)^2}{(\xi - \varrho + 1)^2 + (\varrho - 1)^2 - (2\xi\varrho + 1)} \left( \left( 2 + \sqrt{2} \right)\varrho - 1 < \xi \right). \tag{13}$$ Proof. The inequalities (9) and (10) were proved by Eker and Ece [19]. On the other hand, by using the triangle inequality and the following inequality (see [19]) $$\frac{\Gamma(\xi+1)}{\Gamma(\xi+v)} \le \frac{1}{(\xi+1)^{v-1}},\tag{14}$$ we have $$\left| \frac{\mathbb{E}_{\xi,\varrho}(\rho)}{\rho} \right| \le \frac{\xi - 2\varrho + 1}{\xi - \varrho + 1} (2\varrho - 1 < \xi) \tag{15}$$ and $$\left| \rho \mathbb{E}'_{\xi,\varrho}(\rho) \right| = \left| \rho + \sum_{v=2}^{\infty} \frac{v \Gamma(\xi+1) \varrho^{v-1}}{\Gamma(\xi+v)} \rho^{v} \right|$$ $$\leq 1 + \sum_{v=2}^{\infty} \frac{v \Gamma(\xi+1) \varrho^{v-1}}{\Gamma(\xi+v)}$$ $$\leq 1 + \sum_{v=2}^{\infty} v \left( \frac{\varrho}{\xi+1} \right)^{v-1} = \left( \frac{\xi+1}{\xi-\varrho+1} \right)^{2} (\varrho - 1 < \xi).$$ (16) Thus, from (15) and (16), we obtain $$\left|\frac{\rho \mathbb{E}_{\xi,\varrho}'(\rho)}{\mathbb{E}_{\xi,\varrho}(\rho)}\right| \leq \frac{(\xi+1)^2}{(\xi-\varrho+1)(\xi-2\varrho+1)}(2\varrho-1<\xi).$$ Using the inequality (14), it follows that $$\left| \rho \mathbb{E}_{\xi,\varrho}^{"}(\rho) \right| = \left| \sum_{v=2}^{\infty} \frac{v(v-1)\Gamma(\xi+1)\varrho^{v-1}}{\Gamma(\xi+v)} \rho^{v-1} \right|$$ $$\leq \sum_{v=2}^{\infty} \frac{v(v-1)\Gamma(\xi+1)\varrho^{v-1}}{\Gamma(\xi+v)}$$ $$\leq 1 + \sum_{v=2}^{\infty} v(v-1) \left( \frac{\varrho}{\xi+1} \right)^{v-1} = \frac{(\xi-\varrho+1)^3 + 2\varrho(\xi+1)^2}{(\xi-\varrho+1)^3}$$ (17) for $\varrho - 1 < \xi$ . Finally, applying reverse triangle inequality, we conclude that $$\left| \mathbb{E}'_{\xi,\varrho}(\rho) \right| = \left| 1 + \sum_{v=2}^{\infty} \frac{v\Gamma(\xi+1)\varrho^{v-1}}{\Gamma(\xi+v)} \rho^{v-1} \right| \\ \geq 1 - \sum_{v=2}^{\infty} \frac{v\Gamma(\xi+1)\varrho^{v-1}}{\Gamma(\xi+v)} \\ \geq 1 - \sum_{v=2}^{\infty} v \left( \frac{\varrho}{\xi+1} \right)^{v-1} = \frac{(\xi-\varrho+1)^2 - 2\xi\varrho - 2\varrho + \varrho^2}{(\xi-\varrho+1)^2} \tag{18}$$ for $(2+\sqrt{2})\varrho-1<\xi$ . Next, by combining the inequalities (17) with (18), we can easily see that $$\left|\frac{\rho \mathbb{E}_{\xi,\varrho}''(\rho)}{\mathbb{E}_{\xi,\varrho}'(\rho)}\right| \leq \frac{(\xi-\varrho+1)^3 + 2\varrho(\xi+1)^2}{(\xi-\varrho+1)^2 + (\varrho-1)^2 - (2\xi\varrho+1)} \Big(\Big(2+\sqrt{2}\Big)\varrho - 1 < \xi\Big).$$ This completes the proof. $\Box$

Theorem 1. Let v = 1, 2, ..., n, $\xi_v > -1$ , $\varrho > 0$ and $(2 + \sqrt{2})\varrho - 1 < \xi_v$ . Also, let $\gamma, \beta, \eta_v$ and $\zeta_v$ be in $\mathbb C$ such that $$\Re(\gamma) > 0, |\beta| \le 1(\beta \ne -1).$$ Assume that these numbers satisfy the following inequality: $$|\beta| + \frac{1}{|\gamma|} \left( \frac{(\xi - \varrho + 1)^3 + 2\varrho(\xi + 1)^2}{(\xi - \varrho + 1)^2 + (\varrho - 1)^2 - (2\xi\varrho + 1)} \sum_{v=1}^n |\eta_v| + \frac{\varrho(\xi + 1)}{(\xi - \varrho + 1)(\xi - 2\varrho + 1)} \sum_{v=1}^n |\zeta_v| \right) \le 1,$$ where $\xi = \min\{\xi_1, \xi_2, \dots, \xi_n\}$ . Then the function $\mathcal{J}_{\eta_1, \eta_2, \dots, \eta_n; \overline{\zeta}_1, \overline{\zeta}_2, \dots, \zeta_n; n; \gamma}^{\xi_1, \xi_2, \dots, \xi_n; \varrho}$ defined by (5) is in the class S. Proof. Let us define the function $\varphi$ as follows: $$\varphi(\rho) = \mathcal{J}_{\eta_1,\eta_2,\dots,\eta_n;\zeta_1,\zeta_2,\dots,\zeta_n;n;1}^{\xi_1,\xi_2,\dots,\xi_n;\rho}(\rho) = \int_0^\rho \prod_{v=1}^n \left(\mathbb{E}'_{\xi_v,\varrho}(t)\right)^{\eta_v} \left(\frac{\mathbb{E}_{\xi_v,\varrho}(t)}{t}\right)^{\zeta_v} dt. \tag{19}$$ First of all, we observe that $\varphi(\rho) = \varphi'(\rho) - 1 = 0$ , since $\mathbb{E}_{\xi_v,\varrho} \in \mathcal{A}$ for all $v = 1, 2, \ldots, n$ . However, we also have $$\varphi'(\rho) = \prod_{v=1}^{n} \left( \mathbb{E}'_{\xi_{v},\varrho}(\rho) \right)^{\eta_{v}} \left( \frac{\mathbb{E}_{\xi_{v},\varrho}(\rho)}{\rho} \right)^{\zeta_{v}}. \tag{20}$$ Taking the logarithmic derivative of both sides of (20), we get $$\frac{\rho \varphi''(\rho)}{\varphi'(\rho)} = \sum_{v=1}^{n} \eta_{v} \frac{\rho \mathbb{E}_{\xi_{v},\rho}''(\rho)}{\mathbb{E}_{\xi_{v},\rho}'(\rho)} + \sum_{v=1}^{n} \zeta_{v} \left( \frac{\rho \mathbb{E}_{\xi_{v},\rho}'(\rho)}{\mathbb{E}_{\xi_{v},\rho}(\rho)} - 1 \right)$$ (21) and, from (10) and (13), we have $$\left| \frac{\rho \varphi''(\rho)}{\varphi'(\rho)} \right| \leq \sum_{v=1}^{n} \left( |\eta_{v}| \left| \frac{\rho \mathbb{E}''_{\xi_{v},\varrho}(\rho)}{\mathbb{E}'_{\xi_{v},\varrho}(\rho)} \right| + |\zeta_{v}| \left| \frac{\rho \mathbb{E}'_{\xi_{v},\varrho}(\rho)}{\mathbb{E}'_{\xi_{v},\varrho}(\rho)} - 1 \right| \right) \\ \leq \sum_{v=1}^{n} \left( |\eta_{v}| \frac{(\xi_{v} - \varrho + 1)^{3} + 2\varrho(\xi_{v} + 1)^{2}}{(\xi_{v} - \varrho + 1)^{2} + (\varrho - 1)^{2} - (2\xi_{v}\varrho + 1)} + |\zeta_{v}| \frac{\varrho(\xi_{v} + 1)}{(\xi_{v} - \varrho + 1)(\xi_{v} - 2\varrho + 1)} \right) \\ \leq \sum_{v=1}^{n} \left( |\eta_{v}| \frac{(\xi - \varrho + 1)^{3} + 2\varrho(\xi + 1)^{2}}{(\xi - \varrho + 1)^{2} + (\varrho - 1)^{2} - (2\xi\varrho + 1)} + |\zeta_{v}| \frac{\varrho(\xi + 1)}{(\xi - \varrho + 1)(\xi - 2\varrho + 1)} \right), \tag{22}$$ $$\left(\rho \in \mathbb{D}; \left(2+\sqrt{2}\right)\varrho - 1 < \xi, \xi_v < \varrho_0, (v=1,2,\ldots,n)\right)$$ where $\varrho_0 = \frac{1}{3} \left[ -3 + \varrho \left( 7 + \frac{49}{\left( 370 - 3\sqrt{2139} \right)^{1/3}} + \left( 370 - 3\sqrt{2139} \right)^{1/3} \right) \right]$ . Here, we have also used the fact that the functions $$\Theta_1, \Theta_2: \left(\left(2+\sqrt{2}\right)\varrho-1, \varrho_0\right) \longrightarrow \mathbb{R},$$ defined by $$\Theta_1(x) = \frac{(x - \varrho + 1)^3 + 2\varrho(x + 1)^2}{(x - \varrho + 1)^2 + (\varrho - 1)^2 - (2x\varrho + 1)} \text{ and } \Theta_2(x) = \frac{\varrho(x + 1)}{(x - \varrho + 1)(x - 2\varrho + 1)}$$ are decreasing and, consequently, we have $$\frac{(\xi_v - \varrho + 1)^3 + 2\varrho(\xi_v + 1)^2}{(\xi_v - \varrho + 1)^2 + (\varrho - 1)^2 - (2\xi_v \varrho + 1)} < \frac{(\xi - \varrho + 1)^3 + 2\varrho(\xi + 1)^2}{(\xi - \varrho + 1)^2 + (\varrho - 1)^2 - (2\xi\varrho + 1)}$$ (23) and $$\frac{\varrho(\xi_v+1)}{(\xi_v-\varrho+1)(\xi_v-2\varrho+1)}<\frac{\varrho(\xi+1)}{(\xi-\varrho+1)(\xi-2\varrho+1)}.$$ Therefore, from hypothesis of theorem we obtain $$\begin{split} & \left| \beta |\rho|^{2\gamma} + \left( 1 - |\rho|^{2\gamma} \right) \frac{\rho \varphi''(\rho)}{\gamma \varphi'(\rho)} \right| \\ & \leq |\beta| + \frac{1}{|\gamma|} \left( \frac{(\xi - \varrho + 1)^3 + 2\varrho(\xi + 1)^2}{(\xi - \varrho + 1)^2 + (\varrho - 1)^2 - (2\xi\varrho + 1)} \sum_{v=1}^n |\eta_v| + \frac{\varrho(\xi + 1)}{(\xi - \varrho + 1)(\xi - 2\varrho + 1)} \sum_{v=1}^n |\zeta_v| \right) \\ & \leq 1, \end{split}$$ which imply that the function $\mathcal{J}_{\eta_1,\eta_2,...,\eta_n;\tilde{\zeta}_1,\zeta_2,...,\zeta_n;n;\gamma}^{\tilde{\zeta}_1,\tilde{\zeta}_2,...,\tilde{\zeta}_n;\tilde{\varrho}} \in \mathcal{S}$ by Lemma 1.

Theorem 2. Let the parameters $\varrho$ , $\gamma$ , $\eta_v$ , $\xi_v$ and $\zeta_v$ (v = 1, 2, ..., n) be as in Theorem 1. Suppose that $\xi = \min\{\xi_1, \xi_2, ..., \xi_n\}$ and that the following inequality holds true: $$\Re(\gamma) \ge \sum_{v=1}^{n} |\eta_v| \frac{(\xi - \varrho + 1)^3 + 2\varrho(\xi + 1)^2}{(\xi - \varrho + 1)^2 + (\varrho - 1)^2 - (2\xi\varrho + 1)} + \sum_{v=1}^{n} |\zeta_v| \frac{\varrho(\xi + 1)}{(\xi - \varrho + 1)(\xi - 2\varrho + 1)}.$$ Then the function $\mathcal{J}_{\eta_1,\eta_2,\ldots,\eta_n;\zeta_1,\zeta_2,\ldots,\zeta_n;n;\gamma}^{\xi_1,\xi_2,\ldots,\xi_n;\theta}$ defined by (5) is in the class $\mathcal{S}$ . Proof. Let us consider the function $\varphi$ as in (19). From (22) and hypothesis of theorem, we get $$\frac{1 - |\rho|^{2\Re(\gamma)}}{\Re(\gamma)} \left| \frac{\rho \varphi''(\rho)}{\varphi'(\rho)} \right| \\ \leq \frac{1 - |\rho|^{2\Re(\gamma)}}{\Re(\gamma)} \sum_{v=1}^{n} \left( |\eta_{v}| \frac{(\xi_{v} - \varrho + 1)^{3} + 2\varrho(\xi_{v} + 1)^{2}}{(\xi_{v} - \varrho + 1)^{2} + (\varrho - 1)^{2} - (2\xi_{v}\varrho + 1)} + |\zeta_{v}| \frac{\varrho(\xi_{v} + 1)}{(\xi_{v} - \varrho + 1)(\xi_{v} - 2\varrho + 1)} \right) \\ \leq \frac{1}{\Re(\gamma)} \sum_{v=1}^{n} \left( |\eta_{v}| \frac{(\xi - \varrho + 1)^{3} + 2\varrho(\xi + 1)^{2}}{(\xi - \varrho + 1)^{2} + (\varrho - 1)^{2} - (2\xi\varrho + 1)} + |\zeta_{v}| \frac{\varrho(\xi + 1)}{(\xi - \varrho + 1)(\xi - 2\varrho + 1)} \right) \\ < 1 \tag{24}$$ By Lemma 2, the inequality (24) imply that the function $\mathcal{J}_{\eta_1,\eta_2,\dots,\eta_n;\zeta_1,\zeta_2,\dots,\zeta_n;n;\gamma}^{\xi_1,\xi_2,\dots,\xi_n;n} \in \mathcal{S}$ .

Theorem 3. Let the parameters $\varrho$ , $\eta_v$ , $\xi_v$ and $\zeta_v$ (v = 1, 2, ..., n) be as in Theorem 1. Suppose that $\xi = \min\{\xi_1, \xi_2, ..., \xi_n\}$ and that the following inequality holds true: $$0 < \sum_{v=1}^{n} \left( |\eta_v| \frac{(\xi - \varrho + 1)^3 + 2\varrho(\xi + 1)^2}{(\xi - \varrho + 1)^2 + (\varrho - 1)^2 - (2\xi\varrho + 1)} + |\zeta_v| \frac{\varrho(\xi + 1)}{(\xi - \varrho + 1)(\xi - 2\varrho + 1)} \right) \le 1.$$ Then the function $\mathcal{J}_{\eta_1,\eta_2,\ldots,\eta_n;\zeta_1,\zeta_2,\ldots,\zeta_n;n;1}^{\xi_1,\xi_2,\ldots,\xi_n;\varrho}$ defined by (5) with $\gamma=1$ is convex of order $\delta$ given by $$\delta = 1 - \sum_{v=1}^{n} \left( |\eta_v| \frac{(\xi - \varrho + 1)^3 + 2\varrho(\xi + 1)^2}{(\xi - \varrho + 1)^2 + (\varrho - 1)^2 - (2\xi\varrho + 1)} + |\zeta_v| \frac{\varrho(\xi + 1)}{(\xi - \varrho + 1)(\xi - 2\varrho + 1)} \right).$$ Proof. The inequality (22) and hypothesis of theorem show that $$\left| \frac{\rho \varphi''(\rho)}{\varphi'(\rho)} \right| \leq \sum_{v=1}^{n} |\eta_v| \frac{(\xi - \varrho + 1)^3 + 2\varrho(\xi + 1)^2}{(\xi - \varrho + 1)^2 + (\varrho - 1)^2 - (2\xi\varrho + 1)} + \sum_{v=1}^{n} |\zeta_v| \frac{\varrho(\xi + 1)}{(\xi - \varrho + 1)(\xi - 2\varrho + 1)}$$ $$= 1 - \delta.$$ As a result, the function $\varphi$ is convex of order $$\delta = 1 - \sum_{v=1}^{n} \left( |\eta_v| \frac{(\xi - \varrho + 1)^3 + 2\varrho(\xi + 1)^2}{(\xi - \varrho + 1)^2 + (\varrho - 1)^2 - (2\xi\varrho + 1)} + |\zeta_v| \frac{\varrho(\xi + 1)}{(\xi - \varrho + 1)(\xi - 2\varrho + 1)} \right).$$ In Theorem 1 with n = 1, $\xi_1 = 1$ and $\varrho = 1/2$ , we can write the following corollary.

Corollary 1. Let γ, β, η and ζ be in C such that <(γ) > 0 and |β| ≤ 1(β 6= −1). If the inequality $$|\beta| + \frac{1}{|\gamma|} \left( \frac{59}{4} |\eta| + \frac{2}{3} |\zeta| \right) \le 1$$ holds true, then the function $$\left[\gamma \int_0^\rho t^{-\zeta+\gamma-1} \left(e^{t/2}\right)^{\eta} \left(2e^{t/2}-2\right)^{\zeta} dt\right]^{1/\gamma}$$ is in the class S.

Corollary 2. Let γ, β, η and ζ be in C such that <(γ) > 0 and |β| ≤ 1(β 6= −1). If the inequality $$|\beta| + \frac{1}{|\gamma|} \left( \frac{59}{2} |\eta| + \frac{2}{3} |\zeta| \right) \leqq 1$$ holds, then the function $$\left[ 6^{\eta} 3^{\zeta} \gamma \int_{0}^{\rho} t^{-3\eta - 3\zeta + \gamma - 1} \big( t e^{t} - 2 e^{t} + t + 2 \big)^{\eta} \Big( 2 e^{t} - t^{2} - 2t - 2 \Big)^{\zeta} dt \right]^{1/\gamma}$$ is in the class S. From Theorem 3 with n = 1, ξ¹ = 1 and \$ = 1/2, we can get result below.$

Corollary 3. Let η and ζ be complex numbers such that $$\frac{59}{4}|\eta| + \frac{2}{3}|\zeta| \leqq 1.$$ Then the function $$\int_0^\rho t^{-\zeta} \left(e^{t/2}\right)^\eta \left(2e^{t/2}-2\right)^\zeta dt$$ is convex of order δ given by $$\delta = 1 - \frac{59}{4} |\eta| - \frac{2}{3} |\zeta|.$$ Let n = 1, ξ¹ = 3 and \$ = 1 in the Theorem 3, then we get following result.$

Corollary 4. Let η and ζ be complex numbers such that $$0 < \frac{59}{2}|\eta| + \frac{2}{3}|\zeta| \le 1.$$ Then the function $$6^{\eta}3^{\zeta}\int_{0}^{\rho}t^{-3\eta-3\zeta}\left(te^{t}-2e^{t}+t+2\right)^{\eta}\left(2e^{t}-t^{2}-2t-2\right)^{\zeta}dt$$ is convex of order δ given by $$\delta = 1 - \frac{59}{2} |\eta| - \frac{2}{3} |\zeta|.$$

Theorem 4. Let v = 1, 2, . . . , n, ξ^v > −1, \$ > 0 and 2\$ − 1 < ξv. Also, let β, η^v and ζ^v be in C such that $$|\beta| \leq 1(\beta \neq -1) \ \text{and} \ \Re \Bigg( 1 + \sum_{v=1}^n \eta_v \Bigg) > 0.$$ Assume that these numbers satisfy the following inequality: $$|\beta| + \frac{1}{|1 + \sum_{v=1}^{n} \eta_v|} \sum_{v=1}^{n} \left[ |\eta_v| \frac{(\xi + 1)^2}{(\xi - \varrho + 1)(\xi - 2\varrho + 1)} + |\zeta_v| \left( \frac{\xi + 1}{\xi - \varrho + 1} \right)^2 \right] \le 1,$$ where ξ = min{ξ1, ξ2, . . . , ξn}. Then the function K ξ1 ,ξ2,...,ξn;\$ η1 ,η2,...,ηn;ζ¹ ,ζ2,...,ζn;n defined by (6) is in the class S. Proof. Let us define the functions ψ by $$\psi(\rho) = \int_0^\rho \prod_{v=1}^n \left( \mathbb{E}_{\xi_v, \varrho_v}(t) \right)^{\eta_v} \left( e^{\mathbb{E}_{\xi_v, \varrho_v}(t)} \right)^{\zeta_v} dt. \tag{25}$$ Then ψ(0) = ψ 0 (0) − 1 = 0. Differentiating both sides of (25) logarithmically, we get $$\frac{\rho\psi''(\rho)}{\psi'(\rho)} = \sum_{v=1}^n \eta_v \left( \frac{\rho \mathbb{E}'_{\xi_v,\varrho}(\rho)}{\mathbb{E}_{\xi_v,\varrho}(\rho)} \right) + \sum_{v=1}^n \zeta_v \left( \rho \mathbb{E}'_{\xi_v,\varrho}(\rho) \right)$$ and, from (11) and (12) in Lemma 3, we obtain $$\left| \frac{\rho \psi''(\rho)}{\psi'(\rho)} \right| \leq \sum_{v=1}^{n} \left( \left| \eta_{v} \right| \left| \frac{\rho \mathbb{E}'_{\xi_{v},\varrho}(\rho)}{\mathbb{E}_{\xi_{v},\varrho}(\rho)} \right| + \left| \zeta_{v} \right| \left| \rho \mathbb{E}'_{\xi_{v},\varrho}(\rho) \right| \right) \\ \leq \sum_{v=1}^{n} \left[ \left| \eta_{v} \right| \frac{(\xi_{v}+1)^{2}}{(\xi_{v}-\varrho+1)(\xi_{v}-2\varrho+1)} + \left| \zeta_{v} \right| \left( \frac{\xi_{v}+1}{\xi_{v}-\varrho+1} \right)^{2} \right] \\ \leq \sum_{v=1}^{n} \left[ \left| \eta_{v} \right| \frac{(\xi+1)^{2}}{(\xi-\varrho+1)(\xi-2\varrho+1)} + \left| \zeta_{v} \right| \left( \frac{\xi+1}{\xi-\varrho+1} \right)^{2} \right]. \\ (\rho \in \mathbb{D}; \xi, \xi_{v} > 2\varrho - 1, (v = 1, 2, \dots, n)). \tag{26}$$ Here, since the functions $$\Theta_3, \Theta_4: (2\varrho - 1, \infty) \longrightarrow \mathbb{R}$$ defined by $$\Theta_3(x) = \frac{(x+1)^2}{(x-\varrho+1)(x-2\varrho+1)}$$ and $\Theta_4(x) = \left(\frac{x+1}{x-\varrho+1}\right)^2$ , are decreasing, the inequalities $$\frac{(\xi_v + 1)^2}{(\xi_v - \varrho + 1)(\xi_v - 2\varrho + 1)} < \frac{(\xi + 1)^2}{(\xi - \varrho + 1)(\xi - 2\varrho + 1)}$$ and $$\left(\frac{\xi_v + 1}{\xi_v - \varrho + 1}\right)^2 < \left(\frac{\xi + 1}{\xi - \varrho + 1}\right)^2 \tag{27}$$ holds. Thus, we have $$\left| \beta |\rho|^{2(1+\sum_{v=1}^{n} \eta_{v})} + \left(1 - |\rho|^{2(1+\sum_{v=1}^{n} \eta_{v})}\right) \frac{\rho \psi''(\rho)}{(1+\sum_{v=1}^{n} \eta_{v})\psi'(\rho)} \right| \leq |\beta| + \left| \frac{\rho \psi''(\rho)}{(1+\sum_{v=1}^{n} \eta_{v})\psi'(\rho)} \right| \leq |\beta| + \frac{1}{|1+\sum_{v=1}^{n} \eta_{v}|} \sum_{v=1}^{n} \left[ |\eta_{v}| \frac{(\xi+1)^{2}}{(\xi-\varrho+1)(\xi-2\varrho+1)} + |\zeta_{v}| \left(\frac{\xi+1}{\xi-\varrho+1}\right)^{2} \right] \leq 1.$$ (28) Using Lemma 1 with $$\alpha = 1 + \sum_{v=1}^{n} \eta_v,$$ the inequality (28) imply that the function $\mathcal{K}_{\eta_1,\eta_2,\ldots,\eta_n;\zeta_1,\zeta_2,\ldots,\zeta_n;n}^{\xi_1,\xi_2,\ldots,\xi_n;\varrho} \in \mathcal{S}.$$

Theorem 5. Let the parameters $\varrho$ , $\eta_v$ , $\xi_v$ and $\zeta_v$ (v = 1, 2, ..., n) be as in Theorem 4. Suppose that $\xi = \min\{\xi_1, \xi_2, ..., \xi_n\}$ and that the following inequality holds true: $$\sum_{v=1}^{n} \left[ |\eta_v| \frac{(\xi+1)^2}{(\xi-\varrho+1)(\xi-2\varrho+1)} + |\zeta_v| \left( \frac{\xi+1}{\xi-\varrho+1} \right)^2 \right] \le 1.$$ Then the function $\mathcal{K}_{\eta_1,\eta_2,...,\eta_n;\zeta_1,\zeta_2,...,\zeta_n;n}^{\xi_1,\xi_2,...,\xi_n;\varrho}$ defined by (6) is in the normalized univalent function class $\mathcal{S}$ . Proof. Let us consider the function $\psi$ as in (25). Therefore, from (26) and hypothesis of theorem we can easily see that $$\left(1 - |\rho|^2\right) \left| \frac{\rho \psi''(\rho)}{\psi'(\rho)} \right| \leq \sum_{v=1}^n \left[ |\eta_v| \frac{(\xi+1)^2}{(\xi-\varrho+1)(\xi-2\varrho+1)} + |\zeta_v| \left(\frac{\xi+1}{\xi-\varrho+1}\right)^2 \right] < 1.$$ (29) By Lemma 2, with $\omega=1$ and $\alpha=1+\sum_{v=1}^n\eta_v$ , the inequality (29) imply that the function $\mathcal{K}_{\eta_1,\eta_2,\ldots,\eta_n;\zeta_1,\zeta_2,\ldots,\zeta_n;n}^{\xi_1,\xi_2,\ldots,\xi_n;\varrho}\in\mathcal{S}$ . $\square$

Theorem 6. Let $v=1,2,\ldots,n$ , $\xi_v>-1$ , $\varrho>0$ and $2\varrho-1<\xi_v$ . Also, let $\eta_v$ and $\zeta_v$ be in $\mathbb C$ such that $$\Re\left(1+\sum_{v=1}^n\eta_v\right)>0.$$ Moreover, suppose that the following inequality holds true: $$0 < \sum_{v=1}^n \left[ |\eta_v| \frac{(\xi+1)^2}{(\xi-\varrho+1)(\xi-2\varrho+1)} + |\zeta_v| \left( \frac{\xi+1}{\xi-\varrho+1} \right)^2 \right] \le 1,$$ where $\xi = \min\{\xi_1, \xi_2, \dots, \xi_n\}$ . Then the function $\mathcal{K}_{\eta_1, \eta_2, \dots, \eta_n; \xi_1, \xi_2, \dots, \xi_n; n}^{\xi_1, \xi_2, \dots, \xi_n; \varrho}$ defined by (6), is convex of order $\delta$ given by $$\delta = 1 - \sum_{v=1}^{n} \left[ |\eta_v| \frac{(\xi+1)^2}{(\xi-\varrho+1)(\xi-2\varrho+1)} + |\zeta_v| \left( \frac{\xi+1}{\xi-\varrho+1} \right)^2 \right].$$ Proof. By using (26) we conclude that $$\left| \frac{\rho \psi''(\rho)}{\psi'(\rho)} \right| \leq \sum_{v=1}^{n} \left[ |\eta_{v}| \left| \frac{\rho \mathbb{E}'_{\xi_{v},\varrho}(\rho)}{\mathbb{E}_{\xi_{v},\varrho}(\rho)} \right| + |\zeta_{v}| \left| z \mathbb{E}'_{\xi_{v},\varrho}(\rho) \right| \right] \\ \leq \sum_{v=1}^{n} \left[ |\eta_{v}| \frac{(\xi+1)^{2}}{(\xi-\varrho+1)(\xi-2\varrho+1)} + |\zeta_{v}| \left( \frac{\xi+1}{\xi-\varrho+1} \right)^{2} \right] \\ = 1 - \delta$$ This show that, the function $\psi$ is convex of order $$\delta = 1 - \sum_{v=1}^{n} \left[ |\eta_v| \frac{(\xi+1)^2}{(\xi-\varrho+1)(\xi-2\varrho+1)} + |\zeta_v| \left(\frac{\xi+1}{\xi-\varrho+1}\right)^2 \right].$$ From Theorem 4 with n = 1, $\xi_1 = 2$ and $\varrho = 1/2$ , we can get the following result.

Corollary 5. Let $\beta$ , $\eta$ and $\zeta$ be in $\mathbb C$ such that $\Re(1+\eta)>0$ and $|\beta|\leq 1(\beta\neq -1)$ . If these numbers satisfy the inequality: $$|\beta| + \frac{1}{|1+\eta|} \left( \frac{9}{5} |\eta| + \frac{36}{25} |\zeta| \right) \le 1,$$ then the function $$\left[2^{\eta}(1+\eta)\int_{0}^{\rho} \left(\frac{4\left(2e^{t/2}-t-2\right)}{t}\right)^{\eta} e^{\frac{4\zeta\left(2e^{t/2}-t-2\right)}{t}} dt\right]^{1/(1+\eta)} \tag{30}$$ is in the class S. Example 2. From Corollary 5 with $\eta = 1$ and $\zeta = 1/8$ , we have $$\hbar_2(\rho)=4\Bigg[\int_0^\rho\Bigg(\frac{2e^{t/2}-t-2}{t}\Bigg)e^{\frac{2e^{t/2}-t-2}{2t}}dt\Bigg]^{1/2}\in\mathcal{S}.$$ In reality, by a simple calculation, we get $$1 + \frac{\rho h_2''(\rho)}{h_2'(\rho)} + \frac{\rho h_2'(\rho)}{h_2(\rho)} = 1 + \frac{2e^{\rho}(\rho - 2) + e^{\rho/2}(\rho^2 - 4\rho + 8) + 2\rho - 4}{2\rho(2e^{\rho/2} - \rho - 2)} = g(\rho).$$ It also holds true that $\Re(g(\rho)) > 0$ for all $\rho \in \mathbb{D}$ (see Figure 2). Therefore, $\hbar_2$ is a 1/2-convex function [[35], Vol. I, p. 142]. Thus it follows from [[35], Vol. I, p. 142] that $\hbar_2$ belongs to the class S.  Figure 2. Image of $\mathbb{D}$ under g. From Theorem 4 with n = 1, $\xi_1 = 3$ and $\varrho = 1$ , we can get result below.

Corollary 6. Let $\beta$ , $\eta$ and $\zeta$ be in $\mathbb C$ such that $\Re(1+\eta)>0$ and $|\beta|\leq 1(\beta\neq -1)$ . If these numbers satisfy the following inequality $$|\beta| + \frac{1}{|1+\eta|} \left( \frac{8}{3} |\eta| + \frac{16}{9} |\zeta| \right) \le 1,$$ then the function $$\left[3^{\eta}(1+\eta)\int_{0}^{\rho} \left(\frac{2e^{t}-t^{2}-2t-2}{t^{2}}\right)^{\eta} e^{\zeta\left(\frac{6e^{t}-3t^{2}-6t-6}{t^{2}}\right)} dt\right]^{1/(1+\eta)}$$ (31) is in the class S. From Theorem 6 with n = 1, $\xi_1 = 2$ and $\varrho = 1/2$ , we can get the following result.

Corollary 7. Let $\eta$ and $\zeta$ be a complex numbers such that $$\Re(1+\eta)>0 \text{ and } 0<\frac{9}{5}|\eta|+\frac{36}{25}|\zeta|\leqq 1.$$ Then the function defined by (30) is convex of order δ given by $$\delta = 1 - \frac{9}{5}|\eta| - \frac{36}{25}|\zeta|.$$ From Theorem 6 with n = 1, ξ¹ = 3 and \$ = 1, we can get the following result.$

Corollary 8. Let η and ζ be a complex numbers such that $$\Re(1+\eta) > 0$$ and $0 < \frac{8}{3}|\eta| + \frac{16}{9}|\zeta| \le 1$ . Then the function defined by (31) is convex of order δ given by $$\delta = 1 - \frac{8}{3}|\eta| - \frac{16}{9}|\zeta|.$$

Theorem 7. Let ^v ⁼ 1, 2, . . . , ⁿ, ^ξ^v ^&gt; ⁻1, \$ ^&gt; ⁰ and 2 + √ 2 \$ − 1 < ξv. Also, let γ, β, η^v and ζ^v be in C such that $$\Re(\gamma) > 0, |\beta| \le 1(\beta \ne -1).$$ Assume that these numbers satisfy the following inequality: $$|\beta| + \frac{1}{|\gamma|} \left\{ \sum_{v=1}^{n} |\eta_v| \frac{(\xi - \varrho + 1)^3 + 2\varrho(\xi + 1)^2}{(\xi - \varrho + 1)^2 + (\varrho - 1)^2 - (2\xi\varrho + 1)} + \sum_{v=1}^{n} |\zeta_v| \left( \frac{\xi + 1}{\xi - \varrho + 1} \right)^2 \right\} \le 1$$ where ξ = min{ξ1, ξ2, . . . , ξn}. Then the function L ξ1 ,ξ2,...,ξn;\$ η1 ,η2,...,ηn;ζ¹ ,ζ2,...,ζn;n;γ defined by (7) is in the class S. Proof. Let us define the function φ by $$\phi(\rho) := \mathcal{L}_{\eta_1, \eta_2, \dots, \eta_n; \zeta_1, \zeta_2, \dots, \zeta_n; n; 1}^{\xi_1, \xi_2, \dots, \xi_n; \rho}(\rho) = \int_0^\rho \prod_{v=1}^n \left( \mathbb{E}'_{\xi_v, \varrho}(t) \right)^{\eta_v} \left( e^{\mathbb{E}_{\xi_v, \varrho}(t)} \right)^{\zeta_v} dt, \tag{32}$$ so that, obviously, $$\phi'(\rho) = \prod_{v=1}^{n} \left( \mathbb{E}'_{\xi_{v,\varrho}}(\rho) \right)^{\eta_v} \left( e^{\mathbb{E}_{\xi_{v,\varrho}}(\rho)} \right)^{\zeta_v} \tag{33}$$ and φ(ρ) = φ 0 (ρ) − 1 = 0. Now we differentiate (33) logarithmically and multiply by ρ, we obtain $$\frac{\rho\phi''(\rho)}{\phi'(\rho)} = \sum_{v=1}^n \eta_v \frac{\rho \mathbb{E}''_{\xi_v,\varrho}(\rho)}{\mathbb{E}'_{\xi_v,\varrho}(\rho)} + \sum_{v=1}^n \zeta_v \rho \mathbb{E}'_{\xi_v,\varrho}(\rho).$$ Furthermore, by (11), (13), (23) and (27) we obtain $$\left| \frac{\rho \phi''(\rho)}{\phi'(\rho)} \right| \leq \sum_{v=1}^{n} \left( |\eta_{v}| \left| \frac{\rho \mathbb{E}''_{\xi_{v},\varrho}(\rho)}{\mathbb{E}'_{\xi_{v},\varrho}(\rho)} \right| + |\zeta_{v}| \left| \rho \mathbb{E}'_{\xi_{v},\varrho}(\rho) \right| \right) \\ \leq \sum_{v=1}^{n} \left[ |\eta_{v}| \frac{(\xi_{v} - \varrho + 1)^{3} + 2\varrho(\xi_{v} + 1)^{2}}{(\xi_{v} - \varrho + 1)^{2} + (\varrho - 1)^{2} - (2\xi_{v}\varrho + 1)} + |\zeta_{v}| \left( \frac{\xi_{v} + 1}{\xi_{v} - \varrho + 1} \right)^{2} \right] \\ \leq \sum_{v=1}^{n} \left[ |\eta_{v}| \frac{(\xi - \varrho + 1)^{3} + 2\varrho(\xi + 1)^{2}}{(\xi - \varrho + 1)^{2} + (\varrho - 1)^{2} - (2\xi\varrho + 1)} + |\zeta_{v}| \left( \frac{\xi + 1}{\xi - \varrho + 1} \right)^{2} \right]. \tag{34}$$ Hence, from (34) we have $$\begin{split} & \left| \beta |\rho|^{2\gamma} + \left( 1 - |\rho|^{2\gamma} \right) \frac{\rho \phi''(\rho)}{\gamma \phi'(\rho)} \right| \\ & \leq |\beta| + \left| \frac{\rho \phi''(\rho)}{\phi'(\rho)} \right| \\ & \leq |\beta| + \frac{1}{|\gamma|} \left\{ \sum_{v=1}^{n} |\eta_v| \frac{(\xi - \varrho + 1)^3 + 2\varrho(\xi + 1)^2}{(\xi - \varrho + 1)^2 + (\varrho - 1)^2 - (2\xi\varrho + 1)} + \sum_{v=1}^{n} |\zeta_v| \left( \frac{\xi + 1}{\xi - \varrho + 1} \right)^2 \right\} \\ & \leq 1, \end{split}$$ which, in view of Lemma 1, implies that $\mathcal{L}_{\eta_1,\eta_2,\ldots,\eta_n;\zeta_1,\zeta_2,\ldots,\zeta_n;n;\gamma}^{\xi_1,\xi_2,\ldots,\xi_n;\eta}\in\mathcal{S}.$$

Theorem 8. Let the parameters $\varrho$ , $\gamma$ , $\eta_v$ , $\xi_v$ and $\zeta_v$ (v = 1, 2, ..., n) be as in Theorem 7. Suppose that $\xi = \min\{\xi_1, \xi_2, ..., \xi_n\}$ and that the following inequality holds true: $$\Re(\gamma) \ge \left\{ \sum_{v=1}^{n} |\eta_v| \frac{(\xi - \varrho + 1)^3 + 2\varrho(\xi + 1)^2}{(\xi - \varrho + 1)^2 + (\varrho - 1)^2 - (2\xi\varrho + 1)} + \sum_{v=1}^{n} |\zeta_v| \left(\frac{\xi + 1}{\xi - \varrho + 1}\right)^2 \right\}.$$ Then the function $\mathcal{L}_{\eta_1,\eta_2,...,\zeta_n;\zeta_1,\zeta_2,...,\zeta_n;n;\gamma}^{\xi_1,\xi_2,...,\xi_n;\eta}$ defined by (7) is in the class S. Proof. By using (34) we obtain $$\begin{split} &\frac{1-|\rho|^{2\Re(\gamma)}}{\Re(\gamma)} \left| \frac{\rho \phi''(\rho)}{\phi'(\rho)} \right| \\ &\leq \frac{1-|\rho|^{2\Re(\gamma)}}{\Re(\gamma)} \left\{ \sum_{v=1}^{n} |\eta_v| \frac{(\xi-\varrho+1)^3 + 2\varrho(\xi+1)^2}{(\xi-\varrho+1)^2 + (\varrho-1)^2 - (2\xi\varrho+1)} + \sum_{v=1}^{n} |\zeta_v| \left( \frac{\xi+1}{\xi-\varrho+1} \right)^2 \right\} \\ &\leq \frac{1}{\Re(\gamma)} \left\{ \sum_{v=1}^{n} |\eta_v| \frac{(\xi-\varrho+1)^3 + 2\varrho(\xi+1)^2}{(\xi-\varrho+1)^2 + (\varrho-1)^2 - (2\xi\varrho+1)} + \sum_{v=1}^{n} |\zeta_v| \left( \frac{\xi+1}{\xi-\varrho+1} \right)^2 \right\} \\ &\leq 1, \end{split}$$ which, in view of Lemma 2, implies that $\mathcal{L}_{\eta_1,\eta_2,\ldots,\eta_n;\zeta_1,\zeta_2,\ldots,\zeta_n;n;\gamma}^{\xi_1,\xi_2,\ldots,\xi_n;n}\in\mathcal{S}.$

Theorem 9. Let v = 1, 2, ..., n, $\xi_v > -1$ , $\varrho > 0$ and $(2 + \sqrt{2})\varrho - 1 < \xi_v$ . Also, let $\gamma, \eta_v$ and $\zeta_v$ be in $\mathbb C$ such that $\Re(\gamma) > 0$ . Assume that these numbers satisfy the following inequality: $$0 < \sum_{v=1}^{n} |\eta_v| \frac{(\xi - \varrho + 1)^3 + 2\varrho(\xi + 1)^2}{(\xi - \varrho + 1)^2 + (\varrho - 1)^2 - (2\xi\varrho + 1)} + \sum_{v=1}^{n} |\zeta_v| \left(\frac{\xi + 1}{\xi - \varrho + 1}\right)^2 \le 1,$$ where $\xi = \min\{\xi_1, \xi_2, \dots, \xi_n\}$ . Then the function $\mathcal{L}_{\eta_1, \eta_2, \dots, \eta_n; \xi_1, \xi_2, \dots, \xi_n; n; 1}^{\xi_1, \xi_2, \dots, \xi_n; \rho}$ , defined by (7) with $\gamma = 1$ , is convex of order $\delta$ given by $$\delta = 1 - \sum_{v=1}^{n} |\eta_v| \frac{(\xi - \varrho + 1)^3 + 2\varrho(\xi + 1)^2}{(\xi - \varrho + 1)^2 + (\varrho - 1)^2 - (2\xi\varrho + 1)} - \sum_{v=1}^{n} |\zeta_v| \left(\frac{\xi + 1}{\xi - \varrho + 1}\right)^2.$$ Proof. From (34) and hypothesis of theorem, we obtain $$\left| \frac{\rho \phi''(\rho)}{\phi'(\rho)} \right| \leq \sum_{v=1}^{n} |\eta_v| \frac{(\xi - \varrho + 1)^3 + 2\varrho(\xi + 1)^2}{(\xi - \varrho + 1)^2 + (\varrho - 1)^2 - (2\xi\varrho + 1)} + \sum_{v=1}^{n} |\zeta_v| \left( \frac{\xi + 1}{\xi - \varrho + 1} \right)^2 \\ = 1 - \delta.$$ Therefore, the function $\phi$ is convex of order $\delta$ . From Theorem 7 with n = 1, $\xi_1 = 1$ and $\varrho = 1/2$ , we can get following result.

Corollary 9. Let $\gamma$ , $\beta$ , $\eta$ and $\zeta$ be in $\mathbb{C}$ such that $\Re(\gamma) > 0$ , $|\beta| \le 1(\beta \ne -1)$ . If these numbers satisfy the inequality: $$|\beta| + \frac{1}{|\gamma|} \left( \frac{59}{4} |\eta| + \frac{16}{9} |\zeta| \right) \le 1$$ then the function $$\left[\gamma \int_0^\rho t^{\gamma-1} \left(e^{t/2}\right)^{\eta} \left(e^{2e^{t/2}-2}\right)^{\zeta} dt\right]^{1/\gamma}$$ is in the normalized univalent function class S. Example 3. From Corollary 9 with $\beta = 0$ , $\gamma = 1$ , $\eta = 0.01$ and $\zeta = 0.1$ , we obtain $$h_3(\rho) = \int_0^{\rho} \left(e^{t/200}\right) \left(e^{\frac{e^{t/2}-1}{5}}\right) dt \in \mathcal{S}.$$ From Theorem 7 with n = 1, $\xi_1 = 3$ and $\varrho = 1$ , we can get result below.

Corollary 10. Let $\gamma$ , $\beta$ , $\eta$ and $\zeta$ be in $\mathbb C$ such that $\Re(\gamma) > 0$ and $\beta \neq -1$ . If these numbers satisfy the inequality $$|\beta| + \frac{1}{|\gamma|} \left( \frac{59}{2} |\eta| + \frac{16}{9} |\zeta| \right) \le 1$$ then the function $$\left[6^{\eta}\gamma \int_{0}^{\rho} t^{-3\eta+\gamma-1} (te^{t}-2e^{t}+t+2)^{\eta} e^{3\zeta \left(\frac{2e^{t}-t^{2}-2t-2}{t^{2}}\right)} dt\right]^{1/\gamma}$$ is in the normalized univalent function class S. From Theorem 9 with n = 1, $\xi_1 = 1$ and $\varrho = 1/2$ , we have following result.

Corollary 11. Let $\eta$ and $\zeta$ be complex numbers such that $$\frac{59}{4}|\eta| + \frac{16}{9}|\zeta| \le 1.$$ Then the function $$\int_0^{\rho} \left(e^{t/2}\right)^{\eta} \left(e^{2e^{t/2}-2}\right)^{\zeta} dt$$ is convex of order $\delta$ given by $$\delta = 1 - \frac{59}{4} |\eta| - \frac{16}{9} |\zeta|.$$ From Theorem 9 with n = 1, $\xi_1 = 3$ and $\varrho = 1$ , we have following result.

Corollary 12. Let $\eta$ and $\zeta$ be complex numbers such that $$\frac{59}{2}|\eta| + \frac{16}{9}|\zeta| \le 1.$$ Then the function $$6^{\eta} \int_{0}^{\rho} t^{-3\eta} \left( te^{t} - 2e^{t} + t + 2 \right)^{\eta} e^{3\zeta \left( \frac{2e^{t} - t^{2} - 2t - 2}{t^{2}} \right)} dt$$ is convex of order $\delta$ given by $$\delta = 1 - \frac{59}{2} |\eta| - \frac{16}{9} |\zeta|.$$ Example 4. From Corollary 12 with $\eta = 0.01$ and $\zeta = 0.1$ , we get $$\hbar_4(\rho) = 6^{1/100} \int_0^{\rho} t^{-3/100} \left( t e^t - 2e^t + t + 2 \right)^{1/100} e^{3 \left( \frac{2e^t - t^2 - 2t - 2}{10t^2} \right)} dt$$ is convex of order $\delta = 949/1800$ .

Lemma 1. (see [58]). If p is a function with a positive real part and $$p(z) = 1 + \sum_{l=1}^{\infty} c_l z^l,$$ then $$|c_l| \leq 2$$ . The problem of finding bounds for the coefficients has always been a key concern in geometric function theory. The size of their coefficients can determine a number of properties of analytic functions, including univalency, rate of growth and distortion. Many scholars have used a variety of methods to overcome the aforementioned issues. Similar to univalent functions, bi-univalent function coefficient estimation has received a lot of interest lately. As a result of the significance of studying the coefficient problems described above, in this section, we utilize the (varrho, q)-fractional derivative operator and the Fabor polynomial technique to obtain coefficient estimates for $|a_1|$ and discuss the unpredictable behavior of the initial coefficient bounds for $|a_2|$ and $|a_3|$ . We also investigate the Fekete–Szegö problem and give some examples. We also demonstrate how some of the previously published results would be improved and generalized as a result of our primary findings as well as their corollaries and consequences.

Theorem 1. If h has the series representation stated in (1) and belongs to the class $\mathcal{K}_{\Sigma}(q, \alpha, \varrho)$ , and if $a_i = 0$ and $2 \le i \le l - 1$ , then $$|a_l| \le \frac{\Gamma_q(2)\Gamma_q(l+1-\varrho)(2(1-\alpha)+l)}{[l]_a\Gamma_a(2-\varrho)\Gamma_a(l+1)} \qquad (l \ge 3).$$ Proof. For $h \in \mathcal{K}_{\Sigma}(q, \alpha, \varrho)$ , there exists a function g. The FPE for $\frac{D_q(\Omega_q^{\varrho}h(z))}{g(z)}$ is given by $$\frac{D_q\left(\Omega_q^{\varrho}h(z)\right)}{g(z)} = 1 + \sum_{l=2}^{\infty} \left[ \begin{array}{c} \left( [l]_q \frac{\Gamma_q(2-\varrho)\Gamma_q(l+1)}{\Gamma_q(2)\Gamma_q(l+1-\varrho)} a_l - b_l \right) \sum_{l=1}^{l-2} \mathfrak{q}_l^{-1}(b_2, b_3, \cdots, b_{l+1}) \\ \cdot \left( \left( [l]_q - l \right) \frac{\Gamma_q(2-\varrho)\Gamma_q(l+1)}{\Gamma_q(2)\Gamma_q(l+1-\varrho)} a_{l-l} - b_{l-l} \right) \end{array} \right] z^{l-1}.$$ (9) Additionally, regarding the inverse maps $\gamma = h^{-1}$ and $\delta = g^{-1}$ , we obtain $$\frac{D_q\left(\Omega_q^{\varrho}\alpha(\vartheta)\right)}{\delta(\vartheta)} = 1 + \sum_{l=2}^{\infty} \left[ \begin{array}{c} \left( [l]_q \frac{\Gamma_q(2-\varrho)\Gamma_q(l+1)}{\Gamma_q(2)\Gamma_q(l+1-\varrho)} A_l - B_l \right) \sum_{l=1}^{l-2} \mathfrak{q}_l^{-1}(B_2, B_3, \cdots, B_{l+1}) \\ \cdot \left( \left( [l]_q - l \right) \frac{\Gamma_q(2-\varrho)\Gamma_q(l+1)}{\Gamma_q(2)\Gamma_q(l+1-\varrho)} A_{l-l} - B_{l-l} \right) \end{array} \right] \vartheta^{l-1}.$$ (10) As opposed to that, since $$\Re\left(\frac{D_q\left(\Omega_q^\varrho h(z)\right)}{g(z)}\right) > \alpha \qquad (z \in \mathbb{E}),$$ there must exist a function p(z) given by $$p(z) = 1 + \sum_{l=1}^{\infty} c_l z^l$$ such that $$\frac{D_q\left(\Omega_q^\varrho h(z)\right)}{g(z)} = 1 + (1 - \alpha)p(z)$$ $$= 1 + (1 - \alpha)\sum_{l=1}^{\infty} c_l z^l. \tag{11}$$ Similarly, since $$\Re\left(\frac{D_q\left(\mathbf{\Omega}_q^\varrho\gamma(\vartheta)\right)}{\delta(\vartheta)}\right) > \alpha \qquad (0 \leq \alpha < 1; \ z \in \mathbb{E}),$$ there must exist a function t given by $$\mathfrak{r}(\vartheta) = 1 + \sum_{l=1}^{\infty} d_l \vartheta^l$$ such that $$\frac{D_q \left(\Omega_q^{\varrho} \gamma(\vartheta)\right)}{\delta(\vartheta)} = 1 + (1 - \alpha)q(\vartheta)$$ $$= 1 + (1 - \alpha) \sum_{l=1}^{\infty} d_l \vartheta^l. \tag{12}$$ For each $l \ge 2$ , evaluating the coefficients of the Equations (9) and (11), we obtain $$\left\{ \begin{array}{l} \left( [l]_{q} \frac{\Gamma_{q}(2-\varrho)\Gamma_{q}(l+1)}{\Gamma_{q}(2)\Gamma_{q}(l+1-\varrho)} a_{l} - b_{l} \right) \sum_{l=1}^{l-2} \mathfrak{q}_{l}^{-1}(b_{2}, b_{3}, \cdots, b_{l+1}) \\ \cdot \left( \left( [l]_{q} - l \right) \frac{\Gamma_{q}(2-\varrho)\Gamma_{q}(l+1)}{\Gamma_{q}(2)\Gamma_{q}(l+1-\varrho)} a_{l-l} - b_{l-l} \right) \end{array} \right\} = (1-\alpha)c_{l-1}.$$ (13) Additionally, by evaluating the coefficients of the Equations (10) and (12), for any $l \ge 2$ , we have $$\left\{ \begin{pmatrix} \left[ [l]_{q} \frac{\Gamma_{q}(2-\varrho)\Gamma_{q}(l+1)}{\Gamma_{q}(2)\Gamma_{q}(l+1-\varrho)} A_{l} - B_{l} \right) \sum_{l=1}^{l-2} \mathfrak{q}_{l}^{-1}(B_{2}, B_{3}, \cdots, B_{l+1}) \\ \cdot \left( \left( [l]_{q} - l \right) \frac{\Gamma_{q}(2-\varrho)\Gamma_{q}(l+1)}{\Gamma_{q}(2)\Gamma_{q}(l+1-\varrho)} A_{l-l} - B_{l-l} \right) \end{pmatrix} = (1-\alpha)d_{l-1}.$$ (14) Using the Equations (13) and (14), we derive the following for the particular case when l = 2: $$\frac{[2]_q \Gamma_q(2-\varrho) \Gamma_q(3)}{\Gamma_q(2) \Gamma_q(3-\varrho)} a_2 - b_2 = (1-\alpha)c_1$$ $$\frac{[2]_q \Gamma_q(2-\varrho) \Gamma_q(3)}{\Gamma_q(2) \Gamma_q(3-\varrho)} A_2 - B_2 = (1-\alpha)d_1$$ and $$a_2 = \frac{\Gamma_q(2)\Gamma_q(3-\varrho)}{[2]_q\Gamma_q(2-\varrho)\Gamma_q(3)}((1-\alpha)c_1 + b_2)$$ $$A_2 = \frac{\Gamma_q(2)\Gamma_q(3-\varrho)}{[2]_q\Gamma_q(2-\varrho)\Gamma_q(3)}((1-\alpha)d_1 + B_2).$$ We now solve for a^l and apply Lemma 1 and the moduli, so that $$|a_2| \leq \frac{2[\Gamma_q(2)\Gamma_q(3-\varrho)}{[2]_q\Gamma_q(2-\varrho)\Gamma_q(3)}(2-\alpha).$$ However, assuming that 2 5 k 5 l − 1 and a^k = 0 are true, the following results are obtained. $$A_l = -a_l$$ and $$\frac{[l]_q \Gamma_q(2-\varrho) \Gamma_q(l+1)}{\Gamma_q(2) \Gamma_q(l+1-\varrho)} a_l - b_l = (1-\alpha) c_{l-1},$$ $$-\frac{[l]_q \Gamma_q(2-\varrho) \Gamma_q(l+1)}{\Gamma_q(2) \Gamma_q(l+1-\varrho)} a_l - B_l = (1-\alpha) d_{l-1}$$ and $$a_{l} = \frac{\Gamma_{q}(2)\Gamma_{q}(l+1-\varrho)}{[l]_{q}\Gamma_{q}(2-\varrho)\Gamma_{q}(l+1)}((1-\alpha)c_{l-1}+b_{l})$$ $$-a_{l} = \frac{\Gamma_{q}(2)\Gamma_{q}(l+1-\varrho)}{[l]_{q}\Gamma_{q}(2-\varrho)\Gamma_{q}(l+1)}((1-\alpha)d_{l-1}+B_{l})$$ By solving for a^l and using Lemma 1 and the moduli, we can derive $$|a_l| \leq \frac{\Gamma_q(2)\Gamma_q(l+1-\varrho)(2(1-\alpha)+l)}{[l]_q\Gamma_q(2-\varrho)\Gamma_q(l+1)},$$ upon noticing that $$|b_l| \leq l$$ and $|B_l| \leq l$ . This completes the proof of Theorem 1. The following corollaries can be obtained by putting different values of the parameters involved.

Corollary 1. If the function h has the series representation stated in (1) and belongs to the class KΣ(q, 0, 1), and if $$a_{\mathfrak{i}}=0 \ (2 \leq \mathfrak{i} \leq l-1),$$ then $$|a_l| \le \frac{\Gamma_q(2)\Gamma_q(l)(2+l)}{[l]_q\Gamma_q(l+1)} \qquad (l \ge 3).$$

Corollary 2. If the function h has the series representation stated in (1) and belongs to KΣ(q, α, 1), and if aⁱ = 0 (2 5 i 5 l − 1), then $$|a_l| \leq \frac{\Gamma_q(2)\Gamma_q(l)(2(1-\alpha)+l)}{[l]_q\Gamma_q(l+1)} \qquad (l \geq 3).$$

Corollary 3. If the function h has the series representation stated in (1) and belongs to the class $\mathcal{K}_{\Sigma}(q \to 1-, \alpha, \varrho)$ , and if $a_i = 0 \ (2 \le i \le l-1)$ , then $$|a_l| \le \frac{\Gamma(2)\Gamma(l+1-\varrho)(2(1-\alpha)+l)}{l\Gamma(2-\varrho)\Gamma(l+1)} \qquad (l \ge 3).$$

Corollary 4. If the h has the series representation stated in (1) and belongs to the class $\mathcal{K}_{\Sigma}(q \to 1-, \alpha, 1)$ , and if $a_i = 0 \ (2 \le i \le l-1)$ , then $$|a_l| \le \frac{\Gamma(2)\Gamma(l)(2(1-\alpha)+l)}{l\Gamma(l+1)}$$ $(l \ge 3).$ The following known consequence of Theorem 1 for $\varrho=0$ and $q\to 1-$ was demonstrated in [27].

Corollary 5. (see [27]). Let $h \in \mathcal{K}_{\Sigma}(\alpha)$ . If $a_{i+1} = 0 \ (1 \le i \le l)$ , then $$|a_l| \le 1 + \frac{2(1-\alpha)}{l}$$ $(l \ge 3)$ .

Corollary 6. (see [56]). If the function h has the series representation stated in (1) and belongs to the class $K_{\Sigma}(q \to 1-, 0, \varrho)$ , and if $a_i = 0$ $(2 \le i \le l-1)$ , then $$|a_l| \le \frac{(2+l)\Gamma(l+1-\varrho)}{l\Gamma(2-\varrho)\Gamma(l+1)} \qquad (l \ge 3).$$ As a special form of Theorem 1, our next result (Theorem 2 below) provides estimates for the initial coefficients $|a_2|$ and $|a_3|$ , and also for the Fekete–Szegö-type functional involved in $|a_3 - a_2^2|$ for functions in the class $\mathcal{K}_{\Sigma}(m, \alpha, q)$ .

Theorem 2. Let the function $h \in \mathcal{K}_{\Sigma}(q, \alpha, \varrho)$ be given by (1). Then, $$|a_2| \leq \begin{cases} \sqrt{\frac{2\Gamma_q(2)\Gamma_q(3-\varrho)\Gamma_q(4-\varrho)(1-\alpha)}{\Gamma_q(2-\varrho)\left([3]_q\Gamma_q(4)\Gamma_q(3-\varrho)-[2]_q\Gamma_q(3)\Gamma_q(4-\varrho)\right)}} \\ (0 \leq \alpha < 1-\phi(q,\varrho)) \\ \frac{2\Gamma_q(2)\Gamma_q(3-\varrho)(1-\alpha)}{[2]_q\Gamma_q(2-\varrho)\Gamma_q(3)-\Gamma_q(2)\Gamma_q(3-\varrho)} \\ (1-\phi(q,\varrho) \leq \alpha < 1) \end{cases}$$ and $$\begin{split} |a_3| & \leq \frac{2\Gamma_q(2)\Gamma_q(4-\varrho)(1-\alpha)}{[3]_q\Gamma_q(4)\Gamma_q(2-\varrho) - \Gamma_q(2)\Gamma_q(4-\varrho)} \\ & \cdot \frac{[2]_q\Gamma_q(2-\varrho)\Gamma_q(3) - \Gamma_q(2)\Gamma_q(3-\varrho) + 2(1-\alpha)\Gamma_q(2)\Gamma_q(3-\varrho)}{[2]_q\Gamma_q(2-\varrho)\Gamma_q(3) - \Gamma_q(2)\Gamma_q(3-\varrho)}, \end{split}$$ where $$\phi(q,\varrho) := \frac{\Gamma_q(2)\Gamma_q(4-\varrho)\big\{[2]_q\Gamma_q(2-\varrho)\Gamma_q(3) - \Gamma_q(2)\Gamma_q(3-\varrho)\big\}^2}{2\Gamma_q(3-\varrho)\Gamma_q(2)\big([3]_q\Gamma_q(4)\Gamma_q(2-\varrho)\Gamma_q(3-\varrho) - [2]_q\Gamma_q(3)\Gamma_q(2-\varrho)\Gamma_q(4-\varrho)\big)}.$$ Furthermore, it is asserted that $$\left|a_3 - a_2^2\right| \le \frac{2\Gamma_q(2)\Gamma_q(4 - \varrho)(1 - \alpha)}{[3]_q\Gamma_q(2 - \varrho)\Gamma_q(4) - \Gamma_q(2)\Gamma_q(4 - \varrho)}.$$ Proof. Taking a function $g(z) = \Omega_q^{\varrho} h(z)$ in the proof of Theorem 1, we obtain $a_l = -b_l$ . For l = 2, the Equations (13) and (14), respectively, yield $$a_2 \left( \frac{[2]_q \Gamma_q(2-\varrho) \Gamma_q(3)}{\Gamma_q(2) \Gamma_q(3-\varrho)} - 1 \right) = (1-\alpha)c_1,$$ $$a_2 \left( \frac{-[2]_q \Gamma_q(2-\varrho) \Gamma_q(3)}{\Gamma_q(2) \Gamma_q(3-\varrho)} + 1 \right) = (1-\alpha)d_1;$$ and $$\begin{split} a_2 &= \frac{\Gamma_q(2)\Gamma_q(3-\varrho)}{[2]_q\Gamma_q(2-\varrho)\Gamma_q(3)-\Gamma_q(2)\Gamma_q(3-\varrho)}(1-\alpha)c_1,\\ -a_2 &= \frac{\Gamma_q(2)\Gamma_q(3-\varrho)}{[2]_q\Gamma_q(2-\varrho)\Gamma_q(3)-\Gamma_q(2)\Gamma_q(3-\varrho)}(1-\alpha)d_1. \end{split}$$ If we use moduli of either of these two equations, we obtain $$|a_2| \leq \frac{2\Gamma_q(2)\Gamma_q(3-\varrho)(1-\alpha)}{[2]_q\Gamma_q(2-\varrho)\Gamma_q(3)-\Gamma_q(2)\Gamma_q(3-\varrho)}.$$ For l = 3, the Equations (13) and (14), respectively, yield $$\left(\frac{[3]_q \Gamma_q(2-\varrho) \Gamma_q(4)}{\Gamma_q(2) \Gamma_q(4-\varrho)} - 1\right) a_3 - \left(\frac{[2]_q \Gamma_q(2-\varrho) \Gamma_q(3)}{\Gamma_q(2) \Gamma_q(3-\varrho)} - 1\right) a_2^2 = (1-\alpha)c_2 \tag{15}$$ and $$\left(2a_2^2 - a_3\right) \left(\frac{[3]_q \Gamma_q(2 - \varrho) \Gamma_q(4)}{\Gamma_q(2) \Gamma_q(4 - \varrho)} - 1\right) - \left(\frac{[2]_q \Gamma_q(2 - \varrho) \Gamma_q(3)}{\Gamma_q(2) \Gamma_q(3 - \varrho)} - 1\right) a_2^2 = (1 - \alpha) d_2.$$ (16) By combining the two equations mentioned above, we obtain $$2a_{2}^{2} \left( \frac{[3]_{q} \Gamma_{q}(2-\varrho) \Gamma_{q}(4)}{\Gamma_{q}(2) \Gamma_{q}(4-\varrho)} - 1 \right) - 2 \left( \frac{[2]_{q} \Gamma_{q}(2-\varrho) \Gamma_{q}(3)}{\Gamma_{q}(2) \Gamma_{q}(3-\varrho)} - 1 \right) a_{2}^{2} = (1-\alpha)(c_{2}+d_{2}),$$ $$2a_{2}^{2} \left( \frac{[3]_{q} \Gamma_{q}(2-\varrho) \Gamma_{q}(4)}{\Gamma_{q}(2) \Gamma_{q}(4-\varrho)} - \frac{[2]_{q} \Gamma_{q}(2-\varrho) \Gamma_{q}(3)}{\Gamma_{q}(2) \Gamma_{q}(3-\varrho)} \right) = (1-\alpha)(c_{2}+d_{2})$$ or $$2a_2^2 = \frac{[3]_q \Gamma_q(3-\varrho) \Gamma_q(2-\varrho) \Gamma_q(4) - [2]_q \Gamma_q(4-\varrho) \Gamma_q(2-\varrho) \Gamma_q(3)}{\Gamma_q(2) \Gamma_q(3-\varrho) \Gamma_q(4-\varrho)} = (1-\alpha)(c_2+d_2)$$ Now, by finding $|a_2|$ , we arrive at $$\left|a_2^2\right| = \frac{\Gamma_q(2)\Gamma_q(3-\varrho)\Gamma_q(4-\varrho)(1-\alpha)|d_2+c_2|}{2\Gamma_q(2-\varrho)\Big\{[3]_q\Gamma_q(4)\Gamma_q(3-\varrho)-[2]_q\Gamma_q(3)\Gamma_q(4-\varrho)\Big\}}.$$ Additionally, by applying Lemma 1, we obtain $$|a_2| \leq \sqrt{\frac{2\Gamma_q(2)\Gamma_q(3-\varrho)\Gamma_q(4-\varrho)(1-\alpha)}{\Gamma_q(2-\varrho)\left([3]_q\Gamma_q(4)\Gamma_q(3-\varrho)-[2]_q\Gamma_q(3)\Gamma_q(4-\varrho)\right)}}.$$ As a result, we obtain the following estimate: $$\begin{split} \sqrt{\frac{2\Gamma_q(2)\Gamma_q(3-\varrho)\Gamma_q(4-\varrho)(1-\alpha)}{\Gamma_q(2-\varrho)\left([3]_q\Gamma_q(4)\Gamma_q(3-\varrho)-[2]_q\Gamma_q(3)\Gamma_q(4-\varrho)\right)}} \\ < \frac{2\Gamma_q(2)\Gamma_q(3-\varrho)\Gamma_q(4-\varrho)(1-\alpha)}{\Gamma_q(2-\varrho)\left([3]_q\Gamma_q(4)\Gamma_q(3-\varrho)-[2]_q\Gamma_q(3)\Gamma_q(4-\varrho)\right)}. \end{split}$$ Upon substituting $$a_2 = \frac{c_1(1-\alpha)\Gamma_q(2)\Gamma_q(3-\varrho)}{[2]_q\Gamma_q(2-\varrho)\Gamma_q(3) - \Gamma_q(2)\Gamma_q(3-\varrho)}$$ into (15), we have $$\begin{split} a_{3} &= \frac{\Gamma_{q}(2)\Gamma_{q}(4-\varrho)(1-\alpha)}{[3]_{q}\Gamma_{q}(4)\Gamma_{q}(2-\varrho) - \Gamma_{q}(2)\Gamma_{q}(4-\varrho)} \\ &\cdot \left(c_{2} + \frac{(1-\alpha)\Gamma_{q}(2)\Gamma_{q}(3-\varrho)}{[2]_{q}\Gamma_{q}(2-\varrho)\Gamma_{q}(3) - \Gamma_{q}(2)\Gamma_{q}(3-\varrho)}c_{1}^{2}\right). \end{split}$$ Taking the moduli on both sides, we find that $$|a_{3}| \leq \frac{\Gamma_{q}(2)\Gamma_{q}(4-\varrho)(1-\alpha)}{[3]_{q}\Gamma_{q}(4)\Gamma_{q}(2-\varrho) - \Gamma_{q}(2)\Gamma_{q}(4-\varrho)} \cdot \left(|c_{2}| + \frac{(1-\alpha)\Gamma_{q}(2)\Gamma_{q}(3-\varrho)}{[2]_{q}\Gamma_{q}(2-\varrho)\Gamma_{q}(3) - \Gamma_{q}(2)\Gamma_{q}(3-\varrho)} |c_{1}^{2}|\right).$$ Applying Lemma 1, we obtain $$\begin{split} |a_3| & \leq \frac{\Gamma_q(2)\Gamma_q(4-\varrho)(1-\alpha)}{[3]_q\Gamma_q(4)\Gamma_q(2-\varrho) - \Gamma_q(2)\Gamma_q(4-\varrho)} \\ & \cdot \left(2 + \frac{4(1-\alpha)\Gamma_q(2)\Gamma_q(3-\varrho)}{[2]_q\Gamma_q(2-\varrho)\Gamma_q(3) - \Gamma_q(2)\Gamma_q(3-\varrho)}\right), \end{split}$$ that is, $$\begin{split} |a_3| & \leq \frac{2\Gamma_q(2)\Gamma_q(4-\varrho)(1-\alpha)}{[3]_q\Gamma_q(4)\Gamma_q(2-\varrho) - \Gamma_q(2)\Gamma_q(4-\varrho)} \\ & \cdot \Bigg( \frac{[2]_q\Gamma_q(2-\varrho)\Gamma_q(3) - \Gamma_q(2)\Gamma_q(3-\varrho) + 2(1-\alpha)\Gamma_q(2)\Gamma_q(3-\varrho)}{[2]_a\Gamma_q(2-\varrho)\Gamma_q(3) - \Gamma_q(2)\Gamma_q(3-\varrho)} \Bigg). \end{split}$$ Lastly, upon subtracting Equation (15) from Equation (16), we have $$\left|a_3 - a_2^2\right| \le \frac{2\Gamma_q(2)\Gamma_q(4-\varrho)(1-\alpha)}{[3]_a\Gamma_a(2-\varrho)\Gamma_a(4) - \Gamma_a(2)\Gamma_a(4-\varrho)}.$$ Our proof of Theorem 2 is thus completed. $\Box$ Several corollaries and consequences of Theorem 2 are presented below.

Corollary 7. Let the function $h \in \mathcal{K}_{\Sigma}(q, \alpha, 1)$ be given by (1). Then, $$|a_2| \leq \begin{cases} \sqrt{\frac{2\Gamma_q(2)\Gamma_q(2)\Gamma_q(3)(1-\alpha)}{\left([3]_q\Gamma_q(4)\Gamma_q(2)-[2]_q\Gamma_q(3)\Gamma_q(3)\right)}} \\ (0 \leq \alpha < 1-\varphi_1(q)) \\ \frac{2\Gamma_q(2)\Gamma_q(2)(1-\alpha)}{[2]_q\Gamma_q(3)-\Gamma_q(2)\Gamma_q(2)} \\ (1-\varphi_1(q) \leq \alpha < 1), \end{cases}$$ $$|a_{3}| \leq \frac{2\Gamma_{q}(2)\Gamma_{q}(3)(1-\alpha)}{[3]_{q}\Gamma_{q}(4) - \Gamma_{q}(2)\Gamma_{q}(3)} \cdot \left(\frac{[2]_{q}\Gamma_{q}(3) - \Gamma_{q}(2)\Gamma_{q}(2) + 2(1-\alpha)\Gamma_{q}(2)\Gamma_{q}(2)}{[2]_{q}\Gamma_{q}(3) - \Gamma_{q}(2)\Gamma_{q}(2)}\right)$$ and $$\left|a_3 - a_2^2\right| \le \frac{2\Gamma_q(2)\Gamma_q(3)(1-\alpha)}{[3]_q\Gamma_q(4) - \Gamma_q(2)\Gamma_q(3)}$$ where $$\varphi_1(q) = \frac{\Gamma_q(2)\Gamma_q(3) ([2]_q \Gamma_q(3) - \Gamma_q(2)\Gamma_q(2))^2}{2\Gamma_q(2)\Gamma_q(2) ([3]_q \Gamma_q(4)\Gamma_q(2) - [2]_q \Gamma_q(3)\Gamma_q(3))}.$$

Corollary 8. Let $h \in \mathcal{K}_{\Sigma}(q, 0, 1)$ be given by (1). Then, $$|a_2| \leq \begin{cases} \sqrt{\frac{2\Gamma_q(2)\Gamma_q(2)\Gamma_q(3)}{[3]_q\Gamma_q(4)\Gamma_q(2) - [2]_q\Gamma_q(3)\Gamma_q(3)}} \\ (0 \leq \alpha < 1 - \varphi_2(q)) \\ \frac{2\Gamma_q(2)\Gamma_q(2)}{[2]_q\Gamma_q(3) - \Gamma_q(2)\Gamma_q(2)} (1 - \varphi_2(q) \leq \alpha < 1), \end{cases}$$ $$\begin{split} |a_3| & \leq \frac{2\Gamma_q(2)\Gamma_q(3)}{[3]_q\Gamma_q(4) - \Gamma_q(2)\Gamma_q(3)} \\ & \cdot \frac{[2]_q\Gamma_q(3) - \Gamma_q(2)\Gamma_q(2) + \Gamma_q(2)\Gamma_q(2)}{[2]_q\Gamma_q(3) - \Gamma_q(2)\Gamma_q(2)} \end{split}$$ and $$|a_3 - a_2^2| \le \frac{2\Gamma_q(2)\Gamma_q(3)}{[3]_q\Gamma_q(4) - \Gamma_q(2)\Gamma_q(3)},$$ where $$\varphi_2(q) = \frac{\Gamma_q(2)\Gamma_q(3)\{[2]_q\Gamma_q(3) - \Gamma_q(2)\Gamma_q(2)\}^2}{2\Gamma_q(2)\Gamma_q(2)\{[3]_q\Gamma_q(4)\Gamma_q(2) - [2]_q\Gamma_q(3)\Gamma_q(3)\}}.$$

Corollary 9. Let $h \in \mathcal{K}_{\Sigma}(q \to 1-, \alpha, \varrho)$ be given by (1). Then, $$|a_2| \leq \begin{cases} \sqrt{\frac{2\Gamma(2)\Gamma(3-\varrho)\Gamma(4-\varrho)(1-\alpha)}{\Gamma(2-\varrho)\{3\Gamma(4)\Gamma(3-\varrho)-2\Gamma(3)\Gamma(4-\varrho)\}}} \\ (0 \leq \alpha < 1-\varphi_3(\varrho)) \\ \frac{2\Gamma(2)\Gamma(3-\varrho)(1-\alpha)}{2\Gamma(2-\varrho)\Gamma(3)-\Gamma(2)\Gamma(3-\varrho)} \\ (1-\varphi_3(\varrho) \leq \alpha < 1), \end{cases}$$ $$\begin{split} |a_3| & \leq \frac{2\Gamma(2)\Gamma(4-\varrho)(1-\alpha)}{3\Gamma(4)\Gamma(2-\varrho) - \Gamma(2)\Gamma(4-\varrho)} \\ & \cdot \frac{2\Gamma(2-\varrho)\Gamma(3) - \Gamma(2)\Gamma(3-\varrho) + 2(1-\alpha)\Gamma(2)\Upsilon(3-\varrho)}{2\Gamma(2-\varrho)\Gamma(3) - \Gamma(2)\Gamma(3-\varrho)} \end{split}$$ and $$\left|a_3 - a_2^2\right| \le \frac{2\Gamma(2)\Gamma(4-\varrho)(1-\alpha)}{3\Gamma(2-\varrho)\Gamma(4) - \Gamma(2)\Gamma(4-\varrho)},$$ where $$\varphi_3(\varrho) = \frac{\Gamma(2)\Gamma(4-\varrho)\{2\Upsilon(2-\varrho)\Gamma(3)-\Gamma(2)\Gamma(3-\varrho)\}^2}{2\Gamma(3-\varrho)\Gamma(2)\{3\Gamma(4)\Gamma(2-\varrho)\Gamma(3-\varrho)-2\Gamma(3)\Gamma(2-\varrho)\Gamma(4-\varrho)\}}.$$ As another application of Theorem 2 for $\varrho=4$ and $q\to 1-$ , we obtain the result given in [27].

Corollary 10. (see [27]). Let $h \in \mathcal{K}_{\Sigma}(q \to 1-, \alpha, 0)$ . Then, $$|a_2| \le \begin{cases} \sqrt{2(1-\alpha)} & \left(0 \le \alpha < \frac{1}{2}\right) \\ 2(1-\alpha) & \left(\frac{1}{2} \le \alpha < 1\right) \end{cases}$$ and $$|a_3| \leq \begin{cases} 2(1-\alpha) & \left(0 \leq \alpha < \frac{1}{2}\right) \\ (1-\alpha)(3-2\alpha) & \left(\frac{1}{2} \leq \alpha < 1\right). \end{cases}$$

Corollary 11. (see [56]). Let $h \in \mathcal{K}_{\Sigma}(q \to 1-, 0, \varrho)$ be given by (1). Then, $$|a_2| \leq \min \left( \begin{array}{c} \sqrt{\frac{2\Gamma(3-\varrho)\Gamma(4-\varrho)}{\Gamma(2-\varrho)\{3\Gamma(4)\Gamma(3-\varrho)-2\Gamma(3)\Gamma(4-\varrho)\}}'} \\ \\ \frac{2\Gamma(2)\Gamma(3-\varrho)}{2\Gamma(2-\varrho)\Gamma(3)-\Gamma(2)\Gamma(3-\varrho)} \end{array} \right),$$ $$|a_3| \leq \frac{2\Gamma_q(4-\varrho)}{3\Gamma(4)\Gamma(2-\varrho) - \Gamma(4-\varrho)} \cdot \left(\frac{2\Gamma(2-\varrho)\Gamma(3) - \Gamma(3-\varrho) + 2\Gamma(3-\varrho)}{2\Gamma(2-\varrho)\Gamma(3) - \Gamma(3-\varrho)}\right)$$ and $$\left|a_3 - a_2^2\right| \le \frac{2\Gamma(2)\Gamma(4-\varrho)}{3\Gamma(2-\varrho)\Gamma(4) - \Gamma(4-\varrho)}.$$ Example 1. For $l \ge 3$ , we will demonstrate that h(z) given by $$h(z) = z + \frac{1 - \alpha}{l - 1} z^l$$ is a bi-close-to-convex function of order $\alpha$ , where $\alpha \in [0,1)$ in $\mathbb{E}$ . Indeed, since the function $$g(z) = z - \frac{1 - \alpha}{l - \alpha} z^l$$ is starlike in $\mathbb{E}$ , we have $$\frac{D_{q} \mathbf{\Omega}_{q}^{\varrho} h(z)}{g(z)} = \frac{1 + \left(\frac{\Psi_{l}(q,\varrho)[l]_{q}(1-\alpha)}{l-1}\right) z^{l-1}}{1 - \left(\frac{1-\alpha}{l-\alpha}\right) z^{l-1}} \\ = 1 + \sum_{j=1}^{\infty} \left(\frac{(1-\alpha)^{j}}{(l-\alpha)^{j}} + \frac{\Psi_{l}(q,\varrho)[l]_{q}(1-\alpha)j}{(l-1)(l-\alpha)^{j-1}}\right) z^{(l-1)k},$$ where $$\Psi_l(q,\varrho) = \frac{\Gamma_q(2-\varrho)\Gamma_q(l+1)}{\Gamma_q(2)\Gamma_q(l+1-\varrho)}.$$ Therefore, we obtain $$\begin{split} \frac{\frac{D_q\Omega_q^\varrho h(z)}{g(z)} - \alpha}{1 - \alpha} &= 1 + \sum_{j=1}^\infty \left( \frac{l\left(\Psi_l(q,\varrho)[l]_q + 1\right) - \Psi_l(q,\varrho)[l]_q \alpha - 1}{(l-1)(l-\alpha)} \right) \\ &\cdot \left( \frac{1 - \alpha}{l - \alpha} \right)^{j-1} z^{(l-1)j}. \end{split}$$ Obviously, we also have $$\Re\left(\frac{D_q\mathbf{\Omega}_q^\varrho h(z)}{g(z)}\right) - \alpha > 0 \qquad (z \in \mathbb{E}).$$ For $\gamma = h^{-1}$ and $\delta = g^{-1}$ , it is easily seen that $$\gamma(\vartheta) = \vartheta - \frac{1-\alpha}{l-1} \,\vartheta^l,$$ and if we set $$\delta(\vartheta) = \vartheta + \frac{1-\alpha}{l-\alpha} \, \vartheta^l$$ which is starlike in $\mathbb{E}$ . As a result, we have $$\begin{split} \frac{\frac{D_q \mathbf{\Omega}_q^o \gamma(\vartheta)}{\delta(z)} - \alpha}{1 - \alpha} &= 1 + \sum_{j=1}^{\infty} (-1)^j \left( \frac{l \big( \Psi_l(q,\varrho)[l]_q + 1 \big) - \Psi_l(q,\varrho)[l]_q \alpha - 1}{(l-1)(l-\alpha)} \right) \\ &\cdot \left( \frac{1 - \alpha}{l-\alpha} \right)^{j-1} \vartheta^{(l-1)j}. \end{split}$$ Thus, clearly, we find that $$\Re\left(\frac{D_q\mathbf{\Omega}_q^\varrho\gamma(\vartheta)}{\delta(\vartheta)}-\alpha\right)>0 \qquad (z\in\mathbb{E}).$$

Lemma 1. ([19]). Let $\Im = \phi + \overline{\psi}$ where $\phi$ and $\psi$ are given by (1) and suppose that $\varrho \geq 0$ , $0 \leq \gamma < 1$ and $$\sum_{\nu=2}^{\infty} \nu [1 + \varrho \left(\nu^2 - 1\right)] |a_{\nu}| + \sum_{\nu=1}^{\infty} \nu [1 + \varrho \left(\nu^2 - 1\right)] |b_{\nu}| \le 1 - \gamma.$$ (6) then $\Im$ is harmonic, sense-preserving univalent functions in $\Xi$ and $\Im \in \mathcal{HF}(\varrho, \gamma)$ . Moreover, if $\Im \in \mathcal{HF}(\varrho, \gamma)$ , then $$|a_{\nu}| \le \frac{1 - \gamma}{\nu [1 + \varrho(\nu^2 - 1)]}, \ \nu \ge 2,$$ (7) and $$|b_{\nu}| \le \frac{1 - \gamma}{\nu[1 + \varrho(\nu^2 - 1)]}, \ \nu \ge 1.$$ (8)

Lemma 2. ([6]). Let $\Im = \phi + \overline{\psi}$ where $\phi$ and $\psi$ are given by (2) and suppose that $0 \le \tau < 1$ . Then $\Im \in \mathcal{T}\mathfrak{N}_{H,\mathcal{F}}(\tau)$ if and only if $$\sum_{\nu=2}^{\infty} \nu |a_{\nu}| + \sum_{\nu=1}^{\infty} \nu |b_{\nu}| \le 1 - \tau.$$ (9) Moreover, if $\Im \in \mathcal{T}\mathfrak{N}_{\mathcal{H},\mathcal{F}}(\tau)$ , then $$|a_{\nu}| \leq \frac{1-\tau}{\nu}, \nu \geq 2, \tag{10}$$ and $$|b_{\nu}| \leq \frac{1-\tau}{\nu}, \nu \geq 1. \tag{11}$$

Lemma 3. ([18]). Let $\Im = \phi + \overline{\psi}$ where $\phi$ and $\psi$ are given by (2), and suppose that $0 \le \tau < 1$ . Then $\Im \in \mathcal{T}\mathfrak{R}_{\mathcal{HF}}(\tau)$ if and only if $$\sum_{\nu=2}^{\infty} \nu^2 |a_{\nu}| + \sum_{\nu=1}^{\infty} \nu^2 |b_{\nu}| \le 1 - \tau.$$ (12) Moreover, if $\Im \in \mathcal{T}\mathfrak{R}_{\mathcal{HF}}(\tau)$ , then $$|a_{\nu}| \le \frac{1-\tau}{\nu^2}, \ \nu \ge 2$$ (13) and $$|b_{\nu}| \le \frac{1-\tau}{\nu^2}, \ \nu \ge 1.$$ (14)

Lemma 4. ([5]). If $\Im = \phi + \overline{\psi} \in \mathcal{S}^*_{\mathcal{HF}}$ where $\phi$ and $\psi$ are given by (1) with $b_1 = 0$ , then $$|a_{\nu}| \le \frac{(2\nu+1)(\nu+1)}{6} \text{ and } |b_{\nu}| \le \frac{(2\nu-1)(\nu-1)}{6}.$$ (15)

Lemma 5. ([5]). If $\Im = \phi + \overline{\psi} \in \mathcal{K}_{\mathcal{HF}}$ where $\phi$ and $\psi$ are given by (1) with $b_1 = 0$ , then $$|a_{\nu}| \leq \frac{\nu+1}{2} \text{ and } |b_{\nu}| \leq \frac{\nu-1}{2}.$$ (16) Throughout the sequence, we use the following: $$\chi_{\rho,\epsilon}(\xi) = \xi + \sum_{\nu=2}^{\infty} \frac{\Gamma(\epsilon)}{\Gamma(\rho(\nu-1)+\epsilon)} \xi^{\nu}; \quad \chi_{\rho,\epsilon}(1) = 1 + \sum_{\nu=2}^{\infty} \frac{\Gamma(\epsilon)}{\Gamma(\rho(\nu-1)+\epsilon)}, \tag{17}$$ $$\chi_{\rho,\epsilon}'(\xi) = 1 + \sum_{\nu=2}^{\infty} \frac{\nu\Gamma(\epsilon)}{\Gamma(\rho(\nu-1)+\epsilon)} \xi^{\nu-1}; \quad \chi_{\rho,\epsilon}'(1) - 1 = \sum_{\nu=2}^{\infty} \frac{\nu\Gamma(\epsilon)}{\Gamma(\rho(\nu-1)+\epsilon)}, \tag{18}$$ $$\chi_{\rho,\epsilon}^{"}(1) = \sum_{\nu=2}^{\infty} \frac{\nu(\nu-1)\Gamma(\epsilon)}{\Gamma(\rho(\nu-1)+\epsilon)},\tag{19}$$ $$\chi_{\rho,\epsilon}^{"'}(1) = \sum_{\nu=2}^{\infty} \frac{\nu(\nu-1)(\nu-2)\Gamma(\epsilon)}{\Gamma(\rho(\nu-1)+\epsilon)},\tag{20}$$ and in general, we have $$\chi_{\rho,\epsilon}^{(j)}(1) = \sum_{\nu=2}^{\infty} \frac{\nu(\nu-1)(\nu-2)\cdots(\nu-(j-1))\Gamma(\epsilon)}{\Gamma(\rho(\nu-1)+\epsilon)}, j = 1, 2, \dots$$ (21)

Theorem 1. Let $\varrho \geq 0$ , $\gamma \in [0,1)$ and $\rho, \epsilon, \eta, \delta \notin \mathbb{Z}_0^-$ . If $$\left[2\varrho\left(\chi_{\rho,\epsilon}^{(5)}(1) + \chi_{\eta,\epsilon}^{(5)}(1)\right) + 23\varrho\chi_{\rho,\epsilon}^{(4)}(1) + (67\varrho + 2)\chi_{\rho,\epsilon}^{(3)}(1) + (45\varrho + 9)\chi_{\rho,\epsilon}^{(2)}(1) + 6\chi_{\rho,\epsilon}^{'}(1) + (19\varrho + 3)\chi_{\eta,\epsilon}^{(4)}(1) + (19\varrho + 3)\chi_{\eta,\epsilon}^{(4)}(1)\right] \\ \leq 6(1 - \gamma), \tag{22}$$ then $$\Theta(\mathcal{S}_{\mathcal{U},\mathcal{T}}^*) \subset \mathcal{HF}(\varrho,\gamma).$$ Proof. Let $\Im = \phi + \overline{\psi} \in \mathcal{S}^*_{\mathcal{HF}}$ where $\phi$ and $\psi$ are of the form (1) with $b_1 = 0$ . We need to show that $\Theta(\Im) = \mathfrak{F}(\xi) \in \mathcal{HF}(\varrho, \gamma)$ , which given by (5) with $b_1 = 0$ . In view of Lemma 1, we need to prove that $$Q(\varrho, \epsilon, \delta, \eta) \leq 1 - \gamma$$ where $$Q(\varrho, \epsilon, \delta, \eta) = \sum_{\nu=2}^{\infty} \nu \left( 1 + \varrho \left( \nu^2 - 1 \right) \right) \left| \frac{\Gamma(\epsilon)}{\Gamma(\rho(\nu - 1) + \epsilon)} a_{\nu} \right| + \sum_{\nu=2}^{\infty} \nu \left( 1 + \varrho \left( \nu^2 - 1 \right) \right) \left| \frac{\Gamma(\delta)}{\Gamma(\eta(\nu - 1) + \delta)} b_{\nu} \right|.$$ (23) Using the inequalities (15) of Lemma 4, we get $$Q(\varrho, \epsilon, \delta, \eta) \leq \frac{1}{6} \left[ \sum_{\nu=2}^{\infty} (2\nu + 1)(\nu + 1) \left( \nu + \varrho \nu \left( \nu^{2} - 1 \right) \right) \frac{\Gamma(\epsilon)}{\Gamma(\rho(\nu - 1) + \epsilon)} \right]$$ $$+ \sum_{\nu=2}^{\infty} (2\nu - 1)(\nu - 1) \left( \nu + \varrho \nu \left( \nu^{2} - 1 \right) \right) \frac{\Gamma(\delta)}{\Gamma(\eta(\nu - 1) + \delta)} \right]$$ $$= \frac{1}{6} \left[ \sum_{\nu=2}^{\infty} \left[ 2\varrho \nu^{5} + 3\varrho \nu^{4} + (2 - \varrho)\nu^{3} + (3 - 3\varrho)\nu^{2} + (1 - \varrho)\nu \right] \frac{\Gamma(\epsilon)}{\Gamma(\rho(\nu - 1) + \epsilon)} \right]$$ $$+ \sum_{\nu=2}^{\infty} \left[ 2\varrho \nu^{5} - 3\varrho \nu^{4} + (2 - \varrho)\nu^{3} + (3\varrho - 3)\nu^{2} + (1 - \varrho)\nu \right] \frac{\Gamma(\delta)}{\Gamma(\eta(\nu - 1) + \delta)} \right]$$ $$(24)$$ Writing $$\nu^2 = \nu(\nu - 1) + \nu,\tag{25}$$ $$\nu^{3} = \nu(\nu - 1)(\nu - 2) + 3\nu(\nu - 1) + \nu, \tag{26}$$ $$\nu^4 = \nu(\nu - 1)(\nu - 2)(\nu - 3) + 6\nu(\nu - 1)(\nu - 2) + 7\nu(\nu - 1) + \nu, \tag{27}$$ and $$\nu^{5} = \nu(\nu - 1)(\nu - 2)(\nu - 3)(\nu - 4) + 10\nu(\nu - 1)(\nu - 2)(\nu - 3) + 25\nu(\nu - 1)(\nu - 2) + 15\nu(\nu - 1) + \nu,$$ (28) in (24), we have $$\begin{split} &Q(\varrho,\epsilon,\delta,\eta)\\ &\leq \frac{1}{6} \left[ \sum_{\nu=2}^{\infty} [2\varrho\nu(\nu-1)(\nu-2)(\nu-3)(\nu-4) + 23\varrho\nu(\nu-1)(\nu-2)(\nu-3) \right. \\ &+ (67\varrho+2)\nu(\nu-1)(\nu-2) + (45\varrho+9)\nu(\nu-1) \\ &+ 6\nu ] \frac{\Gamma(\epsilon)}{\Gamma(\varrho(\nu-1)+\epsilon)} \\ &+ \sum_{\nu=2}^{\infty} [2\varrho\nu(\nu-1)(\nu-2)(\nu-3)(\nu-4) + 17\varrho\nu(\nu-1)(\nu-2)(\nu-3) \\ &+ (31\varrho+2)\nu(\nu-1)(\nu-2) + (9\varrho+3)\nu(\nu-1)] \frac{\Gamma(\delta)}{\Gamma(\eta(\nu-1)+\delta)} \right] \\ &= \frac{1}{6} \left[ 2\varrho\chi_{\varrho,\epsilon}^{(5)}(1) + 23\varrho\chi_{\varrho,\epsilon}^{(4)}(1) + (67\varrho+2)\chi_{\varrho,\epsilon}^{(3)}(1) \\ &+ (45\varrho+9)\chi_{\varrho,\epsilon}^{(5)}(1) + 6\chi_{\varrho,\epsilon}^{\prime}(1) \\ &+ 2\varrho\chi_{\eta,\epsilon}^{(5)}(1) + 17\varrho\chi_{\eta,\epsilon}^{(4)}(1) + (31\varrho+2)\chi_{\eta,\epsilon}^{(3)}(1) + (9\varrho+3)\chi_{\eta,\epsilon}^{(2)}(1) \right]. \end{split}$$ Now Q(\$, e, δ, η) ≤ 1 − γ if (22) holds.$

Theorem 2. Let $\varrho \geq 0$ , $\gamma \in [0,1)$ and $\rho, \epsilon, \eta, \delta \notin \mathbb{Z}_0^-$ . If $$[\varrho\chi_{\rho,\epsilon}^{(4)}(1) + 7\varrho\chi_{\rho,\epsilon}^{(3)}(1) + (9\varrho + 1)\chi_{\rho,\epsilon}^{(2)}(1) + 2\chi_{\rho,\epsilon}'(1) + \varrho\chi_{\eta,\epsilon}^{(4)} + 5\varrho\chi_{\eta,\epsilon}^{(3)}(1) + (5\varrho - 1)\chi_{\eta,\epsilon}^{(2)} + 2(\varrho - 1)\chi_{\eta,\epsilon}'(1)].$$ $$\leq 2(1 - \gamma), \tag{29}$$ then $$\Theta(\mathcal{K}_{\mathcal{H},\mathcal{F}}) \subset \mathcal{H}\mathcal{F}(\varrho,\gamma).$$ Proof. Let $\Im = \phi + \overline{\psi} \in \mathcal{K}_{\mathcal{HF}}$ where $\phi$ and $\psi$ are of the form (2) with $b_1 = 0$ . We need to show that $\Theta(\Im) = \mathfrak{F}(\xi) \in \mathcal{HF}(\varrho, \gamma)$ which given by (5) with $b_1 = 0$ . In view of Lemma 1, we need to prove that $Q(\varrho, \epsilon, \delta, \eta)$ $$Q(\varrho, \epsilon, \delta, \eta) \leq 1 - \gamma$$ where $Q(\varrho, \epsilon, \delta, \eta)$ as given in (23). Using the inequalities (16) of Lemma 5, we get $$\begin{split} Q(\varrho,\epsilon,\delta,\eta) &\leq \frac{1}{2} \left[ \sum_{\nu=2}^{\infty} (\nu+1) \Big( \nu + \varrho \nu \Big( \nu^2 - 1 \Big) \Big) \frac{\Gamma \epsilon)}{\Gamma(\varrho(\nu-1) + \epsilon)} \right. \\ &+ \sum_{\nu=2}^{\infty} (\nu-1) \Big( \nu + \varrho \nu \Big( \nu^2 - 1 \Big) \Big) \frac{\Gamma(\delta)}{\Gamma(\eta(\nu-1) + \delta)} \right] \\ &= \frac{1}{2} \left[ \sum_{\nu=2}^{\infty} \Big[ \varrho \nu^4 + \varrho \nu^3 + (1-\varrho) \nu^2 + (1-\varrho) \nu \Big] \frac{\Gamma(\epsilon)}{\Gamma(\varrho(\nu-1) + \epsilon)} \right. \\ &+ \sum_{\nu=2}^{\infty} \Big[ \varrho \nu^4 - \varrho \nu^3 + (1-\varrho) \nu^2 + (\varrho-1) \nu \Big] \frac{\Gamma(\delta)}{\Gamma(\eta(\nu-1) + \delta)} \right]. \end{split}$$ Using the Equations (25)–(27), we have $$\begin{split} &Q(\varrho,\epsilon,\delta,\eta) \leq \frac{1}{2} \left[ \sum_{\nu=2}^{\infty} [\varrho\nu(\nu-1)(\nu-2)(\nu-3) + 7\varrho\nu(\nu-1)(\nu-2) \right. \\ &\left. + (9\varrho+1)\nu(\nu-1) + 2\nu \right] \frac{\Gamma(\epsilon)}{\Gamma(\varrho(\nu-1)+\epsilon)} \right] \\ &\left. + \frac{1}{2} \sum_{\nu=2}^{\infty} [\varrho\nu(\nu-1)(\nu-2)(\nu-3) + 5\varrho\nu(\nu-1)(\nu-2) + (5\varrho-1)\nu(\nu-1) \right. \\ &\left. + \left. \left( 2(\varrho-1)\nu \right) \frac{\Gamma(\delta)}{\Gamma(\eta(\nu-1)+\delta)} \right. \\ &\left. = \frac{1}{2} [\varrho\chi_{\varrho,\epsilon}^{(4)}(1) + 7\varrho\chi_{\varrho,\epsilon}^{(3)}(1) + (9\varrho+1)\chi_{\varrho,\epsilon}^{(2)}(1) + 2\chi_{\varrho,\epsilon}'(1) + \varrho\chi_{\eta,\epsilon}^{(4)} + 5\varrho\chi_{\eta,\epsilon}^{(3)}(1) + (5\varrho-1)\chi_{\eta,\epsilon}^{(2)} + 2(\varrho-1)\chi_{\eta,\epsilon}'(1) \right]. \end{split}$$ Now $O(\rho, \epsilon, \delta, \eta) < 1 - \gamma$ if (29) holds. $\square$ The connection between $\mathcal{TN}_{\mathcal{HF}}(\tau)$ and $\mathcal{HF}(\varrho,\gamma)$ is given below in the next theorem.

Theorem 3. Let $\varrho \geq 0$ , $\gamma, \tau \in [0,1)$ and $\varrho, \varepsilon, \eta, \delta \notin \mathbb{Z}_0^-$ . If $$(1-\tau)\left[\varrho\left(\chi_{\rho,\epsilon}^{(2)}(1)+\chi_{\eta,\epsilon}^{(2)}(1)\right)+\varrho\left(\chi_{\rho,\epsilon}'(1)+\chi_{\eta,\epsilon}'(1)\right)+(1-\varrho)\left(\chi_{\rho,\epsilon}(1)+\chi_{\eta,\epsilon}(1)-2\right)\right]$$ $$\leq 1-\gamma-|b_{1}|,$$ then $$\Theta(\mathcal{T}\mathfrak{N}_{\mathcal{H}\mathcal{F}}(\tau)) \subset \mathcal{H}\mathcal{F}(\varrho,\gamma).$$ Proof. Let $\Im = \phi + \overline{\psi} \in \mathcal{T}\mathfrak{N}_{\mathcal{HF}}(\tau)$ where $\phi$ and $\psi$ are given by (2). In view of Lemma 1, it is enough to show that $P(\varrho, \epsilon, \delta, \eta) \leq 1 - \gamma$ , where $$P(\varrho, \epsilon, \delta, \eta) = \sum_{\nu=2}^{\infty} \left( \nu + \varrho \nu \left( \nu^2 - 1 \right) \right) \left| \frac{\Gamma(\epsilon)}{\Gamma(\rho(\nu - 1) + \epsilon)} a_{\nu} \right| + |b_1| + \sum_{\nu=2}^{\infty} \left( \nu + \varrho \nu \left( \nu^2 - 1 \right) \right) \left| \frac{\Gamma(\delta)}{\Gamma(\eta(\nu - 1) + \delta)} b_{\nu} \right|.$$ (30) Using the inequalities (10) and (11) of Lemma 2, it follows that $$\begin{split} P(\varrho,\epsilon,\delta,\eta) &\leq (1-\tau) \left[ \sum_{\nu=2}^{\infty} \left( \varrho \nu^2 + 1 - \varrho \right) \frac{\Gamma(\epsilon)}{\Gamma(\rho(\nu-1)+\epsilon)} \right. \\ &+ \sum_{\nu=2}^{\infty} \left( \varrho \nu^2 + 1 - \varrho \right) \frac{\Gamma(\delta)}{\Gamma(\eta(\nu-1)+\delta)} \right] + |b_1| \\ &= (1-\tau) \left[ \sum_{\nu=2}^{\infty} \left[ \varrho \nu(\nu-1) + \varrho \nu + 1 - \varrho \right] \frac{\Gamma(\epsilon)}{\Gamma(\rho(\nu-1)+\epsilon)} \right. \\ &+ \left. \sum_{\nu=2}^{\infty} \left[ \varrho \nu(\nu-1) + \varrho \nu + 1 - \varrho \right] \frac{\Gamma(\delta)}{\Gamma(\eta(\nu-1)+\delta)} \right] + |b_1| \\ &= (1-\tau) \left[ \varrho \chi_{\rho,\epsilon}^{(2)}(1) + \varrho \chi_{\rho,\epsilon}'(1) + (1-\varrho) \left( \chi_{\rho,\epsilon}(1) - 1 \right) \right. \\ &+ \varrho \chi_{\eta,\epsilon}^{(2)}(1) + \varrho \chi_{\eta,\epsilon}'(1) + (1-\varrho) \left( \chi_{\eta,\epsilon}(1) - 1 \right) \right] + |b_1| \\ &\leq 1-\gamma, \end{split}$$ by the given hypothesis. $\Box$ Below we prove that $\Theta(\mathcal{T}\mathfrak{R}_{\mathcal{HF}}(\tau)) \subset \mathcal{HF}(\varrho, \gamma)$ .

Theorem 4. Let $\varrho \geq 0$ , $\gamma, \tau \in [0,1)$ and $\rho, \varepsilon, \eta, \delta \notin \mathbb{Z}_0^-$ . If $$(1-\tau)\left[\varrho\left(\chi_{\rho,\epsilon}'(1)+\chi_{\eta,\epsilon}'(1)\right)+\int_{0}^{1}\frac{\chi_{\rho,\epsilon}(s)}{s}ds+\int_{0}^{1}\frac{\chi_{\eta,\epsilon}(s)}{s}ds\right]$$ $$<1-\delta-|b_{1}|,$$ then $$\Theta(\mathcal{T}\mathfrak{R}_{\mathcal{HF}}(\tau)) \subset \mathcal{HF}(\varrho,\gamma).$$ Proof. Making use of Lemma 1, we need only to prove that $P(\varrho, \epsilon, \delta, \eta) \leq 1 - \gamma$ , where $P(\varrho, \epsilon, \delta, \eta)$ as given in (30). Using the inequalities (13) and (14) of Lemma 3, it follows that $$\begin{split} P(\varrho,\epsilon,\delta,\eta) &= \sum_{\nu=2}^{\infty} \left( \nu + \varrho \nu \left( \nu^2 - 1 \right) \right) \left| \frac{\Gamma(\epsilon)}{\Gamma(\rho(\nu-1)+\epsilon)} a_{\nu} \right| \\ &+ |b_1| + \sum_{\nu=2}^{\infty} \left( \nu + \varrho \nu \left( \nu^2 - 1 \right) \right) \left| \frac{\Gamma(\delta)}{\Gamma(\eta(\nu-1)+\delta)} b_{\nu} \right| \\ &\leq (1-\tau) \left[ \sum_{\nu=2}^{\infty} \left( \varrho \nu + \frac{1-\varrho}{\nu} \right) \frac{\Gamma(\epsilon)}{\Gamma(\rho(\nu-1)+\epsilon)} \right. \\ &+ \sum_{\nu=2}^{\infty} \left( \varrho \nu + \frac{1-\varrho}{\nu} \right) \frac{\Gamma(\delta)}{\Gamma(\eta(\nu-1)+\delta)} \right] \\ &= (1-\tau) \left[ \varrho \chi'_{\rho,\epsilon}(1) + \int_{0}^{1} \frac{\chi_{\rho,\epsilon}(s)}{s} dt + \varrho \chi'_{\eta,\epsilon}(1) + \int_{0}^{1} \frac{\chi_{\eta,\epsilon}(s)}{s} ds \right] + |b_1| \\ &\leq 1-\gamma, \end{split}$$ by given hypothesis. $\square$

Theorem 5. Let $\varrho \geq 0$ , $\gamma, \tau \in [0,1)$ and $\rho, \varepsilon, \eta, \delta \notin \mathbb{Z}_0^-$ . If $$\chi_{\rho,\epsilon}(1) + \chi_{\eta,\epsilon}(1) \le 3 - \frac{|b_1|}{1 - \gamma}$$ then $$\Theta(\mathcal{HF}(\varrho,\gamma)) \subset \mathcal{HF}(\varrho,\gamma).$$ Proof. Using Lemma 1 and the inequalities (7) and (8) of Lemma 1, we obtain $$P(\varrho, \epsilon, \delta, \eta) \leq (1 - \gamma) \left[ \sum_{\nu=2}^{\infty} \frac{\Gamma(\epsilon)}{\Gamma(\rho(\nu - 1) + \epsilon)} + \sum_{\nu=2}^{\infty} \frac{\Gamma(\delta)}{\Gamma(\eta(\nu - 1) + \delta)} \right] + |b_1|$$ $$= (1 - \gamma) \left[ (\chi_{\rho, \epsilon}(1) - 1) + (\chi_{\eta, \epsilon}(1) - 1) \right] + |b_1|$$ $$= (1 - \gamma) \left[ \chi_{\rho, \epsilon}(1) + \chi_{\eta, \epsilon}(1) - 2 \right] + |b_1|$$ $$\leq 1 - \gamma,$$ by the given condition and this completes the proof of the theorem. $\Box$

Corollary 1. Let $\gamma \in [0,1)$ and $\rho, \epsilon, \eta, \delta \notin \mathbb{Z}_0^-$ . If $$2\left(\chi_{\rho,\epsilon}^{(3)}(1) + \chi_{\eta,\epsilon}^{(3)}(1)\right) + 9\chi_{\rho,\epsilon}^{(2)}(1) + 6\chi_{\rho,\epsilon}^{'}(1) + 3\chi_{\eta,\epsilon}^{(2)}(1) \le 6(1-\gamma),$$ then $$\Theta(\mathcal{S}_{\mathcal{UF}}^*) \subset \mathcal{HF}(\gamma).$$

Corollary 2. Let $\gamma \in [0,1)$ and $\rho, \epsilon, \eta, \delta \notin \mathbb{Z}_0^-$ . If $$[\chi_{\rho,\epsilon}^{(2)}(1) - \chi_{\eta,\epsilon}^{(2)}(1) + 2\left(\chi_{\rho,\epsilon}'(1) - \chi_{\eta,\epsilon}'(1)\right)]$$ $$\leq 2(1 - \gamma), \tag{31}$$ then $$\Theta(\mathcal{K}_{\mathcal{HF}}) \subset \mathcal{HF}(\gamma).$$

Corollary 3. Let γ ∈ [0, 1) and ρ, e, η, δ 6∈ Z − 0 . If $$(1-\tau)\left[\left(\chi_{\rho,\epsilon}(1)+\chi_{\eta,\epsilon}(1)\right)-2\right]\leq 1-\gamma-|b_1|,$$ then $$\Theta(\mathcal{T}\mathfrak{N}_{\mathcal{HF}}(\tau)) \subset \mathcal{HF}(\gamma).$$

Corollary 4. Let γ ∈ [0, 1) and ρ, e, η, δ 6∈ Z − 0 . If $$(1-\tau)\left[\int\limits_0^1 \frac{\chi_{\rho,\epsilon}(s)}{s}dt + \int\limits_0^1 \frac{\chi_{\eta,\epsilon}(s)}{s}dt\right] \leq 1-\gamma - |b_1|,$$ then $$\Theta(\mathcal{T}\mathfrak{R}_{\mathcal{HF}}(\tau))\subset\mathcal{HF}(\gamma).$$

Lemma 1. The function $\varphi_{BS}(\xi) = \left(\frac{\xi}{e^{\xi}-1}\right)^2$ satisfies $$\min_{\substack{|\xi|=\ell}} \operatorname{Re} \varphi_{BS}(\xi) = \varphi_{BS}(\ell) = \min_{\substack{|\xi|=\ell}} |\varphi_{BS}(\xi)|$$ $$\max_{\substack{|\xi|=\ell}} \operatorname{Re} \varphi_{BS}(\xi) = \varphi_{BS}(-\ell) = \max_{\substack{|\xi|=\ell}} |\varphi_{BS}(\xi)|,$$ whenever $\ell \in (0,1)$ .

Theorem 1. The class  $S_{RS}^*$  satisfies the following inclusion relations:

- If  $0 \le \lambda \le \frac{1}{(e-1)^2}$ , then  $\mathbf{S}_{BS}^ \subset \mathbf{S}^(\lambda)$ .
- If  $\beta \geq (\frac{e}{e-1})^2$ , then  $\mathbf{S}_{BS}^ \subset \mathbf{RS}^(1/\beta) \subset \mathbf{M}(\beta)$ .  $\mathbf{S}_{BS}^ \subset \mathbf{SS}^(\beta)$ , where  $\beta_0 \leq \beta \leq 1$ , wherein  $\beta_0 = 2h(y_2)/\pi \approx 0.6454469651m$  and h is

Proof. (1) If  $F \in \mathbf{S}_{BS}^*$ , then  $\frac{\xi F'(\xi)}{F(\xi)} \prec (\frac{\xi}{e^{\xi}-1})^2$ . According to Lemma 1, we have

$$\min_{|\xi|=1} Re \left(\frac{\xi}{e^{\xi}-1}\right)^2 < Re \frac{\xi F'(\xi)}{F(\xi)} < \max_{|\xi|=1} Re \left(\frac{\xi}{e^{\xi}-1}\right)^2;$$

therefore,

$$Re\frac{\xi F'(\xi)}{F(\xi)} > \frac{1}{(e-1)^2}.$$

(2) Similarly,

$$Re\frac{\xi F'(\xi)}{F(\xi)} < \frac{e^2}{(e-1)^2}.$$

Thus,  $F \in \mathbf{S}^* \left( \frac{1}{(e-1)^2} \right) \cap \mathbf{M} \left( \frac{e^2}{(e-1)^2} \right)$ . Now, we have the following:

$$Re\frac{F(\xi)}{\xi F'(\xi)} > \left(\frac{e-1}{e}\right)^2.$$

This implies that  $F \in \mathbf{RS}^(\beta)$  for  $\beta \leq \left(\frac{e-1}{e}\right)^2$ . Identically,  $F \in \mathbf{RS}^(1/\beta)$  for  $\beta \geq \left(\frac{e}{e-1}\right)^2$ . Also,  $F \in \mathbf{RS}^*(1/\beta)$  if and only if

$$\left|\frac{\xi F'(\xi)}{F(\xi)} - \frac{\beta}{2}\right| < \frac{\beta}{2},$$

which leads to  $Re(\xi F'(\xi)/F(\xi)) < \beta$ . Therefore,  $\mathbf{S}_{BS}^ \subset \mathbf{RS}^(1/\beta) \subset \mathbf{M}(\beta)$  whenever  $\beta \geq \left(\frac{e}{e-1}\right)^2$ . (3) If  $F \in \mathbf{S}_{RS}^*$ , then

$$\left| \arg \frac{\xi F'(\xi)}{F(\xi)} \right| < \max_{|\xi|=1} \arg \left( \frac{\xi}{e^{\xi} - 1} \right)^{2}$$
$$= \max_{0 \le y < 2\pi} \arctan \left( \frac{V}{U} \right).$$

Let

$$h(y) = \arctan\left(\frac{V}{U}\right),\tag{5}$$

where U and V are given as

$$U = \cos(2y) \left( \frac{\left(e^{\cos(y)}\right)^2 (\cos(\sin(y)))^2 - 2e^{\cos(y)} \cos(\sin(y)) + 1}{-\left(e^{\cos(y)}\right)^2 (\sin(\sin(y)))^2} \right)$$

$$+ \sin(2y) \left( 2\left(e^{\cos(y)}\right)^2 \cos(\sin(y)) \sin(\sin(y)) - 2e^{\cos(y)} \sin(\sin(y)) \right),$$

$$V = \sin(2y) \left( \frac{\left(e^{\cos(y)}\right)^2 (\cos(\sin(y)))^2 - 2e^{\cos(y)} \cos(\sin(y)) + 1}{-\left(e^{\cos(y)}\right)^2 (\sin(\sin(y)))^2} \right)$$

$$- \cos(2y) \left( 2\left(e^{\cos(y)}\right)^2 \cos(\sin(y)) \sin(\sin(y)) - 2e^{\cos(y)} \sin(\sin(y)) \right).$$

Here, h'(y) = 0 has  $y_1 \approx 1.409746460$  and  $y_2 \approx 4.873438847$  roots in  $[0, 2\pi]$ . In addition,  $h''(y_2) \approx -1.0988577$ . Hence,  $\max_{0 \leq y < 2\pi} h(y) = h(y_2) \approx 1.013865722$ , and  $\left|\arg \frac{\xi F'(\xi)}{F(\xi)}\right| \leq \frac{\pi \beta}{2}$ ; that is,  $\beta \geq 0.645186552$ . This implies that  $\mathbf{S}^_{BS} \subset \mathbf{S}^_{\beta}$ .  $\square$ 

Now, we discuss some radii problems for the class  $\mathbf{S}_{BS}^*$ . The following definitions and lemmas are needed to establish the results. The class  $\mathbf{P}$  represents the functions p of the form

$$p(\xi) = 1 + \sum_{n=1}^{\infty} p_n \xi^n \tag{6}$$

that are analytic in E such that  $Rep(\xi) > 0$ ,  $\xi \in \mathbf{E}$ . Let

$$\mathbf{P}_n[a,b] := \left\{ p(\xi) = 1 + \sum_{k=n}^{\infty} c_n \xi^n : p(\xi) \prec \frac{1 + a\xi}{1 + b\xi'}, -1 \le b < a \le 1 \right\}.$$

In particular,  $\mathbf{P}_n(\lambda) := \mathbf{P}_n[1-2\lambda,-1]$ , and  $\mathbf{P}_n := \mathbf{P}_n(0)$ . Let  $\mathbf{S}_n^[a,b] = \mathbf{A}_n \cap \mathbf{S}^[a,b]$ , and  $\mathbf{S}_n^(\lambda) := \mathbf{S}_n^[1-2\lambda,-1]$ . Also, let

$$\mathbf{S}_{BS,n}^ := \mathbf{A}_n \cap \mathbf{S}_{BS}^, \quad \mathbf{S}_n^(\lambda) := \mathbf{A}_n \cap \mathbf{S}^(\lambda), \quad \mathbf{S}_{L,n}^ := \mathbf{A}_n \cap \mathbf{S}_L^.$$

Additionally,

$$\mathbf{S}_n := \{ \mathcal{F} \in \mathbf{A}_n : \mathcal{F}(\xi) / \xi \in \mathbf{P}_n \},$$

and

$$\mathbf{CS}_n(\lambda) := \left\{ F \in \mathbf{A}_n : \frac{F(\xi)}{g(\xi)} \in \mathbf{P}_n, \ g \in \mathbf{S}_n^*(\lambda) \right\};$$

see [24].

Lemma 2. ([25]). If $p \in \mathbf{P}_n(\lambda)$ , then for $|\xi| = \ell$ , $$\left|\frac{\xi p'(\xi)}{p(\xi)}\right| \le \frac{2(1-\lambda)n\ell^n}{(1-\ell^n)(1+(1-2\lambda)\ell^n)}.$$

Lemma 3. ([26]). Let $p \in \mathbf{P}$ . Then, $$|jp_1^3 - kp_1p_2 + lp_3| \le 2|j| + 2|k - 2j| + 2|j - k + l|.$$

Lemma 4. ([27]). If $p \in \mathbf{P}_n[a, b]$ , then for $|\xi| = \ell$ , $$\left| p(\xi) - \frac{1 - ab\ell^{2n}}{1 - b^2\ell^{2n}} \right| \le \frac{(a - b)\ell^n}{1 - b^2\ell^{2n}}$$ If $p \in \mathbf{P}_n(\lambda)$ , then for $|\xi| = \ell$ , $$\left| p(\xi) - \frac{(1 + (1 - 2\lambda))\ell^{2n}}{1 - \ell^{2n}} \right| \le \frac{2(1 - \lambda)\ell^n}{1 - \ell^{2n}}.$$ In the following lemmas, we find disks centered at $(\nu,0)$ and (1,0) of the largest and the smallest radii, respectively, such that $\mho_{BS} := \varphi_{BS}(\mathbf{E})$ lies in the disk with the smallest radius and contains the largest disk.

Lemma 5. Let $\left(\frac{1}{e-1}\right)^2 \le \nu \le \left(\frac{e}{e-1}\right)^2$ . Then $$\{w \in \mathbb{C}: |w-\nu| < \ell_{\nu}\} \subset \mho_{BS} \subset \left\{w \in \mathbb{C}: |w-1| < \left(\frac{e}{e-1}\right)^2\right\},$$ where $$\ell_{\nu} = \begin{cases} \nu - \left(\frac{1}{e-1}\right)^{2}, & \frac{1}{(e-1)^{2}} \leq \nu \leq \frac{e^{2}+1}{2(e-1)^{2}}, \\ \left(\frac{e}{e-1}\right)^{2} - \nu, & \frac{e^{2}+1}{2(e-1)^{2}} \leq \nu \leq \left(\frac{e}{e-1}\right)^{2}. \end{cases}$$ Proof. Let $\cos(y) = \varrho$ and $\sin(y) = \varsigma$ . Then, the square of the distance from the boundary $\mho_{BS}$ to the point $(\nu, 0)$ is given by $$\psi(y) = \left(\frac{\mathbf{A}}{(1 - 2e^{\varrho}\cos(\zeta) + (e^{\varrho})^{2})^{2}} - \nu\right)^{2} + \left(\frac{\mathbf{B}}{(1 - 2e^{\varrho}\cos(\zeta) + (e^{\varrho})^{2})^{2}}\right)^{2},$$ where $$\mathbf{A} = \cos(2y) \Big\{ 1 - 2e^{\varrho} \cos(\varsigma) + (e^{\varrho})^2 \cos(2\varsigma) \Big\} + 2e^{\varrho} \Big\{ e^{\varrho} \cos(\varsigma) - 1 \Big\} \sin(\varsigma) \sin(2y),$$ $$\mathbf{B} = \sin(2y) \Big\{ 1 - 2e^{\varrho} \cos(\zeta) + (e^{\varrho})^2 \cos(2\zeta) \Big\} - 2e^{\varrho} \{ e^{\varrho} \cos(\zeta) - 1 \} \sin(\zeta) \cos(2y).$$ To show that $|w-v|<\ell_{\nu}$ is largest disk contained in $\mho_{BS}$ , it is enough to show that the $\min_{0\leq \leq 2\pi}\psi(y)=\ell_{\nu}$ . Since $\psi(y)=\psi(-y)$ , it is enough to take the range $0\leq y\leq \pi$ . Case 1: When $\frac{1}{(e-1)^2} \le \nu < \frac{e^2}{(2e^2-8e+9)(e-1)^2}$ , then $\psi'(y) = 0$ has 0 and $\pi$ roots. In addition, $\psi'(y) > 0$ for $y \in (0, \pi)$ . Thus, $$\min_{0 \le \ell \le \pi} \psi(y) = \min\{\psi(0), \ \psi(\pi)\} = \psi(0).$$ Hence, $$\ell_{\nu} = \min_{0 \le y \le \pi} \sqrt{\psi(y)} = \sqrt{\psi(0)} = \frac{1}{(e-1)^2} - \nu.$$ Case 2: When $\frac{e^2}{(2e^2-8e+9)(e-1)^2} < \nu \le \frac{e^2}{(e-1)^2}$ , then $\psi'(y) = 0$ has $0, y_{\nu}$ , and $\pi$ roots, where $y_{\nu}$ depends on $\nu$ . In addition, $\psi'(y) > 0$ for $y \in (0, y_{\nu})$ , and $\psi'(y) < 0$ when $y \in (y_{\nu}, \pi)$ . Therefore, $\psi(y)$ has minima at 0 or $\pi$ . We also see that $\psi(0) < \psi(\pi)$ for $\frac{e^2}{(2e^2-8e+9)(e-1)^2} < \nu \le \frac{e^2+1}{2(e-1)^2}$ and $\psi(0) > \psi(\pi)$ for $\frac{e^2+1}{2(e-1)^2} < \nu \le \left(\frac{e}{e-1}\right)^2$ . Thus, the first part of the proof is completed. Now, for the smallest disc that contains $\mho_{BS}$ , the function $\psi(y)$ for $\nu=1$ attains its maximum value at $\pi$ . Thus, the disk with the smallest radius that contains $\mho_{BS}$ has a radius of $\left(\frac{e}{e-1}\right)^2$ . $\square$

Theorem 2. The sharp $R_{\mathbf{S}_{RS,n}^*}$ for $\mathbf{S}_n$ is $$R_{\mathbf{S}_{BS,n}^*}(\mathbf{S}_n) = \left(\frac{e^2 - 2e}{\sqrt{n^2(e-1)^4 + (e^2 - 2e)} + n(e-1)^2}\right)^{1/n}.$$ Proof. Consider a function $h(\xi) \in \mathbf{P}_n$ such that $h(\xi) = F(\xi)/\xi$ . Now, we have the following: $$\frac{\xi F'(\xi)}{F(\xi)} - 1 = \frac{\xi \hbar'(\xi)}{\hbar(\xi)}.$$ From Lemma 2, we have $$\left|\frac{\xi F'(\xi)}{F(\xi)} - 1\right| = \left|\frac{\xi h'(\xi)}{h(\xi)}\right| \le \frac{2n\ell^n}{1 - \ell^{2n}}.$$ From Lemma 4, the map of $|\xi| \le \ell$ under $\xi F'/F$ lies in the $\Im_{BS}$ if the following is satisfied: $$\frac{2n\ell^n}{1-\ell^{2n}} \le 1 - \frac{1}{(e-1)^2}.$$ This is equivalently written as $$((e-1)^2-1)\ell^{2n}+2n(e-1)^2\ell^n+1-(e-1)^2\leq 0.$$ Thus, the $\mathbf{S}_{RS,n}^*$ -radius of the $\mathbf{S}_n$ is the root $\ell \in (0,1)$ of $$((e-1)^2 - 1)\ell^{2n} + 2n(e-1)^2\ell^n + 1 - (e-1)^2 = 0;$$ that is, $$R_{\mathbf{S}_{BS,n}^*}(\mathbf{S}_n) = \left(\frac{e(e-2)}{n(e-1)^2 + \sqrt{1 + (n^2+1)(e-1)^4 - 2(e-1)^2}}\right)^{1/n}.$$ Consider $F_0(\xi) = \xi(1+\xi^n)/1 - \xi^n$ . Then, $\hbar_0(\xi) = F_0(\xi)/\xi = (1+\xi^n)/(1-\xi^n) > 0$ . Thus, $F_0 \in \mathbf{S}_n$ , and $\xi F_0'(\xi)/F_0(\xi) = 1 + 2n\xi^n/(1-\xi^{2n})$ . This is beacuse at $\xi = R_{\mathbf{S}_{BS,n}^*}$ , we have $$\frac{\xi F_0'(\xi)}{F_0(\xi)} - 1 = \frac{2n\xi^n}{1 - \xi^{2n}} = 1 - \frac{1}{(e-1)^2}.$$ Therefore, $F_0$ gives a sharp result. Hence, the proof is completed. $\Box$

Theorem 3. Let $$R_{1} = \left(\frac{4e - e^{2} - 1}{2(e - 1)^{2}(1 - 2\lambda) + e^{2} + 1}\right)^{\frac{1}{2n}},$$ $$R_{2} = \left(\frac{(e - 1)^{2} - 1}{(1 + n - \lambda)(e - 1)^{2} + \sqrt{1 + (2n(1 - \lambda) + \lambda^{2} + n^{2})(e - 1)^{4} - 2(e - 1)^{2}\lambda}}\right)^{\frac{1}{n}},$$ $$R_{3} = \left(\frac{e^{2}}{(1 + n - \lambda)(e - 1)^{2} + \sqrt{1 + (1 + n - \lambda)^{2}(e - 1)^{4} - (1 - 2\lambda)e^{4}}}\right)^{\frac{1}{n}}.$$ Then, a sharp $\mathbf{S}_{BS,n}$ -radius for the class $\mathbf{CS}_n(\lambda)$ is* $$R_{\mathbf{S}_{BS,n}^*}(\mathbf{CS}_n(\lambda)) = \begin{cases} R_2, & \text{if } R_2 \leq R_1, \\ R_3, & \text{if } R_2 > R_1. \end{cases}$$ Proof. Define a function $\hbar(\xi) = F(\xi)/g(\xi)$ , where $g \in \mathbf{S}_n^*(\lambda)$ . Then, $\hbar \in \mathbf{P}_n$ , and $\xi g'(\xi)/g(\xi) \in P_n(\lambda)$ . From the definition of $\hbar$ , we have $$\frac{\xi F'(\xi)}{F(\xi)} = \frac{\xi g'(\xi)}{g(\xi)} + \frac{\xi h'(\xi)}{h(\xi)}.$$ From Lemmas 2 and 3, we see that $$\left| \frac{\xi F'(\xi)}{F(\xi)} - \frac{1 + (1 - 2\lambda)\ell^{2n}}{1 - \ell^{2n}} \right| \le \frac{2(1 + n - \lambda)\ell^n}{1 - \ell^{2n}}.$$ (7) Now, we find the values $R_1$ , $R_2$ and $R_3$ for $0 < \ell < 1$ and $0 \le \lambda < 1$ . Firstly, we find $R_1$ . For $\ell \le R_1$ , this can be found if and only if $$\frac{1 + (1 - 2\lambda)\ell^{2n}}{1 - \ell^{2n}} \le \frac{e^2 + 1}{2(e - 1)^2}.$$ This implies that $$\ell \le \left(\frac{4e - e^2 - 1}{2(e - 1)^2(1 - 2\lambda) + e^2 + 1}\right)^{\frac{1}{2n}}.$$ Now, we obtain $R_2$ . For this, we must have $$\frac{2(1+n-\lambda)\ell^n}{1-\ell^{2n}} \leq \frac{1+(1-2\lambda)\ell^{2n}}{1-\ell^{2n}} - \frac{1}{(e-1)^2}.$$ This implies that $$\ell \leq \frac{(e-1)^2 - 1}{(1+n-\lambda)(e-1)^2 + \sqrt{1 + (-2n\lambda + \lambda^2 + 2n + n^2)(e-1)^4 - 2(e-1)^2\lambda}}$$ For $R_3$ , we have $$\frac{2(1+n-\lambda)\ell^n}{1-\ell^{2n}} \le \left(\frac{e}{e-1}\right)^2 - \frac{1+(1-2\lambda)\ell^{2n}}{1-\ell^{2n}}.$$ This implies that $$\ell \leq \frac{e^2}{(1+n-\lambda)(e-1)^2 + \sqrt{1+(1+n-\lambda)^2(e-1)^4 - (1-2\lambda)e^4}}$$

Theorem 4. The $S_{RS,n}$ -radius for $S_n^[a,b]$ is $$R_{\mathbf{S}_{BS,n}}(\mathbf{S}_n^[a,b]) = \begin{cases} \min\{1;\ell_1\}, & -1 \le b \le 0 < a \le 1, \\ \min\{1;\ell_2\}, & 0 < b < a \le 1, \end{cases}$$ where $$\ell_1 = \left(\frac{2e - 1}{(e - 1)^2 a - be^2}\right)^{1/n},$$ and $$\ell_2 = \left(\frac{e(e-2)}{a(e-1)^2 - b}\right)^{1/n}.$$ Proof. Let $F \in \mathbf{S}_{n}^{*}[a,b]$ . Then, from Lemma 3, we can write $$\left| \frac{\xi F'(\xi)}{F(\xi)} - C \right| \le \frac{(a-b)\ell^n}{1 - b^2 \ell^{2n}},\tag{8}$$ where $$C = \frac{1 - ab\ell^{2n}}{1 - b^2\ell^{2n}}, \ |\xi| = \ell.$$ For b < 0, we see that $C \ge 1$ . Also by using Lemma 4, $F \in \mathbf{S}_{BS,n}^*$ if $$\frac{1 + (a-b)\ell^n - ab\ell^{2n}}{1 - b^2\ell^{2n}} \le \frac{e^2}{(e-1)^2},$$ which is equivalent to $$\ell \le \left(\frac{2e-1}{(e-1)^2a-be^2}\right)^{1/n} = \ell_1.$$ Furthermore, if b = 0, then C = 1. From (8), we have $$\left| \frac{\xi F'(\xi)}{F(\xi)} - 1 \right| \le a\ell^n, \quad (0 < a \le 1).$$ By using Lemma 4 with a=1, this gives $\ell \leq \left(\frac{e(e-2)}{a(e-1)^2}\right)^{1/n}$ for $F \in \mathbf{S}_{BS,n}$ . We see that C < 1 for $0 < b < a \le 1$ . Thus, from Lemma 4 and (8), we have $F \in \mathbf{S}_{BS,n}$ if $$\frac{(a-b)\ell^n}{1-b^2\ell^{2n}} \le \frac{1-ab\ell^{2n}}{1-b^2\ell^{2n}} - \frac{1}{(e-1)^2},$$ or, equivalently, if $$\ell \le \left(\frac{e(e-2)}{a(e-1)^2 - b}\right)^{1/n} = \ell_2.$$ This completes the result. $\Box$

Theorem 5. Let $-1 < b < a \le 1$ . If either (a) $$(1-b) \le (e-1)^2(1-a)$$ and $2(1-b^2) \le (1-ab)(e-1)^2 < (1-b^2)(1+e^2)$ or if (a) $$(1-b) \le (e-1)^2(1-a)$$ and $2(1-b^2) \le (1-ab)(e-1)^2 < (1-b^2)(1+e^2)$ or if (b) $(a+1)(e-1)^2 \le e^2(1+b)$ and $(1-b^2)(1+e^2) \le 2(1-ab)(e-1)^2 \le 2e^2(1-b^2)$ hold, then $\mathbf{S}_n^[a,b] \subset \mathbf{S}_{BS,n}$ . Proof. (a) Let $p(\xi) = \xi F'(\xi) / F(\xi)$ . From Lemma 3, $F \in \mathbf{S}_n^*[a,b]$ if $$\left| p(\xi) - \frac{1 - ab}{1 - b^2} \right| \le \frac{a - b}{1 - b^2}.$$ In connection with Lemma 4, $F \in \mathbf{S}_n^*[a,b]$ if $$\frac{a-b}{1-b^2} \le \frac{1-ab}{1-b^2} - \frac{1}{(e-1)^2},$$ and $$\frac{1}{(e-1)^2} \le \frac{1-ab}{1-b^2} \le \frac{1}{2} \frac{1+e^2}{(e-1)^2},$$ which, upon simplification, reduce to (a). (b) Let $p(\xi) = \xi F'(\xi) / F(\xi)$ . Since $F \in \mathbf{S}_n^*[a,b]$ , thus, in the view of Lemma 3, $$\left| p(\xi) - \frac{1 - ab}{1 - b^2} \right| \le \frac{a - b}{1 - b^2}.$$ By using Lemma 4, we note that $\Gamma \in \mathbf{S}_n^*[a,b]$ if the following is satisfied: $$\frac{a-b}{1-b^2} \le \frac{e^2}{(e-1)^2} - \frac{1-ab}{1-b^2},$$ and $$\frac{1}{2} \left( \frac{1 + e^2}{(e - 1)^2} \right) \le \frac{1 - ab}{1 - b^2} \le \frac{e^2}{(e - 1)^2},$$ which reduced to the conditions (b). $\square$

Theorem 6. The sharp radii for $S_L^*$ , $S_{RL}^*$ , $S_e^*$ , and $S_{lim}^*$ are $$\begin{split} R_{\mathbf{S}_{BS}^*}(\mathbf{S}_{\mathbf{L}}^*) &= \frac{(e-1)^4 - 1}{(e-1)^4} \approx 0.889, \\ R_{\mathbf{S}_{BS}^*}(\mathbf{S}_{\mathbf{RL}}^*) &= \frac{(5+4\sqrt{2})(e-1)^4 + (-6\sqrt{2} - 8)(e-1)^2 + 3 + 2\sqrt{2}}{(5+4\sqrt{2})(e-1)^4 + (8+4\sqrt{2})(e-1)^2 + 2 + 2\sqrt{2}} \approx 0.87193, \\ R_{\mathbf{S}_{BS}^*}(\mathbf{S}_{\mathbf{e}}^*) &= 2 - 2\ln(e-1) \approx 0.917350, \\ R_{\mathbf{S}_{BS}^*}(\mathbf{S}_{\mathbf{lim}}^*) &= \frac{\sqrt{2}(e-2)}{e-1} \approx 0.591174. \end{split}$$ Proof. (1) For $F \in \mathbf{S}_{\mathbf{I}}^*$ , we have $$\frac{\xi F'(\xi)}{F(\xi)} = \sqrt{1 + \omega(\xi)}.$$ By the Schwarz Lemma $|\omega(\xi)| \le |\xi|$ , we thus have $\left|\sqrt{1+\omega(\xi)}-1\right| \le 1-\sqrt{1-\ell}$ . Thus, for $|\xi|=\ell$ , we have $$\left|\frac{\xi F'(\xi)}{F(\xi)} - 1\right| \le 1 - \sqrt{1 - \ell}.$$ By Lemma 4, we have $1-\sqrt{1-\ell} \leq 1-\frac{1}{(e-1)^2}$ . Consider $\digamma_0(\xi)=\frac{4\xi\exp\{2(\sqrt{1+\xi}-1)\}}{(1+\sqrt{1+\xi})^2}$ , which is in $\mathbf{S}^*_{\mathbf{L}}$ and $\frac{\xi\digamma_0'(\xi)}{\digamma_0(\xi)}=\sqrt{1+\xi}=\frac{1}{(e-1)^2}$ at $R_{\mathbf{S}^*_{BS}}(\mathbf{S}^*_{\mathbf{L}})$ . Hence, the sharpness is verified. (2) Let $F \in \mathbf{S}_{\mathbf{RL}}^*$ . Then, for $|\xi| = \ell$ , we have $$\begin{split} \left| \frac{\xi F'(\xi)}{F(\xi)} - 1 \right| &\leq 1 - \sqrt{2} + (\sqrt{2} - 1) \sqrt{\frac{1 + \ell}{1 - 2(\sqrt{2} - 1)\ell}} \\ &\leq 1 - \frac{1}{(e - 1)^2} \end{split}$$ provided that $$\ell \leq \frac{(5+4\sqrt{2})(e-1)^4+(-6\sqrt{2}-8)(e-1)^2+3+2\sqrt{2}}{(5+4\sqrt{2})(e-1)^4+(8+4\sqrt{2})(e-1)^2+2+2\sqrt{2}} = R_{\mathbf{S}_{BS}^*}(\mathbf{S}_{\mathbf{RL}}^*).$$ Consider the function $F_1$ defined by $$F_1(\xi) = \xi \exp\left(\int_0^{\xi} \frac{\varphi_0(t) - 1}{t} dt\right),$$ where $$\varphi_0(\xi) = \sqrt{2} - (\sqrt{2} - 1)\sqrt{\frac{1 - \xi}{1 + 2(\sqrt{2} - 1)\xi}}.$$ At $\xi = R_{\mathbf{S}_{BS}^*}(\mathbf{S}_{\mathbf{RL}}^*)$ , we have $$\frac{\xi F_1'(\xi)}{F_1(\xi)} = \sqrt{2} - (\sqrt{2} - 1)\sqrt{\frac{1 - \xi}{1 + 2(\sqrt{2} - 1)\xi}} = \frac{1}{(e - 1)^2}.$$ Hence, the sharpness is verified. (3) $F \in \mathbf{S}_{\rho}^*$ , so we have $$\left|\frac{\xi F'(\xi)}{F(\xi)} - 1\right| \le e^{\xi} - 1 \le \frac{e^2}{(e-1)^2} - 1.$$ The result is sharp for $F_2$ such that $\frac{\xi F_2'(\xi)}{F_2(\xi)} = e^{\xi}$ . (4) Suppose that $F \in (\mathbf{S}^*_{\lim})$ ; then $\frac{\xi F'(\xi)}{F(\xi)} \prec 1 + \sqrt{2}\xi + \frac{\xi^2}{2}$ . Thus, for $|\xi| = \ell$ , we can it write as $$\left| \frac{\xi F'(\xi)}{F(\xi)} - 1 \right| = \left| 1 + \sqrt{2}\xi + \frac{\xi^2}{2} - 1 \right| \le \sqrt{2}\ell - \frac{\ell^2}{2} \le 1 - \frac{1}{(\ell - 1)^2},$$ which is satisfied for $\ell \leq \frac{\sqrt{2}(e-2)}{e-1}$ . Consider $$F_3(\xi) = \xi \exp \frac{4\sqrt{2}\xi + \xi^2}{4}.$$ Since $\frac{\xi F_3'(\xi)}{F_3(\xi)} = 1 + \sqrt{2}\xi + \frac{\xi^2}{2}$ , it follows that $F_3 \in (\mathbf{S}^*_{\mathbf{lim}})$ and at $\xi = R_{\mathbf{S}^*_{\mathcal{B}S}}(\mathbf{S}^*_{\mathbf{lim}})$ , so we have $\frac{\xi F_3'(\xi)}{F_3(\xi)} = \frac{1}{(e-1)^2}.$ Consider the families: $$\mathbf{F}_1 := \left\{ F \in \mathbf{A}_n : Re\left(\frac{F(\xi)}{g(\xi)}\right) > 0 \text{ and } Re\left(\frac{g(\xi)}{\xi}\right) > 0, g \in \mathbf{A}_n \right\},$$ $$\mathbf{F}_2 := \left\{ F \in \mathbf{A}_n : Re\left(\frac{F(\xi)}{g(\xi)}\right) > 0 \text{ and } Re\left(\frac{g(\xi)}{\xi}\right) > 1/2, g \in \mathbf{A}_n \right\},$$ and $$\mathbf{F}_3 := \left\{ \digamma \in \mathbf{A}_n : \left| \frac{\digamma(\xi)}{g(\xi)} - 1 \right| < 1 \text{ and } Re\left(\frac{g(\xi)}{\xi}\right) > 0, g \in \mathbf{A}_n \right\}.$$

Notation repaired from the source PDF text layer (oa_4396230632_New_Developments_in_Geometric.pdf, p. 145, 146; pdftotext (poppler) 25.02.0). Show original OCR text

Theorem 6. The sharp radii for  $S_L^$ ,  $S_{RL}^$ ,  $S_e^$ , and  $S_{lim}^$  are

$$\begin{split} R_{\mathbf{S}_{BS}^}(\mathbf{S}_{\mathbf{L}}^) &= \frac{(e-1)^4 - 1}{(e-1)^4} \approx 0.889, \\ R_{\mathbf{S}_{BS}^}(\mathbf{S}_{\mathbf{RL}}^) &= \frac{(5+4\sqrt{2})(e-1)^4 + (-6\sqrt{2} - 8)(e-1)^2 + 3 + 2\sqrt{2}}{(5+4\sqrt{2})(e-1)^4 + (8+4\sqrt{2})(e-1)^2 + 2 + 2\sqrt{2}} \approx 0.87193, \\ R_{\mathbf{S}_{BS}^}(\mathbf{S}_{\mathbf{e}}^) &= 2 - 2\ln(e-1) \approx 0.917350, \\ R_{\mathbf{S}_{BS}^}(\mathbf{S}_{\mathbf{lim}}^) &= \frac{\sqrt{2}(e-2)}{e-1} \approx 0.591174. \end{split}$$

Proof. (1) For  $F \in \mathbf{S}_{\mathbf{I}}^*$ , we have

$$\frac{\xi F'(\xi)}{F(\xi)} = \sqrt{1 + \omega(\xi)}.$$

By the Schwarz Lemma  $|\omega(\xi)| \le |\xi|$ , we thus have  $\left|\sqrt{1+\omega(\xi)}-1\right| \le 1-\sqrt{1-\ell}$ . Thus, for  $|\xi|=\ell$ , we have

$$\left|\frac{\xi F'(\xi)}{F(\xi)} - 1\right| \le 1 - \sqrt{1 - \ell}.$$

By Lemma 4, we have  $1-\sqrt{1-\ell} \leq 1-\frac{1}{(e-1)^2}$ . Consider  $\digamma_0(\xi)=\frac{4\xi\exp\{2(\sqrt{1+\xi}-1)\}}{(1+\sqrt{1+\xi})^2}$ , which is in  $\mathbf{S}^_{\mathbf{L}}$  and  $\frac{\xi\digamma_0'(\xi)}{\digamma_0(\xi)}=\sqrt{1+\xi}=\frac{1}{(e-1)^2}$  at  $R_{\mathbf{S}^_{BS}}(\mathbf{S}^*_{\mathbf{L}})$ . Hence, the sharpness is verified.

(2) Let  $F \in \mathbf{S}_{\mathbf{RL}}^*$ . Then, for  $|\xi| = \ell$ , we have

$$\begin{split} \left| \frac{\xi F'(\xi)}{F(\xi)} - 1 \right| &\leq 1 - \sqrt{2} + (\sqrt{2} - 1) \sqrt{\frac{1 + \ell}{1 - 2(\sqrt{2} - 1)\ell}} \\ &\leq 1 - \frac{1}{(e - 1)^2} \end{split}$$

provided that

$$\ell \leq \frac{(5+4\sqrt{2})(e-1)^4+(-6\sqrt{2}-8)(e-1)^2+3+2\sqrt{2}}{(5+4\sqrt{2})(e-1)^4+(8+4\sqrt{2})(e-1)^2+2+2\sqrt{2}} = R_{\mathbf{S}_{BS}^}(\mathbf{S}_{\mathbf{RL}}^).$$

Consider the function  $F_1$  defined by

$$F_1(\xi) = \xi \exp\left(\int_0^{\xi} \frac{\varphi_0(t) - 1}{t} dt\right),$$

where

$$\varphi_0(\xi) = \sqrt{2} - (\sqrt{2} - 1)\sqrt{\frac{1 - \xi}{1 + 2(\sqrt{2} - 1)\xi}}.$$

At  $\xi = R_{\mathbf{S}_{BS}^}(\mathbf{S}_{\mathbf{RL}}^)$ , we have

$$\frac{\xi F_1'(\xi)}{F_1(\xi)} = \sqrt{2} - (\sqrt{2} - 1)\sqrt{\frac{1 - \xi}{1 + 2(\sqrt{2} - 1)\xi}} = \frac{1}{(e - 1)^2}.$$

Hence, the sharpness is verified.

(3)  $F \in \mathbf{S}_{\rho}^*$ , so we have

$$\left|\frac{\xi F'(\xi)}{F(\xi)} - 1\right| \le e^{\xi} - 1 \le \frac{e^2}{(e-1)^2} - 1.$$

The result is sharp for  $F_2$  such that  $\frac{\xi F_2'(\xi)}{F_2(\xi)} = e^{\xi}$ . (4) Suppose that  $F \in (\mathbf{S}^*_{\lim})$ ; then  $\frac{\xi F'(\xi)}{F(\xi)} \prec 1 + \sqrt{2}\xi + \frac{\xi^2}{2}$ . Thus, for  $|\xi| = \ell$ , we can it write as

$$\left| \frac{\xi F'(\xi)}{F(\xi)} - 1 \right| = \left| 1 + \sqrt{2}\xi + \frac{\xi^2}{2} - 1 \right| \le \sqrt{2}\ell - \frac{\ell^2}{2} \le 1 - \frac{1}{(\ell - 1)^2},$$

which is satisfied for  $\ell \leq \frac{\sqrt{2}(e-2)}{e-1}$ . Consider

$$F_3(\xi) = \xi \exp \frac{4\sqrt{2}\xi + \xi^2}{4}.$$

Since  $\frac{\xi F_3'(\xi)}{F_3(\xi)} = 1 + \sqrt{2}\xi + \frac{\xi^2}{2}$ , it follows that  $F_3 \in (\mathbf{S}^_{\mathbf{lim}})$  and at  $\xi = R_{\mathbf{S}^_{\mathcal{B}S}}(\mathbf{S}^*_{\mathbf{lim}})$ , so we have  $\frac{\xi F_3'(\xi)}{F_3(\xi)} = \frac{1}{(e-1)^2}.$ 

Consider the families:

$$\mathbf{F}_1 := \left\{ F \in \mathbf{A}_n : Re\left(\frac{F(\xi)}{g(\xi)}\right) > 0 \text{ and } Re\left(\frac{g(\xi)}{\xi}\right) > 0, g \in \mathbf{A}_n \right\},$$

$$\mathbf{F}_2 := \left\{ F \in \mathbf{A}_n : Re\left(\frac{F(\xi)}{g(\xi)}\right) > 0 \text{ and } Re\left(\frac{g(\xi)}{\xi}\right) > 1/2, g \in \mathbf{A}_n \right\},$$

and

$$\mathbf{F}_3 := \left\{ \digamma \in \mathbf{A}_n : \left| \frac{\digamma(\xi)}{g(\xi)} - 1 \right| < 1 \text{ and } Re\left(\frac{g(\xi)}{\xi}\right) > 0, g \in \mathbf{A}_n \right\}.$$

Theorem 7. The sharp radii for functions in the families $\mathbf{F}_1$ , $\mathbf{F}_2$ , and $\mathbf{F}_3$ respectively, are: $$\begin{split} R_{\mathbf{S}_{BS,n}}(\mathbf{F}_1) &= \left(\frac{e(e-2)}{2n(e-1)^2 + \sqrt{1 + (4n^2 + 1)(e-1)^4 - 2(e-1)^2}}\right)^{1/n}, \\ R_{\mathbf{S}_{BS,n}}(\mathbf{F}_2) &= \left(\frac{2e(e-2)}{3n(e-1)^2 + \sqrt{(9n^2 + 4n + 4)(e-1)^4 - 4(n+2)(e-1)^2 + 4}}\right)^{1/n}, \\ R_{\mathbf{S}_{BS,n}^*}(\mathbf{F}_3) &= \left(\frac{2e(e-2)}{3n(e-1)^2 + \sqrt{(9n^2 + 4n + 4)(e-1)^4 - 4(n+2)(e-1)^2 + 4}}\right)^{1/n}. \end{split}$$ Proof. (1) Let $F \in \mathbf{F}_1$ and define p, $\hbar : \mathbf{E} \to \mathbb{C}$ by $p(\xi) = \frac{g(\xi)}{\xi}$ and $\hbar(\xi) = \frac{F(\xi)}{g(\xi)}$ . Then, clearly, p, $\hbar \in \mathbf{P}_n$ , since $F(\xi) = \xi p(\xi)\hbar(\xi)$ . By Lemma 2, and by combining the above inequalities, we have $$\left|\frac{\xi F'(\xi)}{F(\xi)} - 1\right| \le \frac{4n\ell^n}{1 - \ell^{2n}} \le 1 - \frac{1}{(e-1)^2}.$$ After some simplification, we arrive at $$\ell \leq \left(\frac{e(e-2)}{2n(e-1)^2 + \sqrt{1 + (4n^2 + 1)(e-1)^4 - 2(e-1)^2}}\right)^{1/n} = R_{\mathbf{S}_{BS,n}^*}(\mathbf{F}_1).$$ To verify the sharpness of result, consider the functions defined by $$F_4(\xi) = \xi \left(\frac{1+\xi^n}{1-\xi^n}\right)^2$$ and $g_0(\xi) = \xi \left(\frac{1+\xi^n}{1-\xi^n}\right)$ . Then, clearly $Re\left(\frac{F_4(\xi)}{g_0(\xi)}\right) > 0$ , and $Re\left(\frac{g_0(\xi)}{\xi}\right) > 0$ . Hence, $F_0 \in \mathbf{F}_1$ . We see that at $\xi = R_{\mathbf{S}_{nc...}^*}(\mathbf{F}_1)e^{i\pi/n}$ as follows: $$\frac{\xi F_4'(\xi)}{F_4(\xi)} = 1 + \frac{4n\xi^n}{1 - \xi^{2n}} = \frac{1}{(e - 1)^2}.$$ Hence, the sharpness is satisfied. (2) Let $F \in \mathbf{F}_2$ . Define p, $\hbar : \mathbf{E} \to \mathbb{C}$ by $p(\xi) = \frac{g(\xi)}{\xi}$ and $\hbar(\xi) = \frac{F(\xi)}{g(\xi)}$ . Then, $p \in \mathbf{P}_n$ , and $\hbar \in \mathbf{P}_n(1/2)$ . Since $F(\xi) = \xi p(\xi)\hbar(\xi)$ , then according to Lemma 2, we have $$\left| \frac{\xi F'(\xi)}{F(\xi)} - 1 \right| \le \frac{3n\ell^n + n\ell^{2n}}{1 - \ell^{2n}} \le 1 - \frac{1}{(e-1)^2}$$ which implies that $$\ell \leq \left(\frac{2e(e-2)}{3n(e-1)^2 + \sqrt{9n^2(e-1)^4 + 4[(n+1)(e-1)^2 - 1][e(e-2)]}}\right)^{1/n} = R_{\mathbf{S}_{BS,n}^*}(\mathbf{F}_2).$$ Thus, $F \in \mathbf{S}_{BS,n}$ for $\ell \leq R_{\mathbf{S}_{BS,n}}(\mathbf{F}_2)$ . For sharpness, consider the following: $$F_5(\xi) = \frac{\xi(1+\xi^n)}{(1-\xi^n)^2} \text{ and } g_1(\xi) = \frac{\xi}{1-\xi^n}.$$ Then clearly $Re\left(\frac{\digamma_5(\xi)}{g_1(\xi)}\right) > 0$ , and $Re\left(\frac{g_1(\xi)}{\xi}\right) > \frac{1}{2}$ . Hence, $\digamma \in \mathbf{F}_2$ . Now, at $\xi = R_{\mathbf{S}_{BS,n}^*}(\mathbf{F}_2)$ $$\frac{\xi F_5'(\xi)}{F_5(\xi)} = 1 + \frac{3n\xi^n + n\xi^{2n}}{1 - \xi^{2n}} = \frac{1}{(e - 1)^2}.$$ Hence, the sharpness is satisfied. (3) Let $F \in \mathbf{F}_3$ . Define p, $\hbar : \mathbf{E} \to \mathbb{C}$ by $p(\xi) = \frac{g(\xi)}{\xi}$ and $\hbar(\xi) = \frac{g(\xi)}{F(\xi)}$ . Then, $p \in \mathbf{P}_n$ and $$\left|\frac{1}{\hbar(\xi)}-1\right|<1\iff Re(\hbar(\xi))>1/2;$$ therefore, $\hbar \in \mathbf{P}_n(1/2)$ . Since $\digamma(\xi)\hbar(\xi) = \xi p(\xi)$ , then according to Lemma 2, we have $$\left| \frac{\xi F'(\xi)}{F(\xi)} - 1 \right| \le \frac{3n\ell^n + n\ell^{2n}}{1 - \ell^{2n}} \le 1 - \frac{1}{(e - 1)^2}.$$ This implies that $$\ell \leq \left(\frac{2e(e-2)}{3n(e-1)^2 + \sqrt{(9n^2 + 4n + 4)(e-1)^4 - 4(n+2)(e-1)^2 + 4}}\right)^{1/n} = R_{\mathbf{S}_{BS,n}^*}(\mathbf{F}_3).$$ Thus, $F \in \mathbf{S}_{BS,n}$ for $\ell \leq R_{\mathbf{S}_{RS,n}}(\mathbf{F}_3)$ . For sharpness, consider the following: $$F_6(\xi) = \frac{\xi(1+\xi^n)^2}{1-\xi^n} \text{ and } g_2(\xi) = \frac{\xi(1+\xi^n)}{1-\xi^n}.$$ We see that $$Re\left(\frac{g_2(\xi)}{\digamma_6(\xi)}\right) = Re\left(\frac{1}{1+\xi^n}\right) > \frac{1}{2}$$ and $$Re\left(\frac{F_6(\xi)}{\xi}\right) = Re\left(\frac{1+\xi^n}{1-\xi^n}\right) > 0.$$ Therefore, $F_6 \in \mathbf{F}_3$ . A computation shows that at $\xi = R_{\mathbf{S}_{BS,n}^*}(\mathbf{F}_3)e^{i\pi/n}$ , which comes out to $$\frac{\xi F_6'(\xi)}{F_6(\xi)} - 1 = \frac{3n\xi^n + n\xi^{2n}}{1 - \xi^{2n}} = 1 - \frac{1}{(e-1)^2}.$$ Hence, the sharpness is satisfied. $\Box$

Lemma 6. ([5]). Let $p \in \mathbf{P}$ and be of the form (6). Then for v, a complex number $$|p_2 - vp_1^2| \le 2 \max(1, |2v - 1|).$$

Lemma 7. ([29,30]). Let $p \in \mathbf{P}$ and be of the form (6) such that $|\rho| \le 1$ , and $|\eta| \le 1$ . Then, $$2p_2 = p_1^2 + \rho(4 - p_1^2),$$ $$4p_3 = p_1^3 + 2(4 - p_1^2)p_1\rho - (4 - p_1^2)p_1^2\rho + 2(4 - p_1^2)(1 - |\rho|^2)\eta,$$ (10) $$4p_3 = p_1^3 + 2(4 - p_1^2)p_1\rho - (4 - p_1^2)p_1^2\rho + 2(4 - p_1^2)(1 - |\rho|^2)\eta, \tag{11}$$

Lemma 8. ([31]). Let $\omega \in \mathbf{B}$ be given by $\omega(z) = \sum_{n=0}^{\infty} c_n \xi^n$ , and thus $$\psi(u,v) = \left| c_3 + \mu_1 c_1 c_2 + \mu_2 c_1^3 \right|.$$ Then, $\psi(u,v) \leq |\nu|$ if $(u,v) \in D_6$ , where $$D_6 = \left\{ (u, v) : 2 \le |\mu| \le 4, \quad v \ge \frac{1}{12} (\mu^2 + 8) \right\}.$$

Lemma 9. ([32]). Let $\overline{\mathbf{E}} := \{ \rho \in \mathbb{C} : |\rho| \le 1 \}$ , and, for j, k, and $l \in \mathbb{R}$ , let $$Y(j,k,l) := \max \left\{ |j + k\rho + l\rho^2| + 1 - |\rho|^2 : \rho \in \overline{\mathbf{E}} \right\}. \tag{12}$$ If $jl \geq 0$ , then $$Y(j,k,l) = \begin{cases} |j| + |k| + |l|, & |k| \ge 2(1-|l|), \\ 1 + |j| + \frac{k^2}{4(1-|l|)}, & |k| < 2(1-|l|). \end{cases}$$ If jl < 0, then $$Y(j,k,l) = \begin{cases} 1 - |j| + \frac{k^2}{4(1-|l|)}, & -4jl(l^{-2}-1) \le k^2 \wedge |k| < 2(1-|l|), \\ 1 + |j| + \frac{k^2}{4(1+|l|)}, & k^2 < \min\{4(1+|l|)^2, -4jl(l^{-2}-1)\}. \\ R(j,k,l), & otherwise. \end{cases}$$ In such as case, $$R(j,k,l) = \begin{cases} |j| + |k| - |l|, & |l|(|k| + 4|j|) \le |jk|, \\ 1 + |j| + \frac{k^2}{4(1 + |l|)}, & |jk| < |l|(|k| - 4|j|) \le |jk|. \\ |l| + |j| \sqrt{1 - \frac{k^2}{4jl}}, & otherwise. \end{cases}$$

Theorem 8. Let $F \in \mathbf{S}_{BS}^*$ and be of the form (2). Then, $$|d_2| \le 1$$ , $|d_3| \le \frac{17}{24}$ , $|d_4| \le \frac{29}{72}$ . These bounds are the best possible. Proof. If $F \in \mathbf{S}_{BS}^*$ , then $$\frac{\xi F'(\xi)}{F(\xi)} = \left(\frac{\omega(\xi)}{e^{\omega(\xi)} - 1}\right)^2,$$ where $\omega \in \mathbf{B}$ . The class B consists of Schwarz functions $\omega$ that are analytic in E, with $\omega(0) = 0$ , and $|\omega(\xi)| \le |\xi|$ . Let p be of the form (6). Then, $$\omega(\xi) = \frac{p(\xi) - 1}{p(\xi) + 1}.$$ Now by using (2), we can write out the following: $$\frac{\xi F'(\xi)}{F(\xi)} = 1 + d_2 \xi + (2d_3 - d_2^2) \xi^2 + (3d_4 - 3d_2 d_3 + d_2^3) \xi^3 + (4d_5 - 2d_3^2 - 4d_2 d_4 + 4d_2^2 d_3 - d_2^4) \xi^4 + \cdots$$ (13) In addition, $$\left(\frac{\mu(\xi)}{e^{\mu(\xi)}-1}\right)^{2} = 1 - \frac{1}{2}p_{1}\xi + \left(\frac{29}{96}p_{1}^{2} - \frac{1}{4}p_{2}\right)\xi^{2} + \left(\frac{-109}{576}p_{1}^{3} + \frac{13}{36}p_{1}p_{2} - \frac{1}{6}p_{3}\right)\xi^{3} + \left(\frac{11011}{92160}p_{1}^{4} - \frac{215}{576}p_{2}p_{1}^{2} + \frac{25}{96}p_{1}p_{3} + \frac{23}{192}p_{2}^{2} - \frac{1}{8}p_{4}\right)\xi^{4} + \cdots$$ (14) From (13) and (14), we obtain $$d_2 = \frac{1}{2}p_1, (15)$$ $$d_3 = \frac{29}{96}p_1^2 - \frac{1}{4}p_2, \tag{16}$$ $$d_4 = \frac{-109}{576}p_1^3 + \frac{13}{36}p_1p_2 - \frac{1}{6}p_3. \tag{17}$$ From (15), we have $|d_2| = \frac{1}{2}|p_1| \le 1$ . From (16), we can write out the following: $$|d_3| = \frac{1}{4} \left| p_2 - \frac{29}{24} p_1^2 \right|.$$ An application of Lemma 6 for $v = \frac{29}{24}$ gives the required bound. The function $\omega \in \mathbf{B}$ can be written as a power series: $$\omega(z) = \sum_{n=1}^{\infty} c_n z^n, \quad z \in \mathbf{E}.$$ (18) Since $p \in \mathbf{P}$ , therefore, $$p(z) = \frac{1 + \omega(z)}{1 - \omega(z)}.$$ By comparing the coefficients at powers of z in $$[1 - \omega(z)]p(z) = 1 + \omega(z),$$ we obtain $$p_1 = 2c_1$$ , $p_2 = 2c_2 + 2c_1^2$ , $p_3 = 2c_3 + 4c_1c_2 + 2c_1^3$ . By putting these values in (17), we obtain $$d_4 = -\frac{1}{3} \left( c_3 + \mu_1 c_1 c_2 + \mu_2 c_1^3 \right),$$ where $\mu_1 = -\frac{7}{3}$ , and $\mu_2 = \frac{29}{24}$ . Now, by using Lemma 8, we have $2 \le |\mu_1| \le 4$ , and $\mu_1 - \frac{1}{12}(\mu_2 + 8) = \frac{19}{216}$ ; therefore, $$|d_4| \le \frac{1}{3}|\mu_2| = \frac{29}{72}.$$ The equalities in each coefficient $|d_2|$ , $|d_3|$ , and $|d_4|$ are respectively obtained by taking the following: $$F_1(\xi) = \xi \exp\left(\int_0^{\xi} \frac{(\frac{t}{e^{\ell}-1})^2 - 1}{t} dt\right) = \xi - \xi^2 + \frac{17}{24}\xi^3 - \frac{29}{72}\xi^4 + \cdots$$ (19)

Theorem 9. Let $F \in \mathbf{S}_{BS}^*$ and have the series representation given in (2). Then, $$|d_3 - d_2^2| \le \frac{1}{2}. (20)$$

Theorem 10. Let $F \in \mathbf{S}_{BS}^*$ and have the series representation given in (2). Then, $$H_{2,2}(F) = |d_2 d_4 - d_3^2| \le \frac{521}{576}.$$ (21) The equality is obtained by the $F_1$ given in (19). Proof. Using (15)–(17), we obtain $$H_{2,2}(F) = d_2 d_4 - d_3^2 = -\frac{571}{3072} p_1^4 + \frac{191}{576} p_1^2 p_2 - \frac{1}{12} p_1 p_3 - \frac{1}{16} p_2^2.$$ (22) As we can see that the functional $H_{2,2}(F)$ and the class $\mathbf{S}_{BS}^*$ are rotationally invariant, we may therefore take $p := p_1$ such that $p \in [0,2]$ . Then, by using Lemma 7, and after some computations, we may write out the following: $$H_{2,2}(F) = -\frac{571}{9216}p^4 + \frac{107}{1152}p^2(4-p^2)\rho - \frac{1}{192}(4-p^2)\left(12-7p^2\right)\rho^2 - \frac{1}{24}p(4-p^2)\left(1-|\rho|^2\right)\eta,$$ where $\rho$ and $\eta$ satisfy the relation $|\rho| \leq 1$ and $|\eta| \leq 1$ . Firstly, we consider the case when p=0. Then, $|H_{2,2}(F)|=\left|-\frac{1}{4}\rho^2\right|\leq \frac{1}{4}$ . Next, we assume that p=2; then, $|H_{2,2}(F)|=\frac{521}{576}$ . Now suppose that $p\in(0,2)$ ; then, $$|H_{2,2}(F)| \leq \frac{1}{24}p(4-p^2)\Phi(j,k,l),$$ where $$\Phi(j,k,l) = \left| j + k\rho + l\rho^2 \right| + 1 - |\rho|^2, \qquad \rho \in \overline{\mathbb{E}},$$ with $$j = \frac{521p^3}{384(4-p^2)}$$ , $k = -\frac{107p}{48}$ , and $l = \frac{\left(12-7p^2\right)}{8p}$ ; then clearly, $$jl = \frac{521p^2(12 - 7p^2)}{3072(4 - p^2)} \ge 0, \quad \text{for } p \in \left[\sqrt{\frac{12}{7}}, 2\right).$$ (23) In addition, $$|k| - 2(1 - |l|) = \frac{23p^2 - 96p + 144}{48p} > 0 \quad p \in \left[\sqrt{\frac{12}{7}}, 2\right]$$ so that |k| > 2(1 - |l|), and by applying Lemma 9, we can obtain $$|H_{2,2}(F)| \le \frac{1}{24}p(4-p^2)(|j|+|k|+|l|) = g(p),$$ where $$g(p) = \frac{1}{9216}p^4 + \frac{24}{288}p^2 + \frac{1}{4}. (24)$$ Clearly, g'(p) > 0, and so $$\max g(p) = g(2) = \frac{521}{576}$$ We also see from (23) that $$jl = \frac{521p^2(12 - 7p^2)}{3072(4 - p^2)} < 0, \quad \text{for } p \in \left(0, \sqrt{\frac{12}{7}}\right).$$ Thus, $$k^2 - 4jl\left(l^{-2} - 1\right) = \frac{1}{576} \frac{p^2(889p^2 - 20280)}{7p^2 - 12} < 0, \quad p \in \left(0, \sqrt{\frac{12}{7}}\right).$$ This shows that $-4jl(l^{-2}-1) \le k^2 \wedge |k| < 2(1-|l|)$ . In addition, $$\begin{split} \Phi(p) &= 4(1+|l|)^2 + 4jl \Big(l^{-2} - 1\Big) \\ &= \frac{(7p-6) \left(1295p^5 - 4266p^4 + 2688p^3 + 2073p^2 + 2304p - 13824\right)}{p^2 (12-7p^2)} \end{split}$$ We see that $\Phi(p)>0$ for $p\in(0,0.76032)\cup\left(\frac{6}{7},\sqrt{\frac{12}{7}}\right)$ , and $\Phi(p)<0$ for $p\in\left(0.76032,\frac{6}{7}\right)$ . Hence, we conclude that $$\min\left\{4(1+|l|)^2, -4jl\left(l^{-2}-1\right)\right\} = \begin{cases} -4jl\left(l^{-2}-1\right), & p \in (0,0.76032) \cup \left(\frac{6}{7}, \sqrt{\frac{12}{7}}\right), \\ 4(1+|l|)^2 & (0.76032, \frac{6}{7}). \end{cases}$$ As a result, $$k^2 - 4(1+|l|)^2 = \frac{\left(23p^2 - 96p + 144\right)\left(191p^2 + 96p - 144\right)}{2304} > 0 \text{ for } \left(0.76032, \frac{6}{7}\right).$$ In addition, $$k^2 + 4(1 + |l|)^2 = \frac{p^2(78365p^2 - 96828)}{1152(7p^2 - 12)} < 0 \text{ for } \left(\sqrt{\frac{96828}{78365}}, \sqrt{\frac{12}{7}}\right).$$ This shows that $k^2 < \min\{4(1+|l|)^2, -4jl(l^{-2}-1)\}$ hold for $p \in \left(\sqrt{\frac{96828}{78365}}, \sqrt{\frac{12}{7}}\right)$ . By applying Lemma 9, we arrive at the following: $$|H_{2,2}(F)| \le \frac{1}{24}p(4-p^2)\left(1+|j|+\frac{k^2}{4(1+|l|)}\right) = g_1(p),$$ where $$g_1(p) = \frac{p(127p^4 - 1364p^3 - 1728p^2 - 8064p + 6912)}{6912(6 - 7p)}.$$ This attains its maxima at $p = \sqrt{\frac{12}{7}}$ . Hence, $$|H_{2,2}(F)| \le \frac{\sqrt{21}\left(-4413 + 3034\sqrt{21}\right)}{-148176 + 49392\sqrt{21}} < \frac{521}{576}$$ We are left with the case $p \in \left(0, \sqrt{\frac{96828}{78365}}\right)$ . We also see that $$|l|(|k|+4|j|)-|jk| = \frac{4171p^4 - 55392p^2 + 246528}{18432(p^2-4)} < 0 \qquad p \in \left(0, \sqrt{\frac{96828}{78365}}\right).$$ We conclude that |l|(|k|+4|j|) < |jk|. By applying Lemma 9, we arrive at the following: $$|H_{2,2}(F)| \le \frac{1}{24}p(4-p^2)(|j|+|k|-|l|) = g(p),$$ where g is given in (24), this giving us the required result. The function given in (19) belongs to the $\mathbf{S}_{BS}^*$ , as $d_2 = -1$ , $d_4 = -29/72$ , and $d_3 = 17/24$ , which yields the sharpness of (21). Hence, the proof is done. $\square$

Lemma 1. (see [42]). If $h \in \mathcal{P}$ , then $|c_k| \leq 2$ for each k, where $\mathcal{P}$ is the family of all functions h, analytic in $\mathbb{U}$ , for which $$\Re\{h(z)\} > 0 \qquad (z \in \mathbb{U}),$$ where $$h(z) = 1 + c_1 z + c_2 z^2 + \cdots$$ $(z \in \mathbb{U}).$ We begin by estimating the coefficients $|a_2|$ and $|a_3|$ for functions in the class $\mathcal{B}^{\sigma,m;\wp}_{i,\delta,\Sigma}(\lambda,\Xi)$ . Let P(z) be defined by $$P(z): = \frac{1+\omega(z)}{1-\omega(z)} = 1 + c_1 z + c_2 z^2 + \cdots$$ It is evident that $$\omega(z) = \frac{P(z) - 1}{P(z) + 1}$$ $$= \frac{1}{2} \left[ c_1 z + \left( c_2 - \frac{c_1^2}{2} \right) z^2 + \left( c_3 - c_1 c_2 + \frac{c_1^3}{4} \right) z^3 + \cdots \right]. \quad (17)$$ Since $\omega(z)$ is a Schwarz function, it follows that $\Re(p_1(z)) > 0$ and $p_1(0) = 1$ . Therefore, $$\Psi(\omega(z)) = e^{\left(\frac{P(z)-1}{P(z)+1} + \kappa \frac{\left[\frac{P(z)-1}{P(z)+1}\right]^2}{2}\right)}$$ $$= 1 + \frac{c_1}{2}z + \left(\frac{c_2}{2} + \frac{(\wp - 1)c_1^2}{8}\right)z^2 + \left(\frac{c_3}{2} + (\wp - 1)\frac{c_1c_2}{4} + \frac{(1 - 3\wp)}{48}c_1^3\right)z^3 + \dots$$ (18) Define the functions p(z) and q(z) as follows: $$p(z) := \frac{1 + u(z)}{1 - u(z)} = 1 + p_1 z + p_2 z^2 + \cdots$$ and $$q(z) := \frac{1 + v(z)}{1 - v(z)} = 1 + q_1 z + q_2 z^2 + \cdots$$ or, equivalently, $$u(z) := \frac{p(z) - 1}{p(z) + 1} = \frac{1}{2} \left[ p_1 z + \left( p_2 - \frac{p_1^2}{2} \right) z^2 + \cdots \right]$$ and $$v(z) := \frac{q(z) - 1}{q(z) + 1} = \frac{1}{2} \left[ q_1 z + \left( q_2 - \frac{q_1^2}{2} \right) z^2 + \cdots \right].$$ Subsequently, p(z) and q(z) are analytic in $\mathbb{U}$ with p(0) = 1 = q(0). Furthermore, since both u and v map from $\mathbb{U}$ to $\mathbb{U}$ , the functions p(z) and q(z) exhibit a positive real part in $\mathbb{U}$ , and they satisfy the inequalities: $$|p_i| \le 2 \quad \text{and} \quad |q_i| \le 2. \tag{19}$$ For the scope of our study, we introduce the notation: $$\mathcal{V}_2 = \mathcal{V}_{\vartheta,m}^{a,c}(2) = \left[1 + c^j(\delta)\right]^m \left(\frac{\Gamma(\sigma+2)}{\Gamma(\sigma+1)} \cdot \frac{(\sigma-\lambda)!}{(\sigma+1-\lambda)!}\right),\tag{20}$$ $$\mathcal{V}_{3} = \left[1 + 2c^{j}(\delta)\right]^{m} \left(\frac{\Gamma(\sigma+3)}{\Gamma(\sigma+1)} \cdot \frac{(\sigma-\lambda)!}{(\sigma+2-\lambda)!}\right). \tag{21}$$ In the subsequent theorem, we embark on the initial exploration of the Taylor-Maclaurin coefficients $|a_2|$ and $|a_3|$ for functions belonging to this novel subclass $\mathcal{B}^{\sigma,m;\wp}_{j,\delta,\Sigma}(\lambda,\Xi)$ .

Theorem 1. Let assume that the f function is as in (1) and in the class $\mathcal{B}_{i,\delta,\Sigma}^{\sigma,m;\wp}(\lambda,\Xi)$ . Then $$|a_2| \le \min \begin{cases} \frac{1}{(\lambda+1)\mathcal{V}_2'} \\ \sqrt{\frac{2}{|\{(\lambda-1)(\lambda+2) - (\wp-1)(\lambda+1)^2\}\mathcal{V}_2^2 + 2(\lambda+2)\mathcal{V}_3|}} \end{cases}$$ (22) and $$|a_3| \le \min \left\{ \begin{array}{l} \frac{1}{(\lambda+1)^2 \mathcal{V}_2^2} + \frac{1}{(2+\lambda)\mathcal{V}_3}, \\ \frac{2}{\{(\lambda-1)(\lambda+2) - (\wp-1)(\lambda+1)^2\} \mathcal{V}_2^2 + 2(\lambda+2)\mathcal{V}_3} + \frac{1}{(2+\lambda)\mathcal{V}_3}. \end{array} \right.$$ (23) Proof. It follows from (9) and (10) that $$\left(\frac{z^{1-\lambda}(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}f(z))'}{[\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}f(z)]^{1-\lambda}}\right) = \Xi(u(z))$$ (24) and $$\left(\frac{w^{1-\lambda}(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}g(w))'}{[\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}g(w)]^{1-\lambda}}\right) = \Xi(v(w)),$$ (25) where p(z) and q(w) in P and have the following forms: $$\Xi(u(z)) = 1 + \frac{1}{2}p_1z + \left(\frac{p_2}{2} + \frac{(\wp - 1)p_1^2}{8}\right)z^2 + \cdots$$ (26) and $$\Xi(v(w)) = 1 + \frac{1}{2}q_1w + \left(\frac{q_2}{2} + \frac{(\wp - 1)q_1^2}{8}\right)w^2 + \cdots, \tag{27}$$ respectively. Now, by equating the coefficients in (24) and (25), we have $$(1+\lambda)\mathcal{V}_2 a_2 = \frac{1}{2}p_1,\tag{28}$$ $$\left[\frac{(\lambda - 1)(\lambda + 2)}{2}\mathcal{V}_2^2 a_2^2 + (\lambda + 2)\mathcal{V}_3 a_3\right] = \frac{p_2}{2} + \frac{(\wp - 1)p_1^2}{8},\tag{29}$$ $$-(\lambda + 1)\mathcal{V}_2 a_2 = \frac{1}{2}q_1 \tag{30}$$ and $$\[ \left( 2(\lambda+2)\mathcal{V}_3 + \frac{(\lambda-1)(\lambda+2)}{2}\mathcal{V}_2^2 \right) a_2^2 - (\lambda+2)\mathcal{V}_3 a_3 \] = \frac{q_2}{2} + \frac{(\wp-1)q_1^2}{8}. \tag{31}$$ From (28) and (30), we can determine that $$a_2 = \frac{p_1}{2(1+\lambda)\mathcal{V}_2} = \frac{-q_1}{2(1+\lambda)\mathcal{V}_2},$$ (32) which implies $$p_1 = -q_1 (33)$$ and $$8(\lambda+1)^2 \mathcal{V}_2^2 a_2^2 = p_1^2 + q_1^2. \tag{34}$$ Thus we have $$a_2^2 = \frac{p_1^2 + q_1^2}{8(\lambda + 1)^2 \mathcal{V}_2^2} \tag{35}$$ and $$(\lambda+1)^2 \mathcal{V}_2^2 a_2^2 = \frac{p_1^2 + q_1^2}{8}. (36)$$ By adding (29) and (31), and utilizing (32) as well as (33), we get $$\left[ (\lambda - 1)(\lambda + 2)\mathcal{V}_2^2 + 2(\lambda + 2)\mathcal{V}_3 \right] a_2^2 = \frac{p_2 + q_2}{2} + \frac{(\wp - 1)}{8} (p_1^2 + q_1^2). \tag{37}$$ Thus, by using (36) $$a_2^2 = \frac{p_2 + q_2}{2\left[\left\{(\lambda - 1)(\lambda + 2) - (\wp - 1)(\lambda + 1)^2\right\}\mathcal{V}_2^2 + 2(\lambda + 2)\mathcal{V}_3\right]}$$ (38) Applying Lemma 1 to the coefficients p² and q2, yields the immediate result $$|a_2|^2 \le \frac{2}{|\{(\lambda - 1)(\lambda + 2) - (\wp - 1)(\lambda + 1)^2\}\mathcal{V}_2^2 + 2(\lambda + 2)\mathcal{V}_3|}.$$ (39) Hence, $$|a_2| \le \sqrt{\frac{2}{|\{(\lambda-1)(\lambda+2) - (\wp-1)(\lambda+1)^2\}\mathcal{V}_2^2 + 2(\lambda+2)\mathcal{V}_3|}}$$ This yields the bound on $|a_2|$ as stated in (22). To establish the bound on $|a_3|$ , we subtract (31) from (29), resulting in $$\left[2(\lambda+2)\mathcal{V}_3 a_3 - 2(\lambda+2)\mathcal{V}_3 a_2^2\right] = \frac{1}{2}(p_2 - q_2) + \frac{\wp - 1}{8}(p_1^2 - q_1^2). \tag{40}$$ Using (32), (33) and (40) we can deduce that $$a_3 = a_2^2 + \frac{p_2 - q_2}{4(2+\lambda)\mathcal{V}_3}$$ $$= \frac{p_1^2 + q_1^2}{8(\lambda+1)^2\mathcal{V}_2^2} + \frac{p_2 - q_2}{4(2+\lambda)\mathcal{V}_3}.$$ (41) Applying Lemma 1 once more to for the coefficients $p_2$ , $q_2$ , we immediately obtain $$|a_3| \le \frac{1}{(\lambda+1)^2 \mathcal{V}_2^2} + \frac{1}{(2+\lambda)\mathcal{V}_3}$$ also, $$|a_3| \leq \frac{2}{|\{(\lambda - 1)(\lambda + 2) - (\wp - 1)(\lambda + 1)^2\}\mathcal{V}_2^2 + 2(\lambda + 2)\mathcal{V}_3|} + \frac{1}{(2 + \lambda)\mathcal{V}_3}.$$ This completes the proof of Theorem 1. $\Box$ As a consequence of our results, by appropriately setting the parameter, we present the following corollaries, which are novel and have not been studied for the case of involution numbers involving hybrid-type convolution operators. When we fix $\lambda = 0$ in Theorem 1, the following corollary emerges.

Corollary 1. Let assume that the f function is as in (1) and in the class $S_{j,\delta,\Sigma}^{\sigma,m;\wp}(\Xi)$ . Then $$|a_2| \le \min \left\{ \begin{array}{l} \frac{1}{V_2}, \\ \sqrt{\frac{2}{|4V_3 - (\wp + 1)V_2^2|}} \end{array} \right.$$ (42) and $$|a_3| \le \min \begin{cases} \frac{1}{\mathcal{V}_2^2} + \frac{1}{2\mathcal{V}_3}, \\ \frac{2}{|-(1+\wp)\mathcal{V}_2^2 + 4\mathcal{V}_3|} + \frac{1}{2\mathcal{V}_3}. \end{cases}$$ (43) Fixing $\lambda = 1$ in Theorem 1, we have the following corollary.

Corollary 2. Let assume that the f function is as in (1) and in the class $\mathcal{R}_{j,\delta,\Sigma}^{\sigma,m;\wp}(\Xi)$ . Then $$|a_2| \le \min \begin{cases} \frac{1}{2V_2}, \\ \sqrt{\frac{1}{3V_3 - 2(\wp - 1)V_2^2}} \end{cases}$$ (44) and $$|a_3| \le \min \left\{ \begin{array}{l} \frac{1}{4V_2^2} + \frac{1}{3V_3} \\ \frac{2}{|-4(\wp-1)V_2^2 + 6V_3|} + \frac{1}{3V_3}. \end{array} \right. \tag{45}$$ In the subsequent theorem, we are embarking on the initial exploration of the Taylor-Maclaurin coefficients $|a_2|$ and $|a_3|$ for functions within this novel subclass $\mathcal{M}_{j,\delta,\Sigma}^{\sigma,m;\wp}(\tau,\Xi)$ .

Theorem 2. Let assume that the f function is as in (1) and $f \in \mathcal{M}_{j,\delta,\Sigma}^{\sigma,m;\wp}(\tau,\Xi)$ , $\tau \geq 1$ . Then $$|a_2| \le \min \left\{ \begin{array}{l} \frac{1}{2(2\tau - 1)\mathcal{V}_2}, \\ \frac{1}{\sqrt{|(1+\tau) - 2(2\tau - 1)^2(\wp - 1)|\mathcal{V}_2^2}} \end{array} \right. \tag{46}$$ and $$|a_3| \le \min \left\{ \begin{array}{l} \frac{2}{3(3\tau - 1)\mathcal{V}_3} + \frac{1}{4(2\tau - 1)^2 \mathcal{V}_2^2}, \\ \frac{2}{3(3\tau - 1)\mathcal{V}_3} + \frac{1}{|(1+\tau) - 2(2\tau - 1)^2 (\wp - 1)|\mathcal{V}_2^2}. \end{array} \right.$$ (47) Proof. It follows from (15) and (16) that $$\tau \left( 1 + \frac{(z \mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1} f(z))''}{(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1} f(z))'} \right) + (1 - \tau) \frac{1}{(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1} f(z))'} = \Xi(u(z))$$ (48) and $$\tau \left( 1 + \frac{w(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}g(w))''}{(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}g(w))'} \right) + (1-\tau) \frac{1}{(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}g(w))'} = \Xi(v(w)). \tag{49}$$ From (48) and (49), we have $$1 + 2(2\tau - 1)\mathcal{V}_2 a_2 z + \left[3(3\tau - 1)\mathcal{V}_3 a_3 + 4(1 - 2\tau)\mathcal{V}_2^2 a_2^2\right] z^2 + \cdots$$ $$= 1 + \frac{1}{2}p_1 z + \left(\frac{p_2}{2} + \frac{(\wp - 1)p_1^2}{8}\right) z^2 + \cdots,$$ and $$1 - 2(2\tau - 1)\mathcal{V}_2 a_2 w + \left(2(5\tau - 1)\mathcal{V}_2^2 a_2^2 - 3(3\tau - 1)\mathcal{V}_3 a_3\right) w^2 - \cdots$$ $$= 1 + \frac{1}{2}q_1 w + \left(\frac{q_2}{2} + \frac{(\wp - 1)q_1^2}{8}\right) w^2 + \cdots$$ By equating the coefficients, we obtain $$2(2\tau - 1)\mathcal{V}_2 a_2 = \frac{1}{2}p_1,\tag{50}$$ $$3(3\tau - 1)\mathcal{V}_3 a_3 + 4(1 - 2\tau)\mathcal{V}_2^2 a_2^2 = \frac{p_2}{2} + \frac{(\wp - 1)p_1^2}{8},\tag{51}$$ $$-2(2\tau - 1)\mathcal{V}_2 a_2 = \frac{1}{2}q_1,\tag{52}$$ and $$2(5\tau - 1)\mathcal{V}_2^2 a_2^2 - 3(3\tau - 1)\mathcal{V}_3 a_3 = \frac{q_2}{2} + \frac{(\wp - 1)q_1^2}{8}.$$ (53) Using (50) and (52), we obtain $$p_1 = -q_1 \tag{54}$$ From (50) by using (19), $$|a_2| \leq \frac{1}{2(2\tau - 1)\mathcal{V}_2}. (55)$$ Also $$32(2\tau - 1)^{2}\mathcal{V}_{2}^{2}a_{2}^{2} = p_{1}^{2} + q_{1}^{2}$$ $$a_{2}^{2} = \frac{p_{1}^{2} + q_{1}^{2}}{32(2\tau - 1)^{2}\mathcal{V}_{2}^{2}}.$$ (56) Thus by (19), we get $$|a_2| \le \frac{1}{4(2\tau - 1)\mathcal{V}_2} = \frac{1}{4(2\tau - 1)\mathcal{V}_2}.$$ (57) Now from (51), (53) and using (56), we obtain $$\left(2(1+\tau)-4(2\tau-1)^2(\wp-1)\right)\mathcal{V}_2^2 a_2^2 = \frac{p_2+q_2}{2}.$$ (58) Thus, by (58) we obtain $$a_2^2 = \frac{p_2 + q_2}{4[(1+\tau) - 2(2\tau - 1)^2(\wp - 1)]\mathcal{V}_2^2}$$ $$|a_2| \le \frac{1}{\sqrt{|(1+\tau) - 2(2\tau - 1)^2(\wp - 1)|\mathcal{V}_2^2}}.$$ By using (51) and (53), and then substituting (54), we get $$a_3 = \frac{p_2 - q_2}{6(3\tau - 1)\mathcal{V}_3} + a_2^2. \tag{59}$$ Taking the modulus of both sides, we obtain $$|a_3| \le \frac{2}{3(3\tau - 1)\mathcal{V}_2} + |a_2^2|.$$ (60) Using (55) and (57), we get $$|a_3| \le \frac{2}{3(3\tau - 1)\mathcal{V}_3} + \frac{1}{4(2\tau - 1)^2\mathcal{V}_2^2}.$$ Now by using (58) in (60), $$|a_3| \le \frac{2}{3(3\tau - 1)\mathcal{V}_3} + |a_2^2|$$ $$= \frac{2}{3(3\tau - 1)\mathcal{V}_3} + \frac{1}{|(1+\tau) - 2(2\tau - 1)^2(\wp - 1)|\mathcal{V}_2^2}$$

Corollary 3. Let assume that the f function is as in (1) and $f \in \mathcal{K}^{\sigma,m;\wp}_{j,\delta,\Sigma}(\Xi)$ . Then $$|a_2| \le \min \left\{ \begin{array}{l} \frac{1}{2V_2}, \\ \frac{1}{\sqrt{|2 - 2(\wp - 1)|V_2^2}} \end{array} \right.$$ (61) and $$|a_3| \le \min \begin{cases} \frac{2}{6\mathcal{V}_3} + \frac{1}{4\mathcal{V}_2^2}, \\ \frac{2}{6\mathcal{V}_3} + \frac{1}{|2 - 2(\wp - 1)|\mathcal{V}_2^2}. \end{cases}$$ (62)

Lemma 2. ([44]). Let $k \in \mathbb{R}$ and $z_1, z_2 \in \mathbb{C}$ . If $|z_1| < R$ and $|z_2| < R$ then $$|(k+1)z_1 + (k-1)z_2| \le \begin{cases} 2|k|R, & |k| \ge 1\\ 2R & |k| \le 1. \end{cases}$$ (63)

Lemma 3. ([44]). Let $k, l \in \mathbb{R}$ and $z_1, z_2 \in \mathbb{C}$ . If $|z_1| < R$ and $|z_2| < R$ then $$|(k+l)z_1 + (k-l)z_2| \le \begin{cases} 2|k|R, & |k| \ge |l| \\ 2|l|R, & |k| \le |l|. \end{cases}$$ (64) Now, we obtain Fekete-Szegö inequalities for $f\in\mathcal{B}^{\sigma,m;\wp}_{i.\delta.\Sigma}(\lambda,\Xi)$ :

Theorem 3. For $\aleph \in \mathbb{R}$ , let assume that the f function is as in (1) and $f \in \mathcal{B}_{i,\delta,\Sigma}^{\sigma,m;\wp}(\lambda,\Xi)$ , then $$\left|a_3 - \aleph a_2^2\right| \leq \left\{\begin{array}{ll} \frac{1}{(2+\lambda)\mathcal{V}_3} & ; & 0 \leq |h(\aleph)| \leq \frac{1}{4(2+\lambda)\mathcal{V}_3} \\ 4|h(\aleph)| & ; & |h(\aleph)| \geq \frac{1}{4(2+\lambda)\mathcal{V}_3} \end{array}\right.$$ where $$h(\aleph) = \frac{1 - \aleph}{2[\{(\lambda - 1)(\lambda + 2) - (\wp - 1)(\lambda + 1)^2\}\mathcal{V}_2^2 + 2(\lambda + 2)\mathcal{V}_3]}.$$ (65) Proof. From (41), we have $$a_3 - \aleph a_2^2 = \frac{p_2 - q_2}{4(2+\lambda)\mathcal{V}_3} + (1-\nu)a_2^2. \tag{66}$$ By substituting (38) in (66), we have $$a_{3} - \aleph a_{2}^{2} = \frac{p_{2} - q_{2}}{4(2+\lambda)\mathcal{V}_{3}} + \frac{(1-\aleph)(p_{2} + q_{2})}{2\left[\left\{(\lambda - 1)(\lambda + 2) - (\wp - 1)(\lambda + 1)^{2}\right\}\mathcal{V}_{2}^{2} + 2(\lambda + 2)\mathcal{V}_{3}\right]} = \left(h(\aleph) + \frac{1}{4(2+\lambda)\mathcal{V}_{3}}\right)p_{2} + \left(h(\nu) - \frac{1}{4(2+\lambda)\mathcal{V}_{3}}\right)q_{2},$$ (67) where $$h(\aleph) = \frac{1 - \aleph}{2[\{(\lambda - 1)(\lambda + 2) - (\wp - 1)(\lambda + 1)^2\}\mathcal{V}_2^2 + 2(\lambda + 2)\mathcal{V}_3]}.$$ Thus by taking modulus of (67), we conclude that $$\left| a_3 - \aleph a_2^2 \right| \le \begin{cases} \frac{1}{(2+\lambda)\mathcal{V}_3} & ; \qquad 0 \le |h(\aleph)| \le \frac{1}{4(2+\lambda)\mathcal{V}_3} \\ 4|h(\aleph)| & ; \qquad |h(\aleph)| \ge \frac{1}{4(2+\lambda)\mathcal{V}_3} \end{cases}$$ (68) where $h(\aleph)$ is given by (65). $\square$ By taking $\aleph = 1$ in above Theorem one can easily state the following: Remark 1. Let the function f be assumed by (1) and $f \in \mathcal{B}^{\sigma,m;\wp}_{j,\delta,\Sigma}(\lambda,\Xi)$ . Then $$\left|a_3 - a_2^2\right| \le \frac{1}{(2+\lambda)\mathcal{V}_3}.$$ By taking $\lambda = 0$ and $\lambda = 1$ , we can state the following:

Corollary 4. For $\aleph \in \mathbb{R}$ , let assume that the f function is as in (1) and $f \in \mathcal{S}^{\sigma,m;\wp}_{j,\delta,\Sigma}(\Xi)$ , then $$\left|a_3 - \aleph a_2^2\right| \le \left\{\begin{array}{ll} \frac{1}{2\mathcal{V}_3} & ; & 0 \le |h(\aleph)| \le \frac{1}{8\mathcal{V}_3} \\ 4|h(\aleph)| & ; & |h(\aleph)| \ge \frac{1}{8\mathcal{V}_3} \end{array}\right.$$ where $h(\aleph) = \frac{1-\aleph}{2\left[4\mathcal{V}_3 - (\wp+1)\mathcal{V}_2^2\right]}$ .

Corollary 5. For $\aleph \in \mathbb{R}$ , let assume that the f function is as in (1) and $f \in \mathcal{R}_{j,\delta,\Sigma}^{\sigma,m;\wp}(\Xi)$ , then $$\left|a_3 - \aleph a_2^2\right| \le \begin{cases} \frac{1}{3\mathcal{V}_3} &; \qquad 0 \le |h(\aleph)| \le \frac{1}{12\mathcal{V}_3} \\ 4|h(\aleph)| &; \qquad |h(\aleph)| \ge \frac{1}{12\mathcal{V}_3} \end{cases}$$ where $h(\aleph) = \frac{1-\aleph}{2[6\mathcal{V}_3 - 4(\wp-1)\mathcal{V}_2^2]}$ . Now, we prove Fekete-Szegö inequalities for $f \in \mathcal{M}^{\sigma,m;\wp}_{j,\delta,\Sigma}(\tau,\Xi)$ .

Theorem 4. For $\nu \in \mathbb{R}$ , let assume that the f function is as in (1) and $f \in \mathcal{M}_{i,\delta,\Sigma}^{\sigma,m;\wp}(\tau,\Xi)$ , then $$\left| a_3 - \nu a_2^2 \right| \le \begin{cases} \frac{2}{3(3\tau - 1)\mathcal{V}_3} & ; 0 \le |h(\nu)| \le \frac{1}{6(3\tau - 1)\mathcal{V}_3} \\ 4|h(\nu)| & ; |h(\nu)| \ge \frac{1}{6(3\tau - 1)\mathcal{V}_3} \end{cases}$$ where $$h(\nu) = \frac{1 - \nu}{4[(1 + \tau) - 2(2\tau - 1)^2(\wp - 1)]\mathcal{V}_2^2}.$$ Proof. From (59), we have $$a_3 - \nu a_2^2 = \frac{p_2 - q_2}{6(3\tau - 1)\mathcal{V}_3} + +(1 - \nu)a_2^2. \tag{69}$$ By substituting (58) in (69), we have $$a_{3} - \nu a_{2}^{2} = \frac{p_{2} - q_{2}}{6(3\tau - 1)\mathcal{V}_{3}} + a_{2}^{2} + \frac{(p_{2} + q_{2})(1 - \nu)}{4[(1 + \tau) - 2(2\tau - 1)^{2}(\wp - 1)]\mathcal{V}_{2}^{2}}$$ $$= \left(h(\nu) + \frac{1}{6(3\tau - 1)\mathcal{V}_{3}}\right)p_{2} + \left(h(\nu) - \frac{1}{6(3\tau - 1)\mathcal{V}_{3}}\right)q_{2}, \tag{70}$$ where $$h(\nu) = \frac{1 - \nu}{4[(1 + \tau) - 2(2\tau - 1)^2(\wp - 1)]\mathcal{V}_2^2}.$$ (71) Thus by taking modulus of (70), we get $$\left| a_3 - \nu a_2^2 \right| \le \begin{cases} \frac{2}{3(3\tau - 1)\mathcal{V}_3} & ; 0 \le |h(\nu)| \le \frac{1}{6(3\tau - 1)\mathcal{V}_3} \\ 4|h(\nu)| & ; |h(\nu)| \ge \frac{1}{6(3\tau - 1)\mathcal{V}_3} \end{cases}$$ (72) where h(v) is given by (71). $\square$ By taking $\nu = 1$ in above theorem, we can easily state the following: Remark 2. Let assume that the f function is as in (1) and $f \in \mathcal{M}^{a,b,c}_{\Sigma}(\tau,\Xi)$ . Then $$\left|a_3 - a_2^2\right| \le \frac{2}{3(3\tau - 1)\mathcal{V}_3}.$$

Corollary 6. For $v \in \mathbb{R}$ , let assume that the f function is as in (1) and $f \in \mathcal{K}^{\sigma,m;\wp}_{j,\delta,\Sigma}(\Xi)$ , then $$\left| a_3 - \nu a_2^2 \right| \le \begin{cases} \frac{2}{6\mathcal{V}_3} & ; 0 \le |h(\nu)| \le \frac{1}{12\mathcal{V}_3} \\ 4|h(\nu)| & ; |h(\nu)| \ge \frac{1}{12\mathcal{V}_3} \end{cases}$$ where $$h(\nu) = \frac{1-\nu}{4[2-2(\wp-1)]\mathcal{V}_2^2}$$ .

Theorem 1. If the function s given by (3) belongs to the family $\mathfrak{P}_{\sigma_r}^{\tau}(\eta, \nu, \varphi)$ and $\delta \in \mathbb{R}$ , then $$\frac{|d_{\kappa+1}| \le \frac{|\eta|B_1\sqrt{2B_1}}{\sqrt{|\{M(1+\kappa)+[N\tau(\tau-1)+(1-(1+\kappa)\tau)2\nu\}(1+\kappa)^2\}\eta B_1^2-2L^2B_2|+2L^2B_1}},$$ (9) $$\begin{cases} \frac{B_{1}|\eta|}{M} & ; 0 < B_{1} < \frac{2L^{2}}{|\eta|M(1+\kappa)} \\ \frac{B_{1}|\eta|}{M} + \left(\frac{1+\kappa}{2} - \frac{L^{2}}{|\eta|B_{1}M}\right) \frac{2\eta^{2}B_{1}^{3}}{|\{M(1+\kappa) + [N\tau(\tau-1) + (1-(1+\kappa)\tau)2\nu](1+\kappa)^{2}\}\eta B_{1}^{2} - 2L^{2}B_{2}| + 2L^{2}B_{1}} \\ & ; B_{1} \ge \frac{2L^{2}}{|\eta|M(1+\kappa)}, \end{cases} (10)$$ and $$|d_{2\kappa+1} - \delta d_{\kappa+1}^{2}| \leq \begin{cases} \frac{B_{1}|\eta|}{M} & ; |1 + \kappa - 2\delta| < J\\ \frac{|\eta|^{2}B_{1}^{3}|\kappa - 2\delta + 1|}{|\{M(1+\kappa) + [N\tau(\tau - 1) + (1 - (1+\kappa)\tau)2\nu](1+\kappa)^{2}\}\eta B_{1}^{2} - 2L^{2}B_{2}|}; |1 + \kappa - 2\delta| \geq J, \end{cases}$$ $$(11)$$ where $$J = \left| \frac{\{M(1+\kappa) + [N\tau(\tau-1) + 2\nu(1 - (1+\kappa)\tau)](1+\kappa)^2\}\eta B_1^2 - 2L^2 B_2}{\eta M B_1^2} \right|.$$ (12) $$L = (1 + \kappa)(\tau(1 + \nu\kappa) - \nu), \tag{13}$$ $$M = (\tau(1 + 2\nu\kappa) - \nu)(1 + 2\kappa) \tag{14}$$ and $$N = 1 + \nu \kappa (2 + \kappa). \tag{15}$$ Proof. Let the function s of the form (3) belong to the family $\mathfrak{P}_{\sigma_{\kappa}}^{\tau}(\eta, \nu, \varphi)$ . Then, we have regular functions $\mathfrak{h}, \mathfrak{p} : \mathfrak{D} \longrightarrow \mathfrak{D}, \mathfrak{h}(0) = \mathfrak{p}(0) = 0$ satisfying $$\frac{1}{\eta} \left( \nu \frac{\{(\varsigma s'(\varsigma))'\}^{\tau}}{s'(\varsigma)} + (1 - \nu)(s'(\varsigma))^{\tau} - 1 \right) + 1 = \varphi(\mathfrak{h}(\varsigma)), \tag{16}$$ and $$\frac{1}{\eta} \left( \nu \frac{\{ (\varkappa g'(\varkappa))' \}^{\tau}}{g'(\varkappa)} + (1 - \nu)(g'(\varkappa))^{\tau} - 1 \right) + 1 = \varphi(\mathfrak{p}(\varkappa)). \tag{17}$$ Using (3) in (16) and (17) we obtain: $$\frac{1}{\eta} \left( \nu \frac{\left[ \left\{ (\varsigma s'(\varsigma))' \right\}^{\tau}}{s'(\varsigma)} + (1 - \nu)(s'(\varsigma))^{\tau} - 1 \right) + 1 =$$ $$\frac{1}{\eta} \left\{ L d_{\kappa+1} \varsigma^{\kappa} + \left[ M d_{2\kappa+1} + \left( 1 + \kappa \right)^{2} \left( \frac{N \tau (\tau - 1)}{2} + \nu (1 - (1 + \kappa) \tau) \right) d_{\kappa+1}^{2} \right] \varsigma^{2\kappa} + \cdots \right\} + 1$$ (18) and $$\frac{1}{\eta} \left( \nu \frac{\{(\varkappa g'(\varkappa))'\}^{\tau}}{g'(\varkappa)} + (1 - \nu)(g'(\varkappa))^{\tau} - 1 \right) + 1 =$$ $$\frac{1}{\eta} \left\{ -Ld_{\kappa+1} \varkappa^{\kappa} + \left[ M((1+\kappa)d_{\kappa+1}^{2} - d_{2\kappa+1}) + (1+\kappa)^{2} \left( \frac{N\tau(\tau-1)}{2} + \nu(1-(1+\kappa)\tau) \right) d_{\kappa+1}^{2} \right] \varkappa^{2\kappa} + \cdots \right\} + 1,$$ (19) where L, M, and N are as in (13), (14), and (15), respectively. Comparing (7) and (18), we obtain $$Ld_{\kappa+1} = \eta B_1 h_{\kappa} \tag{20}$$ and $$Md_{2\kappa+1} + \left(\frac{N\tau(\tau-1)}{2} + \nu(1-(1+\kappa)\tau)\right)(1+\kappa)^2 d_{\kappa+1}^2 = \eta[B_1h_{2\kappa} + B_2h_{\kappa}^2].$$ (21) Comparing (8) and (19), we obtain $$-Ld_{\kappa+1} = \eta B_1 p_{\kappa} \tag{22}$$ and $$M((\kappa+1)d_{\kappa+1}^2 - d_{2\kappa+1}) + \left(\frac{N\tau(\tau-1)}{2} + \nu(1 - (1+\kappa)\tau)\right)(1+\kappa)^2 d_{\kappa+1}^2 = \eta[B_1p_{2\kappa} + B_2p_{\kappa}^2],$$ (23) From (20) and (22), we obtain $$h_{\kappa} = -p_{\kappa} \tag{24}$$ and $$2L^2 d_{\kappa+1}^2 = \eta^2 B_1^2 (h_{\kappa}^2 + p_{\kappa}^2). \tag{25}$$ We add (21) and (23) and then use (25) to obtain $$[\{M(1+\kappa) + [N\tau(\tau-1) + (1-(1+\kappa)\tau)2\nu](1+\kappa)^2\}\eta B_1^2 - 2L^2B_2]d_{\kappa+1}^2$$ $$= \eta^2 B_1^3(h_{2\kappa} + p_{2\kappa})$$ (26) By using (6) and (20) in (26) for the coefficients $h_{2\kappa}$ and $p_{2\kappa}$ , we obtain $$[|\{M(1+\kappa) + [N\tau(\tau-1) + (1-(1+\kappa)\tau)2\nu](1+\kappa)^2\}\eta B_1^2 - 2L^2B_2| + 2L^2B_1]|d_{\kappa+1}|^2 \le 2\eta^2B_1^3,$$ (27) which implies (9). We subtract (23) from (21) to find the bound on $|d_{2\kappa+1}|$ : $$d_{2\kappa+1} = \frac{\eta B_1(h_{2\kappa} - p_{2\kappa})}{2M} + \left(\frac{1+\kappa}{2}\right) d_{\kappa+1}^2.$$ (28) In view of (20), (24), (28) and applying (6), we obtain $$|d_{2\kappa+1}| \le \frac{|\eta|B_1}{M} + \left(\frac{1+\kappa}{2} - \frac{L^2}{|\eta|B_1M}\right) \times$$ (29) $$\times \frac{2\eta^2 B_1^3}{|\{M(1+\kappa)+[N\tau(\tau-1)+(1-(1+\kappa)\tau)2\nu](1+\kappa)^2\}\eta B_1^2-2L^2B_2|+2L^2B_1},$$ which obtains (10), the desired assessment. From (26) and (28), for $\delta \in \mathbb{R}$ , we obtain $$d_{2\kappa+1} - \delta d_{\kappa+1}^2 = \frac{\eta B_1}{2} \left[ \left( \mathfrak{Y}(\delta) + \frac{1}{M} \right) h_{2\kappa} + \left( \mathfrak{Y}(\delta) - \frac{1}{M} \right) p_{2\kappa} \right],$$ where $$\mathfrak{Y}(\delta) = \frac{\eta B_1^2(\kappa - 2\delta + 1)}{\{M(1+\kappa) + [N\tau(\tau - 1) + (1 - (1+\kappa)\tau)2\nu](\kappa + 1)^2\}\eta B_1^2 - 2L^2B_2}.$$ In view of (6), we conclude that $$|d_{2\kappa+1} - \delta d_{\kappa+1}^2| \le \begin{cases} \frac{|\eta|B_1}{M} & ; 0 \le |\mathfrak{Y}(\delta)| < \frac{1}{M} \\ |\eta|B_1|\mathfrak{Y}(\delta)| & ; |\mathfrak{Y}(\delta)| \ge \frac{1}{M}, \end{cases}$$ form which we obtain (11) with J as in (12). So the proof is completed. $\Box$ Remark 2. We obtain Corollary 1 of [24] if $\nu = 0$ in Theorem 1. Choosing $\tau = 1$ in $\mathfrak{P}_{\sigma_{\sigma}}^{\tau}(\eta, \nu, \varphi)$ , we have the corollary given below:

Corollary 1. Let $\delta \in \mathbb{R}$ and let the function s given by (3) be in the family $\mathscr{I}_{\sigma_{\kappa}}(\eta, \nu, \varphi)$ . Then, $$\begin{split} |d_{\kappa+1}| &\leq \frac{|\eta|B_1\sqrt{2B_1}}{\sqrt{|\{(1+\kappa)M_1 - 2\nu\kappa(1+\kappa)^2\}\eta B_1^2 - 2L_1^2B_2| + 2L_1^2B_1}},\\ |d_{2\kappa+1}| &\leq \\ &\left\{ \frac{|\eta|B_1}{M_1} & ; \ 0 < B_1 < \frac{2L_1^2}{(1+\kappa)M_1|\eta|} \\ &\frac{|\eta|B_1}{M_1} + \left(\frac{1+\kappa}{2} - \frac{L_1^2}{B_1M_1|\eta|}\right) \frac{2\eta^2B_1^3}{|\{(1+\kappa)M_1 - 2\nu\kappa(1+\kappa)^2\}\eta B_1^2 - 2L_1^2B_2| + 2L_1^2B_1}; B_1 \geq \frac{2L_1^2}{(1+\kappa)M_1|\eta|} \end{split} \right. \end{split}$$ and $$|d_{2\kappa+1} - \delta d_{\kappa+1}^2| \leq \begin{cases} \frac{|\eta|B_1}{M_1} & ; |\kappa - 2\delta + 1| < J_1 \\ \frac{B_1^3|\kappa - 2\delta + 1||\eta|^2}{|\{(\kappa+1)M_1 - 2\nu\kappa(\kappa+1)^2\}\eta B_2^2 - 2L_2^2B_2|} & ; |\kappa - 2\delta + 1| \ge J_1, \end{cases}$$ where $$J_{1} = \left| \frac{\{(1+\kappa)M_{1} - 2\nu\kappa(1+\kappa)^{2}\}\eta B_{1}^{2} - 2L_{1}^{2}B_{2}}{M_{1}B_{1}^{2}\eta} \right|,$$ $$L_{1} = (1+\kappa)((\kappa-1)\nu+1)$$ (30) and $$M_1 = (1 + 2\kappa)((2\kappa - 1)\nu + 1), \tag{31}$$ Remark 3. If $\nu = 0$ and $\eta = 1$ in Corollary 1 are allowed, then the first and second theorems of Tang et al. [19] are obtained. Choosing $\kappa = 1$ in Theorem 1, we have

Corollary 2. If $s \in \mathfrak{P}^{\tau}_{\sigma_1}(\eta, \nu, \varphi)$ is given by (1) and $\delta \in \mathbb{R}$ , then $$\begin{split} |d_2| &\leq \frac{|\eta|B_1\sqrt{B_1}}{\sqrt{|\{M_2+2(N_2\tau(\tau-1)+2\nu(1-2\tau))\}\eta B_1^2-L_2^2B_2|+L_2^2B_1}},\\ |d_3| &\leq \begin{cases} \frac{|\eta|B_1}{M_2} &; \ 0 < B_1 < \frac{L_2^2}{|\eta|M_2} \\ \frac{|\eta|B_1}{M_2} + \left(1 - \frac{L_2^2}{|\eta|B_1M_2}\right) \frac{\eta^2B_1^3}{|\{M_2+2(N_2\tau(\tau-1)+2\nu(1-2\tau))\}\eta B_1^2-L_2^2B_2|+L_2^2B_1}; B_1 \geq \frac{L_2^2}{|\eta|M_2}, \end{cases} \end{split}$$ and $$|d_3 - \delta d_2^2| \le \begin{cases} \frac{|\eta|B_1}{M_2} & ; |1 - \delta| < J_2\\ \frac{|\eta|^2 B_1^3 |1 - \delta|}{|\{M_2 + 2(N_2 \tau(\tau - 1) + 2\nu(1 - 2\tau))\}\eta B_1^2 - L_2^2 B_2|}; |1 - \delta| \ge J_2, \end{cases}$$ where $$J_{2} = \left| \frac{\{M_{2} + 2(N_{2}\tau(\tau - 1) + 2\nu(1 - 2\tau))\}\eta B_{1}^{2} - L_{2}^{2}B_{2}}{\eta M_{2}B_{1}^{2}} \right|,$$ $$L_{2} = 2((1 + \nu)\tau - \nu),$$ (32) $$M_2 = 3((1+2\nu)\tau - \nu) \tag{33}$$ and $$N_2 = 3\nu + 1. (34)$$ Setting $\eta = \tau = 1$ in Corollary 2, we obtain the following.

Corollary 3. If $s \in \mathfrak{P}^1_{\sigma_1}(1, \nu, \varphi)$ is given by (1) and $\delta \in \mathbb{R}$ , then $$\begin{aligned} |d_2| &\leq \frac{B_1\sqrt{B_1}}{\sqrt{|(3-\nu)B_1^2 - 4B_2| + 4B_1}}, \\ |d_3| &\leq \begin{cases} \frac{B_1}{3(\nu+1)} & ; \ 0 < B_1 < \frac{4}{3(\nu+1)} \\ \frac{B_1}{3(\nu+1)} + \left(1 - \frac{4}{3(\nu+1)B_1}\right) \frac{B_1^3}{|(3-\nu)B_1^2 - 4B_2| + 4B_1}; B_1 \geq \frac{4}{3(\nu+1)}, \end{aligned}$$ and $$|d_3 - \delta d_2^2| \le \begin{cases} \frac{B_1}{3(\nu+1)} & ; |1 - \delta| < \left| \frac{(3-\nu)B_1^2 - 4B_2}{3(\nu+1)B_1^2} \right| \\ \frac{B_1^3|1 - \delta|}{|(3-\nu)B_1^2 - 4B_2|} & ; |1 - \delta| \ge \left| \frac{(3-\nu)B_1^2 - 4B_2}{3(\nu+1)B_1^2} \right|. \end{cases}$$ Remark 4. When v = 0 is selected in Corollary 3, we obtain Corollaries 1 and 4 of Tang et al. [19] (also see [30]).

Corollary 4. If the function s given by (3) belongs to the family $\in \mathfrak{Q}_{\sigma_{\kappa}}^{\tau}(\eta, \nu, \varrho)$ and $\delta \in \mathbb{R}$ , then $$\begin{split} |d_{\kappa+1}| &\leq \frac{2\varrho|\eta|}{\sqrt{\varrho|\{M(1+\kappa)+[N\tau(\tau-1)+(1-(1+\kappa)\tau)2\nu](1+\kappa)^2\}\eta-L^2|+L^2}},\\ |d_{2\kappa+1}| &\leq \\ &\left\{ \begin{array}{l} \frac{2\varrho|\eta|}{M} & ; \ 0 < \varrho < \frac{L^2}{M(1+\kappa)|\eta|} \\ \frac{2\varrho|\eta|}{M} + \left(1+\kappa - \frac{L^2}{\varrho M|\eta|}\right) \frac{2\varrho^2\eta^2}{\varrho|\{(1+\kappa)M+[N\tau(\tau-1)+(1-(1+\kappa)\tau)2\nu](1+\kappa)^2\}\eta-L^2|+L^2}; \ \varrho \geq \frac{L^2}{M(1+\kappa)|\eta|}, \end{array} \right. \end{split}$$ and $$|d_{2\kappa+1} - \delta d_{\kappa+1}^2| \leq \begin{cases} \frac{2\varrho|\eta|}{M} & ; |\kappa - 2\delta + 1| < J_3 \\ \frac{2\varrho|\kappa - 2\delta + 1||\eta|^2}{|\{(1+\kappa)M + [\tau(\tau-1)N + (1-(\kappa+1)\tau)2\nu](\kappa+1)^2\}\eta - L^2|}; |\kappa - 2\delta + 1| \geq J_3, \end{cases}$$ where $$J_3 = \left| \frac{\{M(1+\kappa) + [N\tau(\tau-1) + (1-(1+\kappa)\tau)2\nu](1+\kappa)^2\}\eta - L^2}{\eta M} \right|,$$ L, M, and N are as in (13), (14), and (15), respectively Remark 6. For v = 0 in Corollary 4, we obtain Corollary 4 in [24]. Choosing $\tau = 1$ in $\mathfrak{Q}^{\tau}_{\sigma_{\kappa}}(\eta, \nu, \varphi)$ , we obtain the corollary given below:

Corollary 5. If the function s given by (3) belongs to the family $\mathcal{D}_{\sigma_{\kappa}}(\eta, \nu, \varrho)$ and $\delta \in \mathbb{R}$ , then $$\begin{split} |d_{\kappa+1}| &\leq \frac{2\varrho|\eta|}{\sqrt{\varrho|\{M_1(1+\kappa)-2\nu\kappa(1+\kappa)^2\}\eta-L_1^2|+L_1^2}},\\ |d_{2\kappa+1}| &\leq \\ &\int \frac{2\varrho|\eta|}{M_1} &; 0 < \varrho < \frac{L_1^2}{M_1(1+\kappa)|\eta|} \\ &\frac{2\varrho|\eta|}{M_1} + \left(1+\kappa - \frac{L_1^2}{\varrho M_1|\eta|}\right) \frac{2\varrho^2\eta^2}{\varrho|\{M_1(1+\kappa)-2\nu\kappa(1+\kappa)^2\}\eta-L_1^2|+L^2\}}; \varrho \geq \frac{L_1^2}{(1+\kappa)M_1|\eta|} \end{split}$$ and $$|d_{2\kappa+1} - \delta d_{\kappa+1}^2| \leq \begin{cases} \frac{2\varrho|\eta|}{M_1} & ; |\kappa - 2\delta + 1| < J_4 \\ \frac{2\varrho|\kappa - 2\delta + 1||\eta|^2}{|\{M_1(\kappa+1) - 2\nu\kappa(\kappa+1)^2\}\eta - L_1^2|} ; |\kappa - 2\delta + 1| \ge J_4, \end{cases}$$ where $$J_4 = \left| \frac{\{M_1(1+\kappa) - 2\nu\kappa(1+\kappa)^2\}\eta - L_1^2}{\eta M_1} \right|,$$ $L_1$ and $M_1$ are as in (30) and (31), respectively. Corollary 4 yields the following if $\kappa = 1$ :

Corollary 6. If $s \in \mathfrak{Q}^{\tau}_{\sigma_1}(\eta, \nu, \varrho)$ is given by (1) and $\delta \in \mathbb{R}$ , then $$\begin{split} |d_2| &\leq \frac{2|\eta|\varrho}{\sqrt{\varrho|\{2M_2 + 4(N_2\tau(\tau - 1) + 2\nu(1 - 2\tau))\}\eta - L_2^2| + L_2^2}}, \\ |d_3| &\leq \begin{cases} \frac{2|\eta|\varrho}{M_2} &; 0 < \varrho < \frac{L_2^2}{2|\eta|M_2} \\ \frac{2|\eta|\varrho}{M_2} + \left(2 - \frac{L_2^2}{|\eta|\varrho M_2}\right) \frac{2\eta^2\varrho^2}{\varrho|\{2M_2 + 4(N_2\tau(\tau - 1) + 2\nu(1 - 2\tau))\}\eta - L_2^2| + L_2^2}; \varrho \geq \frac{L_2^2}{2|\eta|M_2}, \end{cases} \end{split}$$ and $$|d_3 - \delta d_2^2| \le \begin{cases} \frac{2|\eta|\varrho}{M_2} & ; |1 - \delta| < J_5\\ \frac{2|\eta|^2\varrho|1 - \delta|}{|\{2M_2 + 4(N_2\tau(\tau - 1) + 2\nu(1 - 2\tau))\}\eta - L_2^2|} ; |1 - \delta| \ge J_5, \end{cases}$$ where $$J_5 = \left| \frac{\{2M_2 + 4(N_2\tau(\tau - 1) + 2\nu(1 - 2\tau))\}\eta - L_2^2}{2\eta M_2} \right|,$$ $L_2$ , $M_2$ , and $N_2$ are as in (32), (33), and (34), respectively Corollary 6 would yield the following if $\eta = \tau = 1$ .

Corollary 7. If the function s of the form $(1) \in \mathfrak{Q}^1_{\sigma_1}(1, \nu, \varphi)$ and $\delta \in \mathbb{R}$ , then $$|d_2| \le \frac{\varrho\sqrt{2}}{\sqrt{\varrho(1-\nu)+2}},$$ $$|d_3| \le \begin{cases} \frac{2\varrho}{3(\nu+1)} & ; \ 0 < \varrho < \frac{2}{3(\nu+1)} \\ \frac{2\varrho}{3(\nu+1)} + \left(1 - \frac{2}{3\varrho(\nu+1)}\right) \frac{2\varrho^2}{\varrho(1-\nu)+2} ; \varrho \ge \frac{2}{3(\nu+1)}, \end{cases}$$ and $$|d_3 - \delta d_2^2| \le \begin{cases} \frac{2\varrho}{3(\nu+1)} \ ; \ |1 - \delta| < \frac{1 - \nu}{3(\nu+1)} \\ \frac{\varrho|1 - \delta|}{1 - \nu} \ ; \ |1 - \delta| \ge \frac{1 - \nu}{3(\nu+1)}. \end{cases}$$ Remark 7. Letting v = 0 in Corollary 7, we obtain Corollary 2 of Tang et al. [19]. The estimate obtained here for $|d_3|$ is more accurate when compared to that in Theorem 2 of Srivastava et al. [9].

Corollary 8. Let the function s of the form (3) belong to the class $\mathfrak{X}_{\sigma_{\kappa}}^{\tau}(\eta, \nu, \xi)$ and $\delta \in \mathbb{R}$ . Then, $$|d_{\kappa+1}| \leq \frac{(1-\xi)2|\eta|}{\sqrt{|\{(1+\kappa)M+(1+\kappa)^2[N\tau(\tau-1)+(1-(1+\kappa)\tau)2\nu]\}(1-\xi)\eta-L^2|+L^2}}$$ $$|d_{2\kappa+1}| \leq$$ $$\begin{cases} \frac{(1-\xi)2|\eta|}{M} & ; 1 - \frac{L^2}{(1+\kappa)M|\eta|} < \xi < 1 \\ \frac{(1-\xi)2|\eta|}{M} + \left(1 + \kappa - \frac{L^2}{(1-\xi)M|\eta|}\right) \frac{2(1-\xi)^2|\eta|^2}{|\{(1+\kappa)M + [N\tau(\tau-1) + (1-(1+\kappa)\tau)2\nu](1+\kappa)^2\}(1-\xi)\eta - L^2| + L^2} \\ & ; 0 \le \xi \le 1 - \frac{L^2}{(1+\kappa)M|\eta|} \end{cases}$$ and $$|d_{2\kappa+1} - \delta d_{\kappa+1}^2| \leq \begin{cases} \frac{(1-\xi)2|\eta|}{M} & ; |\kappa - 2\delta + 1| < J_6 \\ \frac{2(1-\xi)^2|\eta|^2|\kappa - 2\delta + 1|}{|\{(1+\kappa)M + [N\tau(\tau-1) + (1-(1+\kappa)\tau)2\nu](1+\kappa)^2\}(1-\xi)\eta - L^2|}; |\kappa - 2\delta + 1| \geq J_6, \end{cases}$$ where $$J_6 = \left| \frac{\{(1+\kappa)M + (1+\kappa)^2[N\tau(\tau-1) + (1-(1+\kappa)\tau)2\nu]\}(1-\xi)\eta - L^2}{M(1-\xi)\eta} \right|.$$ L, M, and N are as in (13), (14), and (15), respectively. Remark 9. We obtain Corollary 7 of Aldawish et al. [24] if $\nu = 0$ in Corollary 8. In addition, we obtain Corollary 11 of Swamy et al. [22] when $\eta = \tau = 1$ .

Corollary 9. Let the function s of the form (3) belong to the class $\mathscr{F}_{\sigma_x}^{\tau}(\eta, \nu, \xi)$ and $\delta \in \mathbb{R}$ . Then, $$\begin{split} |d_{\kappa+1}| &\leq \frac{(1-\xi)2|\eta|}{\sqrt{|\{(1+\kappa)M_1 - 2\nu\kappa(1+\kappa)^2\}(1-\xi)\eta - L_1^2| + L_1^2}},\\ |d_{2\kappa+1}| &\leq \\ &\left\{ \frac{(1-\xi)2|\eta|}{M_1} & ; 1 - \frac{L_1^2}{(\kappa+1)M_1|\eta|} < \xi < 1 \right. \\ &\left. \frac{(1-\xi)2|\eta|}{M_1} + \left(1 + \kappa - \frac{L_1^2}{(1-\xi)M_1|\eta|}\right) \frac{2(1-\xi)^2|\eta|^2}{|\{(1+\kappa)M_1 - 2\nu\kappa(1+\kappa)^2\}\eta(1-\xi) - L_1^2| + L_1^2} \\ & ; 0 \leq \xi \leq 1 - \frac{L_1^2}{(1+\kappa)M_1|\eta|} \end{split}$$ and $$|d_{2\kappa+1} - \delta d_{\kappa+1}^2| \leq \begin{cases} \frac{2|\eta|(1-\xi)}{M_1} & ; |\kappa - 2\delta + 1| < J_7 \\ \frac{2|\eta|^2(1-\xi)^2|\kappa - 2\delta + 1|}{|\{(1+\kappa)M_1 - 2\nu\kappa(1+\kappa)^2\}(1-\xi)\eta - L_1^2|} & ; |\kappa - 2\delta + 1| \ge J_7, \end{cases}$$ where $$J_7 = \left| \frac{\{(1+\kappa)M_1 - 2\nu\kappa(1+\kappa)^2\}(1-\xi)\eta - L_1^2}{M_1(1-\xi)\eta} \right|,$$ $L_1$ and $M_1$ are as in (30) and (31), respectively If we let $\kappa = 1$ in Corollary 8, then we have

Corollary 10. Let the function s of the form (1) belong to the class $\mathfrak{X}_{\sigma_1}^{\tau}(\eta, \nu, \xi)$ and $\delta \in \mathbb{R}$ . Then, $$\begin{split} |d_2| & \leq \frac{2(1-\xi)|\eta|}{\sqrt{|\{2M_2+4(N_2\tau(\tau-1)+2\nu(1-2\tau))\}(1-\xi)\eta-L_2^2|+L_2^2}}, \\ |d_3| & \leq \begin{cases} \frac{2(1-\xi)|\eta|}{M_2} & ; \ 1-\frac{L_2^2}{2M_2|\eta|} < \xi < 1\\ \frac{2(1-\xi)|\eta|}{M_2} + \left(2-\frac{L_2^2}{(1-\xi)M_2|\eta|}\right) \frac{2(1-\xi)^2|\eta|^2}{|\{2M_2+4(N\tau(\tau-1)+2\nu(1-2\tau))\}(1-\xi)\eta-L_2^2|+L_2^2\}}, \\ & ; \ 0 \leq \xi \leq 1-\frac{L_2^2}{2M_2|\eta|} \end{split}$$ and $$|d_3 - \delta d_2^2| \le \begin{cases} \frac{2(1-\xi)|\eta|}{3(\nu+1)} & ; |1-\delta| < J_8\\ \frac{2(1-\xi)^2|\eta|^2|1-\delta|}{|\{2M_2+4(N_2\tau(\tau-1)+2\nu(1-2\tau))\}(1-\xi)\eta-L_2^2|} & ; |1-\delta| \ge J_8, \end{cases}$$ where $$J_8 = \left| \frac{\{2M_2 + 4(N\tau(\tau - 1) + 2\nu(1 - 2\tau))\}(1 - \xi)\eta - L_2^2}{2M_2(1 - \xi)\eta} \right|,$$ $L_2$ , $M_2$ , and $N_2$ are as in (32), (33), and (34), respectively If $\eta = \tau = 1$ in Corollary 10, then we obtain

Corollary 11. If $s \in \mathfrak{X}^1_{\sigma_1}(1, \nu, \varphi)$ is of the form (1) and $\delta \in \mathbb{R}$ , then $$\begin{split} |d_2| &\leq \frac{\sqrt{2}(1-\xi)}{\sqrt{|(3-\nu)(1-\xi)-2|+2}}, \\ |d_3| &\leq \begin{cases} \frac{2(1-\xi)}{3(1+\nu)} & ; \frac{3\nu+1}{3(1+\nu)} < \xi < 1 \\ \frac{2(1-\xi)}{3(1+\nu)} + \left(1 - \frac{2}{3(1-\xi)(1+\nu)}\right) \frac{2(1-\xi)^2}{|(1-\xi)(3-\nu)-2|+2}; 0 \leq \xi \leq \frac{3\nu+1}{3(1+\nu)}, \end{cases} \end{split}$$ and $$|d_3 - \delta d_2^2| \le \begin{cases} \frac{2(1-\xi)}{3(1+\nu)} \; ; \; |1-\delta| < \left| \frac{(1-\xi)(3-\nu)-2}{3(1-\xi)(1+\nu)} \right| \\ \frac{2|1-\delta|(1-\xi)^2}{|(1-\xi)(3-\nu)-2|} ; |1-\delta| \ge \left| \frac{(3-\nu)(1-\xi)-2}{3(1-\xi)(1+\nu)} \right|, \end{cases}$$ Remark 10. Putting v = 0 in Corollary 11, we obtain Corollary 3 of Tang et al. [19]. The estimates obtained here for $|d_2|$ and $|d_3|$ are more accurate when compared to those estimates of Theorem 2 in [9].

Lemma 1. ([28,29]). Let $f \in S$ be given by (1). Then, the coefficients of its inverse map $g = f^{-1}$ are given in terms of the Faber polynomials of f with $$g(w) = f^{-1}(w) = w + \sum_{n=2}^{\infty} \frac{1}{n} K_{n-1}^{-n}(a_2, a_3, \dots, a_n) w^n,$$ where $$K_{n-1}^{-n} = \frac{(-n)!}{(-2n+1)!(n-1)!} a_2^{n-1} + \frac{(-n)!}{(2(-n+1))!(n-3)!} a_2^{n-3} a_3$$ $$+ \frac{(-n)!}{(-2n+3)!(n-4)!} a_2^{n-4} a_4 + \frac{(-n)!}{(2(-n+2))!(n-5)!} a_2^{n-5} \left[ a_5 + (-n+2)a_3^2 \right]$$ $$+ \frac{(-n)!}{(-2n+5)!(n-6)!} a_2^{n-6} \left[ a_6 + (-2n+5)a_3a_4 \right] + \sum_{j \ge 7} a_2^{n-j} V_j,$$ such that $V_j$ $(7 \le j \le n)$ is a homogeneous polynomial in the variables $a_2, a_3, \ldots, a_n$ and the expressions such as (for example) (-m)! are to be interpreted symbolically by $$(-m)! \equiv \Gamma(1-m) := (-m)(-m-1)(-m-2)\dots, \quad m \in \mathbb{N}_0 := \mathbb{N} \cup \{0\}, \ \mathbb{N} := \{1, 2, \dots\}.$$ We see that the initial three terms of $K_{n-1}^{-n}$ are given by $$K_1^{-2} = -2a_2$$ , $K_2^{-3} = 3(2a_2^2 - a_3)$ , and $K_3^{-4} = -4(5a_2^3 - 5a_2a_3 + a_4)$ . Typically, for every real number p, the expansion of $K_n^p$ is given below (see [28] for details; see also [29], p. 349): $$K_n^p = pa_{n+1} + \frac{p(p-1)}{2}D_n^2 + \frac{p!}{(p-3)!3!}D_n^3 + \ldots + \frac{p!}{(p-n)!n!}D_n^n$$

Lemma 2. ([30]). Let $f(z) = z + \sum_{k=n}^{\infty} a_k z^k$ , $n \ge 2$ be a univalent function in $\mathbb D$ and $$f^{-1}(w) = w + \sum_{k=n}^{\infty} b_k w^k \quad \left( |w| < r_0(f), \ r_0(f) \ge \frac{1}{4} \right).$$ Then, $$b_{2n-1} = na_n^2 - a_{2n-1}$$ , and $b_k = -a_k$ for $n \le k \le 2n - 2$ .

Lemma 3. ([31] Exercise 9, p. 172). Assume that $\omega(z) = \sum_{j=1}^{\infty} p_j z^j \in \mathcal{B}$ . Then, $$|p_n| \le 1, \ n \ge 2.$$ This lemma represents a special case of the result in [31] [Exercise 9, p. 172] obtained from this exercise for $p_0 = 0$ .

Lemma 4. Let $u(z) = u_1 z + u_2 z^2 + u_3 z^3 + \cdots \in \mathcal{B}$ and s be a complex number. Then, for all $n \in \mathbb{N}$ , the following inequality holds: $$\left|u_{2n} - su_n^2\right| \le 1 + (|s| - 1)\left|u_n^2\right| \le \max\{1; |s|\}.$$ Moreover, the functions u(z) = z and $u(z) = z^2$ prove that the above inequality is sharp for $|s| \ge 1$ and for |s| < 1, respectively. Proof. For $u(z) = u_1 z + u_2 z^2 + u_3 z^3 + \cdots \in \mathcal{B}$ and a fixed $n \in \mathbb{N}$ , let $$\varepsilon_k := e^{2k\pi i/n}, \quad k \in \{1, 2, \dots, n\}$$ be the nth order complex roots of the unit. If we define the function $v : \mathbb{D} \to \mathbb{C}$ by $$v(z) := \frac{1}{n} \sum_{k=1}^{n} u(\varepsilon_k z), \ z \in \mathbb{D}, \tag{2}$$ using the well-known relation $$\sum_{k=1}^{n} \varepsilon_k^m = \begin{cases} 0, & \text{if } m \in \mathbb{N} \text{ is not a multiple of } n, \\ n, & \text{if } m \in \mathbb{N} \text{ is a multiple of } n, \end{cases}$$ it follows that $$v(z) = u_n z^n + u_{2n} z^{2n} + \dots, \ z \in \mathbb{D}.$$ (3) Since u is an analytic function in $\mathbb{D}$ , from Definition (2), it follows that v is also analytic in $\mathbb{D}$ and v(0) = 0. Moreover, since $u \in \mathcal{B}$ , we have $$|v(z)| \leq \frac{1}{n} \sum_{k=1}^{n} \left| u\left(e^{-2ik\pi/n}z\right) \right| < \frac{n}{n} = 1, z \in \mathbb{D}.$$ Therefore, $v \in \mathcal{B}$ . Because the function $\chi(z) := z^n$ is a surjective endomorphism of the unit disk $\mathbb{D}$ , setting $\zeta := z^n$ in (3) and using the fact that $v \in \mathcal{B}$ , we deduce that the function $\psi : \mathbb{D} \to \mathbb{C}$ by $$\psi(\zeta) := u_n \zeta + u_{2n} \zeta^2 + u_{3n} \zeta^3 + \dots, \ \zeta \in \mathbb{D}$$ belongs to the class $\mathcal{B}$ . Now, using [32] (page 10, inequality (7)) for the function $\psi \in \mathcal{B}$ , we obtain the desired outcome with the aforementioned power series expansion. $\square$ We now prove the following main theorem using the aforementioned lemmas and a new method.

Theorem 1. Let the function $f(z) = z + \sum_{k=n_0}^{\infty} a_k z^k \in \mathcal{N}_{\Sigma}(\lambda, \delta, h)$ , $n_0 \geq 2$ . Then, $$|a_{n_0}| \le \min \left\{ \frac{|B_1|}{1 + (n_0 - 1)(\lambda + n_0 \delta)}; \sqrt{\frac{2|B_1| \max\left\{1; \left|\frac{B_2}{B_1}\right|\right\}}{n_0 \left(1 + (2n_0 - 2)[\lambda + (2n_0 - 1)\delta]\right)}} \right\}, \tag{4}$$ and $$\left| n_0 a_{n_0}^2 - a_{2n_0 - 1} \right| \le \frac{|B_1| \max\left\{1; \left| \frac{B_2}{B_1} \right| \right\}}{1 + (2n_0 - 2)[\lambda + (2n_0 - 1)\delta]}. \tag{5}$$ Proof. If $f(z) = z + \sum_{k=n_0}^{\infty} a_k z^k \in \mathcal{N}_{\Sigma}(\lambda, \delta, h)$ , then there are two functions as defined by the quasi-subordination $u, v \in \mathcal{B}$ of the form $$u(z) = \sum_{k=1}^{\infty} u_k z^k$$ and $v(z) = \sum_{k=1}^{\infty} v_k z^k$ , satisfying $$(1-\lambda)\frac{f(z)}{z} + \lambda f'(z) + \delta z f''(z) = 1 + \sum_{k=n_0}^{\infty} [1 + (k-1)(\lambda + k\delta)] a_k z^{k-1} = h(u(z))$$ (6) and $$(1-\lambda)\frac{g(w)}{w} + \lambda g'(w) + \delta w g''(w) = 1 + \sum_{k=2}^{\infty} [1 + (k-1)(\lambda + k\delta)]b_k w^{k-1} = h(v(w)), \quad (7)$$ respectively, where, according to Lemma 1, $$b_k = \frac{1}{k} K_{n-1}^{-k}(a_2, a_3, \dots, a_k), \ k \ge 2.$$ (8) We have $$h(u(z)) = 1 + B_1 \left( u_1 z + u_2 z^2 + \dots \right) + B_2 \left( u_1 z + u_2 z^2 + \dots \right)^2 + \dots, \tag{9}$$ and, according to (6) and (9), the corresponding coefficients of the power expansions are equal. Hence, we equate these coefficients step by step. First, from (6), we have $a_k = 0$ for $2 \le k \le n_0 - 1$ . Thus, the term containing "z" in (9) is equal to zero, that is, $B_1u_1 = 0$ . Using the fact that $B_1 \ne 0$ , it follows $u_1 = 0$ . Therefore, (9) becomes $$h(u(z)) = 1 + B_1(u_2z^2 + \dots) + B_2(u_2z^2 + \dots)^2 + \dots$$ (10) Secondly, since, in (6), the term containing " $z^2$ " is zero, it follows that, for the corresponding term of (10), we have $B_1u_2=0$ . Since $B_1\neq 0$ , it follows that $u_2=0$ . Hence, (10) becomes $$h(u(z)) = 1 + B_1(u_3z^3 + \dots) + B_2(u_3z^3 + \dots)^2 + \dots$$ We repeat the same method $n_0-2$ times and take into account that from the " $n_0-3$ " step we obtain $$h(u(z)) = 1 + B_1 \left( u_{n_0 - 2} z^{n_0 - 2} + \dots \right) + B_2 \left( u_{n_0 - 2} z^{n_0 - 2} + \dots \right)^2 + \dots$$ (11) Since the coefficient of term containing " $z^{n_0-2}$ " in (6) is zero, we obtain that the relevant coefficient in (11) is $B_1u_{n_0-2}=0$ . Thus, the assumption $B_1\neq 0$ implies $u_{n_0-2}=0$ . Hence, (11) becomes $$h(u(z)) = 1 + B_1 \left( u_{n_0 - 1} z^{n_0 - 1} + \dots \right) + B_2 \left( u_{n_0 - 1} z^{n_0 - 1} + \dots \right)^2 + \dots$$ (12) Now, by comparing the terms in " $z^{n_0-1}$ " in (6) and (12), we obtain that $$B_1 u_{n_0-1} = [1 + (n_0 - 1)(\lambda + n_0 \delta)]a_{n_0},$$ that is, $$a_{n_0} = \frac{B_1 u_{n_0 - 1}}{1 + (n_0 - 1)(\lambda + n_0 \delta)}. (13)$$ On the other hand, since $a_k = 0$ for $2 \le k \le n_0 - 1$ , from (8), we obtain $b_k = 0$ for $2 \le k \le n_0 - 1$ , and from (8), we have $$b_{n_0} = \frac{1}{n_0} K_{n_0-1}^{-n_0}(0,0,\ldots,0,a_{n_0}) = -a_{n_0}.$$ Furthermore, similar to the method described above, from the relation (7), we obtain that the term containing " $w^{n_0-1}$ " is given by $$B_1 v_{n_0-1} = -[1 + (n_0 - 1)(\lambda + n_0 \delta)]a_{n_0}$$ that is, $$a_{n_0} = -\frac{B_1 v_{n_0 - 1}}{1 + (n_0 - 1)(\lambda + n_0 \delta)}. (14)$$ From (13) and (14), using Lemma 3 and considering the previous reasons, we obtain $$|a_{n_0}| = |b_{n_0}| \le \frac{|B_1|}{1 + (n_0 - 1)(\lambda + n_0 \delta)}.$$ (15) Also, equating the terms that contain " $z^{2n_0-2}$ " from (6) for $k=2n_0-1$ and those of (12), we obtain $$\left(1+(2n_0-2)[\lambda+(2n_0-1)\delta]\right)a_{2n_0-1}=B_1u_{2n_0-2}+B_2u_{n_0-1}^2=B_1\left(u_{2n_0-2}+\frac{B_2}{B_1}u_{n_0-1}^2\right).$$ Thus, based on the previous equality and according to the Lemma 4, it follows that $$\left(1+(2n_0-2)[\lambda+(2n_0-1)\delta]\right)\left|a_{2n_0-1}\right| \leq |B_1|\max\left\{1;\left|\frac{B_2}{B_1}\right|\right\}.$$ Hence, $$|a_{2n_0-1}| \le \frac{|B_1| \max\left\{1; \left|\frac{B_2}{B_1}\right|\right\}}{1 + (2n_0 - 2)[\lambda + (2n_0 - 1)\delta]}.$$ (16) From Definition 1, because $f \in \mathcal{N}_{\Sigma}(\lambda, \delta, h)$ implies $g \in \mathcal{N}_{\Sigma}(\lambda, \delta, h)$ and using the above method of proof, we have $$(1-\lambda)\frac{g(w)}{w} + \lambda g'(w) + \delta w g''(w) = 1 + \sum_{k=n_0}^{\infty} [1 + (k-1)(\lambda + k\delta)]b_k w^{k-1} = h(v(w)).$$ Hence, we obtain $$|b_{2n_0-1}| \le \frac{|B_1| \max\left\{1; \left|\frac{B_2}{B_1}\right|\right\}}{1 + (2n_0 - 2)[\lambda + (2n_0 - 1)\delta]}.$$ (17) Furthermore, in view of Lemma 2, using the relations (16) and (17), we deduce that $$|a_{n_0}| \le \sqrt{\frac{|a_{2n_0-1}| + |b_{2n_0-1}|}{n_0}} \le \sqrt{\frac{2|B_1| \max\{1; \left|\frac{B_2}{B_1}\right|\}}{n_0(1 + (2n_0 - 2)[\lambda + (2n_0 - 1)\delta])}},$$ (18) and from (15) and (18), we obtain the inequality (4). In addition, using (17) and Lemma 2, it follows that $$|n_0 a_{n_0}^2 - a_{2n_0 - 1}| = |b_{2n_0 - 1}| \le \frac{|B_1| \max\left\{1; \left|\frac{B_2}{B_1}\right|\right\}}{1 + (2n_0 - 2)[\lambda + (2n_0 - 1)\delta]},$$ which completes our proof. $\Box$ Next, this study shows why this theorem improves and generalizes some previous ones by a suitable choice of parameters. Remark 3. By choosing $\lambda$ , $\delta$ , and h properly, we obtain from Theorem 1 the bounds that are better, in some ranges of the parameters, than the estimates obtained before. 1. If $$h(z) = \frac{1 + (1 - 2\alpha)z}{1 - z}, \ 0 \le \alpha < 1,$$ then the bounds are better than those in [20, Theorem 2]; 2. If $$h(z) = \frac{1 + (1 - 2\alpha)z}{1 - z}, \ 0 \le \alpha < 1,$$ and $\delta = 0$ or $\lambda = 1$ , then the bounds are better than those in [20] [Corollary 3] and [20] [Corollary 4], respectively; 3. If $\delta = 0$ , then the bounds are better than those in [33] [Theorems 3.1] in the case of subordination. In the following part, we emphasize the significance of our present results that improve some previous results concerning different recent subclasses of bi-univalent functions. Remark 4. In the proof of Theorem 4 of [18], assume for convenience that $\rho = 1$ , $\vartheta = 0$ with $f(z) = z + \sum_{k=n}^{\infty} a_k z^k \in \mathcal{S}(0,1;h)$ . By the definition of the subordination, there exist two functions $u, v \in \mathcal{B}$ with $$u(z) = \sum_{k=1}^{\infty} u_k z^n$$ and $v(z) = \sum_{k=1}^{\infty} v_k z^k$ , satisfying $$\frac{zf'(z)}{f(z)} = h(u(z))$$ and $\frac{wg'(w)}{g(w)} = h(v(w))$ , respectively. Since $$\left. \frac{zf'(z)}{f(z)} \right|_{z=0} = 1,$$ it follows that $$\frac{zf'(z)}{f(z)} = 1 + \beta_1 z + \ldots + \beta_n z^n + \ldots, \ z \in \mathbb{D},\tag{19}$$ that is, $$z + na_n z^n + \ldots = (z + a_n z^n + \ldots) (1 + \beta_1 z + \beta_2 z^2 + \ldots), z \in \mathbb{D}.$$ Equating the corresponding coefficients of the above relation, we obtain $$\beta_1 = \beta_2 = \dots = \beta_{n-2} = 0,$$ $\beta_{n-1} = (n-1)a_n,$ and from (19), it follows that $$\frac{zf'(z)}{f(z)} = 1 + (n-1)a_n z^{n-1} + \dots, \ z \in \mathbb{D}.$$ Let us consider again, for convenience, that n = 3. Thus, $$f(z) = z + a_3 z^3 + a_4 z^4 + \dots$$ then $$\frac{zf'(z)}{f(z)} = 1 + \beta_2 z^2 + \beta_3 z^3 + \dots,$$ where $$\beta_2 = 2a_3$$ , $\beta_3 = 3a_4$ , $\beta_4 = 4a_5 - 2a_3^2$ . Consequently, if f has the above form, then it is impossible that $\beta_2 = 2a_3$ and $\beta_4 = 4a_5$ at the same time. We have $\beta_4 = 4a_5$ while $n_0 = 5$ , but in this case, $\beta_2 = 0$ . Therefore, the relation (2.11) of [18] and the Theorem 4 of [18] are not correct. Similarly, for the same reason, Theorem 2.6 of [27] is not correct. Example 1. As an example of Theorem 1, if we consider the analytic function in $\mathbb{D}$ defined by $$f(z) := \frac{1}{\ell} \log \left( \frac{1}{1 - \ell z} \right) = z + \frac{\ell z^2}{2} + \frac{\ell^2 z^3}{3} + \dots, \ z \in \mathbb{D}, \quad with \quad 0 < |\ell| \le 1,$$ then $f \in \mathcal{A}$ and its inverse is $f^{-1}(w) = \frac{e^{\ell w} - 1}{\ell e^{\ell w}}$ , which have an analytic extension in $\mathbb{D}$ denoted as $g(z) = \frac{e^{\ell z} - 1}{\ell e^{\ell z}}$ . $$h(z) := 1 + 0.35z + 0.1z^2 + 0.1z^3, z \in \mathbb{D},$$ (20) like we may see in Figure 3a made with MAPLE™ computer software, we have Re $$h(z) > 0$$ , $z \in \mathbb{D}$ , $B_1 = 0.35 = h'(0) \neq 0$ , $B_2 = 0.1$ , $B_3 = 0.1$ , and $$B_n = 0$$ for $n \ge 4$ . Also, we see that Re $$\frac{zh'(z)}{h(z)-1}$$ = Re $\frac{3z^2+2z+3.5}{z^2+z+3.5} > 0.1 > 0, z \in \mathbb{D}$ . Hence, h is a starlike (univalent) function in D with respect to the point z⁰ = 1. For some "very small" values of the parameter |`| (i.e., close to zero), we have f ∈ NΣ(1.1, 0.15, h) with h given by (20) since the ranges f(D) and g(D) with small neighborhoods of the point w<sup>0</sup> = 1 are included in h(D). According to Theorem 1, the inequalities (4) and (5) reduce to $$|\ell| \le 0.5843487098\dots$$ and $|\ell| \le 0.7156780854\dots$ , respectively. Hence, $$0 < |\ell| \le 0.5843487098\dots \tag{21}$$ (i) Unfortunately, the upper bound of (21) represents a necessary but not sufficient condition for f ∈ NΣ(1.1, 0.15, h) with h given by (20). Let us consider λ = 1.1 and δ = 0.15. Thus, for ` = 0.5843487098 from Figure 3b,c, we see that $$I_{1.1,0.15}[f](\mathbb{D}) \not\subset h(\mathbb{D})$$ and $I_{1.1,0.15}[g](\mathbb{D}) \not\subset h(\mathbb{D})$ , but the reverse inclusions are true. Hence, for ` = 0.5843487098, we have f ∈ N/ <sup>Σ</sup>(1.1, 0.15, h).  Figure 3. Figures for Example 1(i). (ii) As we see in Figure 4a,b, for λ = 1.1, δ = 0.15, and, for example, ` = 0.201, we have the inclusions $$I_{1.1,0.15}[f](\mathbb{D}) \subset h(\mathbb{D})$$ and $I_{1.1,0.15}[g](\mathbb{D}) \subset h(\mathbb{D})$ , and from the fact that h is univalent in D, it follows that both of the subordinations of Definition 1 hold, i.e., f ∈ NΣ(1.1, 0.15, h).  Figure 4. Figures for Example 1(ii).

Lemma 1. ([1]). Let $U_{r_0} = \{z : |z| < r_0\}$ , with $0 < r_0 < 1$ . Let $p(z) = b + b_n z^n + b_{n+1} z^{n+1} + \ldots$ be analytic in U with $n \ge 2$ and $p(z) \not\equiv b$ , and let $q \in Q(b)$ . If there exist points $z_0 = r_0 e^{i\theta_0} \in U$ and $\xi_0 \in \partial U \setminus E(q)$ such that $p(z_0) = q(\xi_0)$ , $p(\overline{U}_{r_0}) \subset q(U)$ , $$\Re \frac{\xi_0 q''(\xi_0)}{q'(\xi_0)} \ge 0$$ , and (9) $$\left| \frac{z p'(z)}{q'(\tilde{z})} \right| \le n \tag{10}$$ where $z \in \overline{\mathcal{U}}_{r_0}$ and $\xi \in \partial \mathcal{U} \setminus E(q)$ , then there exists a real constant $k \geq n \geq 2$ such that $$z_0 p'(z_0) = n\xi_0 q'(\xi_0), \tag{11}$$ $$\Re\left(\frac{z_0 p''(z_0)}{p'(z_0)} + 1\right) \ge n \left[\Re\frac{\xi_0 q''(\xi_0)}{q'(\xi_0)} + 1\right],$$ $$\Re\left(\frac{z_0^2 p'''(z_0)}{p'(z_0)} + 1\right) \ge n^2 \left[\Re\frac{\xi_0^2 q'''(\xi_0)}{q'(\xi_0)}\right] + 1,$$ (12) or $$\Re\left(\frac{z_0^2 p'''(z_0)}{p'(z_0)}\right) \ge n^2 \left[\Re\frac{\xi_0^2 q'''(\xi_0)}{q'(\xi_0)}\right]. \tag{13}$$ Consider a special case when q is univalent in Lemma 1. If $$q(w) = M \frac{Mw + b}{M + \bar{b}w},\tag{14}$$ with |b| < M, then $q(\overline{\mathcal{U}}) = \mathcal{U}_M$ , q(0) = b and $E(q) = \phi$ .

Lemma 2. ([1]). Let $U_{r_0} = \{z : |z| < r_0\}$ , with $0 < r_0 < 1$ . Suppose q given in (14) and $p(z) = b + b_n z^n + b_{n+1} z^{n+1} + \ldots$ is analytic in $\mathcal{U}$ with $n \ge 2$ and $p(z) \not\equiv b$ . If there exist points $z_0 = r_0 e^{i\theta_0} \in \mathcal{U}_M$ and $w_0 \in \partial \mathcal{U}$ such that $p(z_0) = q(w_0)$ , $p(\mathcal{U}_{r_0}) \subset \mathcal{U}_M$ and $$|zp'(z)||[M + \bar{b}e^{i\theta}]|^2 < nM[M^2 - |b|^2]$$ (15) when $z \in \overline{\mathcal{U}}_{r_0}$ and $\theta \in [0, 2\pi]$ , then $$\begin{split} z_0 p'(z_0) &= nq(w_0) \frac{|q(w_0) - b|^2}{|q(w_0)|^2 - |b|^2}, \\ \Re\left(\frac{z_0 p''(z_0)}{p'(z_0)} + 1\right) &\geq n\left(\frac{|q(w_0) - b|^2}{|q(w_0)|^2 - |b|^2}\right), \quad \text{and} \\ \Re\left(\frac{z_0^2 p'''(z_0)}{p'(z_0)}\right) &\geq 6n^2 \frac{\left[|q(w_0) - b|^2\right]^2}{\left[|q(w_0)|^2 - |b|^2\right]^2}. \end{split}$$ Our main objective in this article is to systematically investigate several potentially useful results that are based upon third-order strong differential subordinations and their applications in geometric function theory of complex analysis. Our results give interesting new properties and, together with other papers that appeared in recent years, could emphasize the perspective of the importance of third-order strong differential subordination theory and the generalized Srivastava–Attiya operator. The organization of this article is as follows. In Section 2 below, we derive the notion of third-order strong differential subordination, some definitions and the interesting main results. We consider some suitable classes of admissible functions and investigate several third-order strong differential subordination properties of multivalent functions involving the Srivastava–Attiya operator defined by (2) in Section 3. Some corollaries and consequences of our main results are also presented in Sections 2 and 3. Finally, in the last Section 4, some potential directions for related further research are presented.

Theorem 1. Consider $p \in \mathcal{H}[b, n]$ and $q \in \mathcal{Q}(b)$ fulfills $$\Re \frac{\xi q''(\xi)}{q'(\xi)} \ge 0$$ and $\left| \frac{zp'(z)}{q'(\xi)} \right| \le n$ , (17) where $\xi \in \partial U \setminus E(q)$ . If $\Pi$ is a set in $\mathbb{C}$ , $\Xi \in \Xi_n[\Pi, q]$ and $$\Xi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z, \zeta) \subset \Pi, \tag{18}$$ then $$p(z) \prec q(z)$$ . Proof. If we assume that $p \not\prec q$ , then there exist points $z_0 = r_0 e^{i\theta_0} \in \mathcal{U}$ and $\xi_0 \in \partial \mathcal{U} \setminus E(q)$ such that $p(z_0) = q(\xi_0)$ and $p(\overline{\mathcal{U}}_{r_0}) \subset q(\mathcal{U})$ . From (17), we see that (9) and (10) of Lemma 1 are satisfied when $z \in \mathcal{U}$ and $\xi \in \partial \mathcal{U} \setminus E(q)$ . The conditions of that lemma are satisfied; we conclude that (11)–(13) also follow. Using these conditions with $r_1 = p(z_0), s_1 = z_0 p'(z_0), t_1 = z_0^2 p''(z_0), u_1 = z_0^3 p'''(z_0)$ and $z = z_0$ in Definition 3 leads to $$\Xi(\mathbf{p}(z_0), z_0 \mathbf{p}'(z_0), z_0^2 \mathbf{p}''(z_0), z_0^3 \mathbf{p}'''(z_0); z, \zeta) \notin \Pi$$ which contradicts (18); thus, we have $$p(z) \prec q(z)$$ . In Theorem 1, inequalities (17) and (18) are the most necessary for solving third-order differential subordination. If third-order terms in (18) are missing, then they are not required to satisfy (17). The next result is a special case where the behavior of q on $\partial \mathcal{U}$ is not known in Theorem 1.

Corollary 1. Suppose q is univalent in $\mathcal{U}$ , q(0) = b and set $q_{\rho}(z) \equiv q(\rho z)$ for $\rho \in (0,1)$ . Consider that $p \in \mathcal{H}[b,n]$ and $q_{\rho}$ fulfill $$\Re \frac{\xi \, q_{\rho}''(\xi)}{q_{\rho}'(\xi)} \geq 0$$ and $\left| \frac{z \, p'(z)}{q_{\rho}'(\xi)} \right| \leq n$ , when $\xi \in \partial U \backslash E(q)$ . If $\Pi$ is a set in $\mathbb{C}$ and $\Xi \in \Xi_n[\Pi, q_o]$ , then $$\Xi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z, \zeta) \subset \Pi$$ implies $$p(z) \prec q(z)$$ . Proof. Given $q_{\rho}$ is univalent in $\partial \mathcal{U}$ , and hence $E(q_{\rho}) = \phi$ and $q_{\rho} \in \mathcal{Q}(b)$ . Since the class $\Xi_n[\Pi, q_{\rho}]$ is an admissible functions and from Theorem 1 we obtain $p \prec q_{\rho}$ . Since $q_{\rho} \prec q_{\rho}$ here we conclude that $p \prec q$ . $\square$ In Definition 3, there are no specific conditions on $\Pi$ . When $\Pi \neq \mathbb{C}$ is a simply connected domain and there is a conformal mapping h of $\mathcal{U}$ onto $\Pi$ , we denote the class $\Xi_n[h(\mathcal{U}), \mathbf{q}]$ by $\Xi_n[h, \mathbf{q}]$ . The next two results are directly from Theorem 1 and Corollary 1.

Theorem 2. Consider $p \in \mathcal{H}[b,n]$ and $q \in \mathcal{Q}(b)$ and that they fulfill $$\Re \frac{\xi \, q''(\xi)}{q'(\xi)} \geq 0$$ and $\left| \frac{z p'(z)}{q'(\xi)} \right| \leq n$ , where $\xi \in \partial \mathcal{U} \setminus E(q)$ . If $\Xi \in \Xi_n[h, q]$ and $\Xi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z, \zeta)$ is analytic in $\mathcal{U}$ , then $$\Xi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z, \zeta) \prec \prec h(z)$$ implies $$p(z) \prec q(z)$$ .

Corollary 2. Suppose q is univalent in $\mathcal{U}$ , with q(0) = b, and set $q_{\rho}(z) \equiv q(\rho z)$ for $\rho \in (0,1)$ . Consider that $p \in \mathcal{H}[b,n]$ and $q_{\rho}$ fulfill $$\Re \frac{\xi \, q_{\rho}''(\xi)}{q_{\rho}'(\xi)} \ge 0$$ and $\left| \frac{zp'(z)}{q_{\rho}'(\xi)} \right| \le n$ where $\xi \in \partial U \setminus E(q)$ . If $\Xi \in \Xi_n[h, q_\rho]$ and $\Xi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z, \zeta)$ is analytic in U, then $$\Xi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z, \zeta) \prec \prec h(z)$$ implies $$p(z) \prec q(z)$$ . We next specify the connection between the best dominant of a strong differential subordination and the solution of a corresponding differential equation.

Theorem 3. Consider $p \in \mathcal{H}[b,n]$ , $\Xi : \mathbb{C}^4 \times \mathcal{U} \times \overline{\mathcal{U}} \longrightarrow \mathbb{C}$ and that $$\Xi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z, \zeta)$$ is analytic in U. Suppose h is univalent in U and the differential equation $$\Xi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z, \zeta) = h(z)$$ (19) has a solution $q \in \mathcal{Q}(b)$ and $$\Re \frac{\xi \, q''(\xi)}{q'(\xi)} \geq 0$$ and $\left| \frac{z p'(z)}{q'(\xi)} \right| \leq n$ , where $\xi \in \partial U \setminus E(q)$ . If $\Xi \in \Xi_n[h, q]$ , then $$\Xi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z, \zeta) \prec \prec h(z)$$ (20) implies that $$p(z) \prec q(z)$$ and q is the best dominant. Proof. From Theorem 1, we have that q is a dominant of (20). Again, q fulfills (19) and it is a solution of (20). Thus, q will be dominated by all dominants of (20). Therefore, q is the best dominant. $\Box$ We further pursue the family of admissible functions and theorems, when $q(\mathcal{U})$ is a disc. Since q is given by (14), the class denoted by $\Xi_n[\Pi, M, b]$ . When $\Pi = \Delta$ , the class denoted by $\Xi_n[M, b]$ . Since $q(w) = Me^{i\theta}$ with $0 \le \theta \le 2\pi$ when |w| = 1, from Lemma 2 we derived the following.

Theorem 4. Consider that the q given in (14) and p ∈ H[b, n] satisfy $$|zp'(z)||M+\bar{b}e^{i\theta}|^2\leq Mn\left[M^2-|b|^2\right],$$ where z ∈ U and 0 ≤ θ ≤ 2π. If Ξ ∈ Ξn[Π, M, b], then $$\Xi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z, \zeta) \subset \Pi$$ implies $$p(z) \prec q(z)$$ . Next, we obtain the following corollary when b = 0 in Theorem 4.

Corollary 3. Consider that q(w) = Mw and p ∈ H[0, n] fulfill $$|zp'(z)| \leq Mn$$ when z ∈ U. If Π is a set in C and Ξ ∈ Ξn[Π, M, 0] as characterized by (22), then $$\Xi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z, \zeta) \subset \Pi$$ implies $$p(z) \prec Mz$$ . In this particular case, Theorem 4 becomes

Theorem 5. Consider that the q given in (14) and p ∈ H[b, n] satisfy (17). If Π is a set in C and (i) Ξ ∈ Ξn[Π, M, b], then $$\Xi(p(z),zp'(z),z^2p''(z),z^3p'''(z);z,\zeta)\subset\Pi\Longrightarrow|p(z)|< M.$$ (ii) If Ξ ∈ Ξn[M, b], then $$|\Xi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z, \zeta)| < M \Longrightarrow |p(z)| < M.$$

Theorem 6. Consider $\mathcal{I}_{p,\delta}^{\lambda}f(z)\in\mathcal{H}[0,p]$ with $p\geq 2$ , $q\in\mathcal{Q}(0)$ and that they satisfy $$\Re \frac{\xi q''(\xi)}{q'(\xi)} \ge 0 \quad and \quad \left| \frac{z(\mathcal{I}_{p,\delta}^{\lambda} f(z))'}{q'(\xi)} \right| \le k, \tag{23}$$ when $\xi \in \partial U \setminus E(q)$ and $k \geq p$ . If $\Pi$ is a set in $\mathbb{C}$ , $\Theta \in \Theta_I[\Pi, q]$ and $f(z) \in \mathcal{A}(p)$ satisfies $$\Theta(\mathcal{I}_{p,\delta}^{\lambda}f(z),\mathcal{I}_{p,\delta}^{\lambda-1}f(z),\mathcal{I}_{p,\delta}^{\lambda-2}f(z),\mathcal{I}_{p,\delta}^{\lambda-3}f(z);z,\zeta)\subset\Pi,\tag{24}$$ then $$\mathcal{I}_{p,\delta}^{\lambda}f(z) \prec q(z).$$ Proof. Let $$g(z) := \mathcal{I}_{p,\delta}^{\lambda} f(z). \tag{25}$$ Differentiating (25) with respect to z, and using the identity (3), we obtain $$\mathcal{I}_{p,\delta}^{\lambda-1}f(z) = \frac{zg'(z) + \delta g(z)}{p+\delta}.$$ (26) Again, by differentiating (26), we have $$\mathcal{I}_{p,\delta}^{\lambda-2}f(z) = \frac{z^2 g''(z) + (1+2\delta)zg'(z) + \delta^2 g(z)}{(p+\delta)^2}.$$ (27) Further computations show that $$\mathcal{I}_{p,\delta}^{\lambda-3}f(z) = \frac{z^3 g'''(z) + 3(1+\delta)z^2 g''(z) + (1+3\delta+3\delta^2)zg'(z) + \delta^3 g(z)}{(p+\delta)^3}.$$ (28) Set the transformations from $\mathbb{C}^4$ to $\mathbb{C}$ by $$\alpha = r_1, \quad \beta = \frac{s_1 + \delta r_1}{p + \delta}, \quad \gamma = \frac{t_1 + (1 + 2\delta)s_1 + \delta^2 r_1}{(p + \delta)^2},$$ $$\eta = \frac{u_1 + 3(1 + \delta)t_1 + (1 + 3\delta + 3\delta^2)s_1 + \delta^3 r_1}{(p + \delta)^3}.$$ (29) Let $\Xi(r_1, s_1, t_1, u_1; z, \zeta) = \Theta(\alpha, \beta, \gamma, \eta; z, \zeta)$ $$=\Theta\left(r_{1},\frac{s_{1}+\delta r_{1}}{p+\delta},\frac{t_{1}+(1+2\delta)s_{1}+\delta^{2}r_{1}}{(p+\delta)^{2}},\frac{u_{1}+3(1+\delta)t_{1}+(1+3\delta+3\delta^{2})s_{1}+\delta^{3}r_{1}}{(p+\delta)^{3}};z,\zeta\right). \tag{30}$$ Using Equations (25)–(28), and from (30), we obtain $$\Xi(g(z),zg'(z),z^2g''(z),z^3g'''(z);z,\zeta)=\Theta(\mathcal{I}^{\lambda}_{p,\delta}f(z),\mathcal{I}^{\lambda-1}_{p,\delta}f(z),\mathcal{I}^{\lambda-2}_{p,\delta}f(z),\mathcal{I}^{\lambda-3}_{p,\delta}f(z);z,\zeta).$$ Therefore, the inclusion (24) leads to $$\Xi(g(z), zg'(z), z^2g''(z), z^3g'''(z); z, \zeta) \in \Pi.$$ Now, $$\frac{t_1}{s_1} + 1 = \frac{(p+\delta)^2 \gamma - \delta^2 \alpha}{(p+\delta)\beta - \delta \alpha} - 2\delta$$ and $$\frac{u_1}{s_1} = \frac{(p+\delta)^2(\eta(p+\delta) - 3\gamma(1+\delta)) + (3+2\delta)\delta^2\alpha}{(p+\delta)\beta - \delta\alpha} + 2 + 3(2+\delta)\delta.$$ Hence, the admissibility condition in Definition 5 for $\Theta \in \Theta_I[\Pi, q]$ is equivalent to Definition 3. Thus, by use of (23) and applying Theorem 1, we obtain $$g(z) \prec q(z)$$ or $$\mathcal{I}_{v,\delta}^{\lambda}f(z) \prec q(z).$$ The hypothesis of Theorem 6 requires that the behavior of q on the boundary is not known.

Corollary 4. Consider q to be univalent in $\mathcal{U}$ , with q(0)=0, and set $q_{\rho}(z)\equiv q(\rho z)$ for $\rho\in(0,1)$ . Let $\mathcal{I}^{\lambda}_{p,\delta}f(z)\in\mathcal{H}[0,p]$ for $p\geq 2$ and let $\mathcal{I}^{\lambda}_{p,\delta}f(z)$ and $q_{\rho}$ satisfy (23). If $\Pi$ is a set in $\mathbb{C}$ and $\Theta\in\Theta_{I}[\Pi,q_{\rho}]$ and $f(z)\in\mathcal{A}(p)$ fulfill $$\Theta(\mathcal{I}^{\lambda}_{p,\delta}f(z),\mathcal{I}^{\lambda-1}_{p,\delta}f(z),\mathcal{I}^{\lambda-2}_{p,\delta}f(z),\mathcal{I}^{\lambda-3}_{p,\delta}f(z);z,\zeta)\subset\Pi,$$ then $$\mathcal{I}_{p,\delta}^{\lambda}f(z) \prec q(z).$$ Proof. Proof of the corollary is an immediate consequence of using Theorem 6, and we obtain $$\mathcal{I}_{p,\delta}^{\lambda}f(z) \prec \mathsf{q}_{\varrho}(z).$$ Since $q_o \prec q$ , we conclude that $$\mathcal{I}_{p,\delta}^{\lambda}f(z) \prec q(z).$$ In Definition 5, there are no special conditions on $\Pi$ . When $\Pi \neq \mathbb{C}$ , then there is some conformal mapping h of $\mathcal{U}$ onto $\Pi$ . Let it be denoted by $\Theta_I[h,q]$ . We then obtain the results that are an immediate consequence of Theorem 6 and Corollary 4.

Theorem 7. Consider that $\mathcal{I}^{\lambda}_{p,\delta}f(z)\in\mathcal{H}[0,p]$ with $p\geq 2$ and $q\in\mathcal{Q}(0)$ and that they satisfy (23). If $\Pi$ is a set in $\mathbb{C}$ , $\Theta\in\Theta_I[\Pi,q]$ , $f(z)\in\mathcal{A}(p)$ and $$\Theta(\mathcal{I}_{p,\delta}^{\lambda}f(z),\mathcal{I}_{p,\delta}^{\lambda-1}f(z),\mathcal{I}_{p,\delta}^{\lambda-2}f(z),\mathcal{I}_{p,\delta}^{\lambda-3}f(z);z,\zeta)$$ is analytic in U, then $$\Theta(\mathcal{I}^{\lambda}_{p,\delta}f(z),\mathcal{I}^{\lambda-1}_{p,\delta}f(z),\mathcal{I}^{\lambda-2}_{p,\delta}f(z),\mathcal{I}^{\lambda-3}_{p,\delta}f(z);z,\zeta)\prec\prec h(z)$$ implies $$\mathcal{I}_{p,\delta}^{\lambda}f(z) \prec q(z).$$

Corollary 5. Consider q to be univalent in $\mathcal{U}$ , with q(0)=0, and set $q_{\rho}(z)\equiv q(\rho z)$ for $\rho\in(0,1)$ . Let $\mathcal{I}^{\lambda}_{p,\delta}f(z)\in\mathcal{H}[0,p]$ for $p\geq 2$ and let $\mathcal{I}^{\lambda}_{p,\delta}f(z)$ and $q_{\rho}$ satisfy (23). If $\Pi$ is a set in $\mathbb{C},\Theta\in\Theta_I[\Pi,q_{\rho}]$ , $f(z)\in\mathcal{A}(p)$ and $$\Theta(\mathcal{I}^{\lambda}_{p,\delta}f(z),\mathcal{I}^{\lambda-1}_{p,\delta}f(z),\mathcal{I}^{\lambda-2}_{p,\delta}f(z),\mathcal{I}^{\lambda-3}_{p,\delta}f(z);z,\zeta)$$ is analytic in U, then $$\Theta(\mathcal{I}^{\lambda}_{p,\delta}f(z),\mathcal{I}^{\lambda-1}_{p,\delta}f(z),\mathcal{I}^{\lambda-2}_{p,\delta}f(z),\mathcal{I}^{\lambda-3}_{p,\delta}f(z);z,\zeta)\prec\prec h(z)$$ implies $$\mathcal{I}_{p,\delta}^{\lambda}f(z) \prec q(z).$$ We next indicate the connection between the best dominant and the solution of a strong differential subordination.

Theorem 8. Consider that $\mathcal{I}_{p,\delta}^{\lambda}f(z)\in\mathcal{H}[0,p]$ with $p\geq 2$ , $\Theta:\mathbb{C}^4\times\mathcal{U}\times\overline{\mathcal{U}}\longrightarrow\mathbb{C}$ and that $$\Theta(\mathcal{I}^{\lambda}_{p,\delta}f(z),\mathcal{I}^{\lambda-1}_{p,\delta}f(z),\mathcal{I}^{\lambda-2}_{p,\delta}f(z),\mathcal{I}^{\lambda-3}_{p,\delta}f(z);z,\zeta)$$ is analytic in $\mathcal{U}$ . Suppose h is univalent in $\mathcal{U}$ and $q \in \mathcal{Q}(0)$ is a solution of the following differential equation $$\Theta(\mathcal{I}_{p,\delta}^{\lambda}f(z), \mathcal{I}_{p,\delta}^{\lambda-1}f(z), \mathcal{I}_{p,\delta}^{\lambda-2}f(z), \mathcal{I}_{p,\delta}^{\lambda-3}f(z); z, \zeta) = h(z)$$ (31) and satisfies (23). If $\Pi$ is a set in $\mathbb{C}$ , $\Theta \in \Theta_I[h, q]$ and $f(z) \in \mathcal{A}(p)$ fulfills $$\Theta(\mathcal{I}_{p,\delta}^{\lambda}f(z), \mathcal{I}_{p,\delta}^{\lambda-1}f(z), \mathcal{I}_{p,\delta}^{\lambda-2}f(z), \mathcal{I}_{p,\delta}^{\lambda-3}f(z); z, \zeta) \prec \prec h(z), \tag{32}$$ then $$\mathcal{I}_{v,\delta}^{\lambda}f(z) \prec q(z)$$ and q is the best dominant. Proof. From Theorem 6, we conclude that q is a dominant of (32). Since q satisfies (31), q is a solution of (32). Thus, q is dominated by all dominants of (32). Therefore, q is the best dominant. $\Box$ Our next outcomes are for the specialized case of q being a disc, where q is given by (14) and the class $\Theta_I[\Pi, M, b]$ . Also, we denote the class $\Theta_I[M, b]$ , when $\Pi = \Delta$ . And $q(w)=Me^{i\theta}$ with $0\leq\theta\leq 2\pi$ when |w|=1. Notably, the case q(z)=Mz, M>0 denotes the admissible functions class $\Theta_I[\Pi,M]$ .

Corollary 6. Consider q(z) = Mz and $\mathcal{I}_{n,\delta}^{\lambda} f(z) \in \mathcal{H}[0,p]$ with $p \geq 2$ to satisfy $$|z(\mathcal{I}_{p,\delta}^{\lambda}f(z))'| \leq Mk$$ , when $z \in \mathcal{U}$ and $k \geq p$ . If $\Theta \in \Theta_I[\Pi, M]$ , $f(z) \in \mathcal{A}(p)$ satisfies $$\Theta(\mathcal{I}_{p,\delta}^{\lambda}f(z),\mathcal{I}_{p,\delta}^{\lambda-1}f(z),\mathcal{I}_{p,\delta}^{\lambda-2}f(z),\mathcal{I}_{p,\delta}^{\lambda-3}f(z);z,\zeta)\subset\Pi,$$ then $$\mathcal{I}_{n,\delta}^{\lambda}f(z) \prec q(z).$$

Corollary 7. Consider q(z) = Mz and $\mathcal{I}^{\lambda}_{p,\delta} f(z) \in \mathcal{H}[0,p]$ with $p \geq 2$ . If $\Pi$ is a set in $\mathbb{C}$ and (i) $\Theta \in \Theta_I[\Pi,M]$ , $f(z) \in \mathcal{A}(p)$ satisfies $$\Theta(\mathcal{I}_{p,\delta}^{\lambda}f(z),\mathcal{I}_{p,\delta}^{\lambda-1}f(z),\mathcal{I}_{p,\delta}^{\lambda-2}f(z),\mathcal{I}_{p,\delta}^{\lambda-3}f(z);z,\zeta)\subset\Pi\Longrightarrow|p(z)|< M.$$ (ii) If $f(z) \in \mathcal{A}(p)$ and $\Theta \in \Theta_I[M]$ , it satisfies $$|\Theta(\mathcal{I}_{p,\delta}^{\lambda}f(z),\mathcal{I}_{p,\delta}^{\lambda-1}f(z),\mathcal{I}_{p,\delta}^{\lambda-2}f(z),\mathcal{I}_{p,\delta}^{\lambda-3}f(z);z,\zeta)| < M \Longrightarrow |p(z)| < M.$$

Definition 1. A function $f \in A$ is said to be in the class $\mathcal{GS}^*_{\mathcal{F}\psi}(m,\beta)$ , where $m \in [-1,1]$ and $\beta \geq 1$ , if the following subordination is satisfied $$Re \frac{\mathcal{G}\left(\mathcal{F}'_{\psi}\right)(z)}{\mathcal{F}_{\psi}(z)} \prec \sqrt{1+z}$$ (22) with the condition $\mathcal{G}_{\psi}(0)=0$ . The function $\mathcal{F}_{\psi}(z)$ , defined by (18), is a convolution of (7) and (17). The class of functions $\mathcal{G}S_{\mathcal{F}_{\leftarrow}}^*(m,\beta)$ , defined above, is investigated in the next section in relation to the Carathéodory function p(z), the modified sigmoid function and Bell numbers by means of the subordination principle, and initial coefficient estimates are obtained. Furthermore, those results are used for investigating the Fekete–Szegö problem.

Definition 1. If the following subordinations are met, a function $f \in \Sigma$ given by (5) is said to belong to the class $\mathfrak{G}^t_{\Sigma}(x, p, q, \lambda, \beta, \gamma)$ : $$(1-\gamma)\frac{\mathbb{P}_{\lambda}f(\xi)}{\xi} + \gamma(\mathbb{P}_{\lambda}f(\xi))' + \beta\xi(\mathbb{P}_{\lambda}f(\xi))'' \prec \Omega(x,\xi) + 1 - a \tag{11}$$ and $$(1-\gamma)\frac{\mathbb{P}_{\lambda}f(w)}{w} + \gamma(\mathbb{P}_{\lambda}f(w))' + \beta w(\mathbb{P}_{\lambda}f(w))'' \prec \Omega(x,w) + 1 - a, \tag{12}$$ where $\xi, w \in \mathfrak{U}$ , $\gamma, \beta \geq 0$ , $x \in \mathbb{R}$ , and the function $g = f^{-1}$ is given by (6). Example 1. For $\beta = 0$ , we have, $\mathfrak{G}^t_{\Sigma}(x, p, q, \lambda, 0, \gamma) = \mathfrak{G}^t_{\Sigma}(x, p, q, \lambda, \gamma)$ , in which $\mathfrak{G}^t_{\Sigma}(x, p, q, \lambda, \gamma)$ indicates the group of functions $f \in \Sigma$ given by (5) and satisfying the criterion below. $$(1 - \gamma) \frac{\mathbb{P}_{\lambda} f(\xi)}{\xi} + \gamma (\mathbb{P}_{\lambda} f(\xi))' \prec \Omega(x, \xi) + 1 - a \tag{13}$$ and $$(1-\gamma)\frac{\mathbb{P}_{\lambda}f(w)}{w} + \gamma(\mathbb{P}_{\lambda}f(w))' \prec \Omega(x,w) + 1 - a, \tag{14}$$ where $\xi, w \in \mathfrak{U}$ , $\gamma \geq 0$ , $x \in \mathbb{R}$ , and the function $g = f^{-1}$ is given by (6). Example 2. For $\beta=0$ and $\gamma=1$ , we have, $\mathfrak{G}^t_{\Sigma}(x,p,q,\lambda,1)=\mathfrak{G}^t_{\Sigma}(x,p,q,\lambda)$ , in which $\mathfrak{G}^t_{\Sigma}(x,p,q,\lambda)$ denotes the class of functions $f\in\Sigma$ given by (5) and satisfying the following condition $$\left(\mathbb{P}_{\lambda} f(\xi)\right)' \prec \Omega(x, \xi) + 1 - a \tag{15}$$ and $$\left(\mathbb{P}_{\lambda} f(w)\right)' \prec \Omega(x, w) + 1 - a,\tag{16}$$ where $\xi, w \in \mathfrak{U}$ , $x \in \mathbb{R}$ , and the function $g = f^{-1}$ is given by (6). Example 3. For $\beta = 0$ and $\gamma = 0$ , we have, $\mathfrak{G}^t_{\Sigma}(x, p, q, \lambda, 0, 0) = \mathfrak{G}^t_{\Sigma}(x, p, q, \lambda, 0)$ , in which $\mathfrak{G}^t_{\Sigma}(x, p, q, \lambda, 0)$ indicates the group of functions $f \in \Sigma$ given by (5) and satisfying the criterion below. $$\frac{\mathbb{P}_{\lambda}f(\xi)}{\xi} \prec \Omega(x,\xi) + 1 - a \tag{17}$$ and $$\frac{\mathbb{P}_{\lambda}f(w)}{w} \prec \Omega(x, w) + 1 - a,\tag{18}$$ where $\xi$ , $w \in \mathfrak{U}$ , $x \in \mathbb{R}$ , and the function $g = f^{-1}$ is given by (6). Example 4. For $\lambda = 1$ , we have, $\mathfrak{G}^t_{\Sigma}(x, p, q, 1, \beta, \gamma) = \mathfrak{G}^t_{\Sigma}(x, p, q, \beta, \gamma)$ , in which $\mathfrak{G}^t_{\Sigma}(x, p, q, \beta, \gamma)$ indicates the group of functions $f \in \Sigma$ given by (5) and satisfying the criterion below. $$(1-\gamma)\frac{\mathbb{P}_1 f(\xi)}{\xi} + \gamma (\mathbb{P}_1 f(\xi))' + \beta \xi (\mathbb{P}_1 f(\xi))'' \prec \Omega(x,\xi) + 1 - a \tag{19}$$ and $$(1 - \gamma) \frac{\mathbb{P}_1 f(w)}{w} + \gamma (\mathbb{P}_1 f(w))' + \beta w (\mathbb{P}_1 f(w))'' \prec \Omega(x, w) + 1 - a, \tag{20}$$ where $\xi, w \in \mathfrak{U}$ , $\gamma \geq 0$ , $x \in \mathbb{R}$ , and the function $g = f^{-1}$ is given by (6). First, we give the coefficient estimates for the class $\mathfrak{G}^t_{\Sigma}(x, p, q, \lambda, \beta, \gamma)$ given in Definition 1.

Definition 1. (See [2].) Let $f = z|z|^{2\beta}h(z)\overline{g(z)}$ be a univalent log-harmonic mapping. We say that f is a starlike log-harmonic mapping of order $\alpha$ if $$\frac{\partial \arg f(re^{i\theta})}{\partial \theta} = \operatorname{Re} \frac{zf_z - \overline{z}f_{\overline{z}}}{f} > \alpha, \qquad 0 \le \alpha < 1$$ for all $z \in E$ . Denote by $ST_{LH}(\alpha)$ the class of all starlike log-harmonic mappings. By taking $\beta = 0$ and g(z) = 1 in Definition 1, we obtain the class of starlike analytic functions in $\mathcal{A}$ , which we denote by $S^*(\alpha)$ . The following lemma shows the relationship of the classes $ST_{LH}(\alpha)$ and $S^*(\alpha)$ .

Definition 2. Let $f(z) = z|z|^{2\beta}h(z)\overline{g(z)}$ be a univalent log-harmonic mapping on E, with $0 \notin hg(E)$ . Then, we say that f is an $\alpha - spirallike$ log-harmonic mapping of order $\rho$ $(0 \le \rho < 1)$ if $$\operatorname{Re}\left(e^{-i\alpha}\frac{zf_z-\overline{z}f_{\overline{z}}}{f(z)}\right) > \rho\cos\alpha \qquad (z \in E)$$ for some real $\alpha(|\alpha| < \frac{\pi}{2})$ . The class of these functions is denoted by $S_{LH}^{\alpha}(\rho)$ . Furthermore, we define $S_{LH}^{\alpha}(1) = \bigcap_{0 < \rho < 1} S_{LH}^{\alpha}(\rho)$ . Additionally, we denote by $S^{\alpha}(\rho)$ the subclass of all $f \in \mathcal{A}$ such that f is $\alpha$ -spiralike of order $\rho$ and $S^{\alpha}(1) = \bigcap_{0 \le \rho \le 1} S^{\alpha}(\rho)$ .

Definition 1. A real-valued function f, defined on the interval [0,a] (a>0), is said to be a q-completely convex function if $f \in C_a^{\infty}[0,a]$ and $$(-1)^n D_{q-1}^{2n} f(aq^k) \ge 0 \quad (\text{for all } \{n,k\} \subset \mathbb{N}_0).$$ (28) Example 1. The functions $f(x) = Sin_q \xi_1 x$ , defined in (9), are q-completely convex on the interval [0,1]. Indeed, one can verify that $$(-1)^n D_{q^{-1}}^{2n} f(x) = (-1)^n D_{q^{-1}}^{2n} Sin_q \xi_1 x = \xi_1^{2n} Sin_q(\xi_1 x) > 0,$$ (29) for all $x \in [0,1]$ and $n \in \mathbb{N}_0$ . In the following, we prove certain properties of q-completely convex functions.

Definition 2. A real-valued function $f \in C_q^{\infty}[0,1]$ is a minimal q-completely convex on [0,1] if it is q-completely convex in the interval [0,1], and if the function $g(x) = f(x) - \epsilon \operatorname{Sin}_q \xi_1 x$ is not q-completely convex for any $\epsilon > 0$ . For example, the function $f(x) = \operatorname{Sin}_q x$ is a minimal q-completely convex in $0 \le x \le 1$ while the function $f(x) = \operatorname{Sin}_q \xi_1 x$ is not because for any $0 < \epsilon < 1$ and $x \in (0,1)$ , $$(-1)^n D_{q^{-1}}^{2n} \left( \operatorname{Sin}_q \xi_1 x - \epsilon \operatorname{Sin}_q \xi_1 x \right) = (1 - \epsilon) \xi_1^{2n} \operatorname{Sin}_q (\xi_1 x) > 0.$$

Definition 1. Suppose that $0 \le \eta \le 1$ , $0 \le \lambda \le 1$ , $0 \le \delta \le 1$ and h is analytic in $\mathbb{D}$ , h(0) = 1. $f \in \Sigma$ is in the class $\mathcal{W}_{\Sigma}(\eta, \delta, \lambda, \sigma, \theta, \alpha, \beta, p, q; h)$ if it provides the subordinations: $$\left(\frac{z\Big(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\Big)'}{W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)}\right)^{\eta}\left[(1-\delta)\frac{z\Big(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\Big)'}{W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)}+\delta\left(1+\frac{z\Big(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\Big)''}{\Big(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\Big)'}\right)\right]^{\lambda} \prec h(z)$$ and $$\left(\frac{w\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\right)'}{W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)}\right)^{\eta}\left[(1-\delta)\frac{w\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\right)'}{W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)}+\delta\left(1+\frac{w\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\right)''}{\left(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\right)'}\right)\right]^{\lambda}\prec h(w),$$ where $f^{-1}$ is given by (2).

Definition 2. Suppose that $0 \le \xi \le 1$ , $0 \le \rho < 1$ and h is analytic in $\mathbb{D}$ , h(0) = 1. $f \in \Sigma$ is in the class $\mathcal{K}_{\Sigma}(\xi, \rho, \sigma, \theta, \alpha, \beta, p, q; h)$ if it provides the subordinations: $$\begin{split} &(1-\xi)\frac{z\Big(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\Big)'}{(1-\rho)W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)+\rho z\Big(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\Big)'}\\ +&\xi\Bigg(\frac{\Big(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\Big)'+z\Big(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\Big)''}{\Big(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\Big)'+\rho z\Big(W_{\alpha,\beta,p,q}^{\sigma,\theta}f(z)\Big)''}\Bigg)\prec h(z) \end{split}$$ and $$\begin{split} &(1-\xi)\frac{w\Big(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\Big)'}{(1-\rho)W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)+\rho w\Big(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\Big)'}\\ +&\xi\Bigg(\frac{\Big(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\Big)'+w\Big(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\Big)''}{\Big(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\Big)'+\rho w\Big(W_{\alpha,\beta,p,q}^{\sigma,\theta}f^{-1}(w)\Big)''}\Bigg) \prec h(w), \end{split}$$ where $f^{-1}$ is given by (2).

Definition 1. The $[\varkappa]_q$ denotes the basic (or q-) number, where 0 < q < 1 is defined as follows: $$[\varkappa]_q = \left\{ \begin{array}{ll} (1-q^{\varkappa})(1-q)^{-1} & , \ \varkappa \in \mathbb{C}\backslash\{0\} \\ \\ 0 & , \ \varkappa = 0 \\ \\ q^{\mathbb{k}-1} + q^{\mathbb{k}-2} + \dots + q^2 + q + 1 = \sum\limits_{i=0}^{\mathbb{k}-1} q^i & , \ \varkappa = \mathbb{k} \in \mathbb{N}. \end{array} \right.$$ It is obvious from Definition 1 that $\lim_{q\to 1^-} [\mathbb{k}]_q = \lim_{q\to 1^-} \frac{1-q^{\mathbb{k}}}{1-q} = \mathbb{k}$ .

Definition 2. ([51]). The q-difference operator (or q-derivative) of a function f is defined by $$\partial_q(\Phi(\eta)) = \left\{ \begin{array}{ll} \frac{\Phi(\eta) - \Phi(q\eta)}{\eta - q\eta} & \eta \in \mathbb{C} \setminus \{0\} \\ 1 & \eta = 0. \end{array} \right.$$ We note that $\lim_{q\to 1^-}\partial_q\Phi(\eta)=\Phi'(\eta)$ if $\Phi$ is differentiable for all $\eta\in\mathbb{C}$ . One can easily see that $$\partial_{q} \left\{ \sum_{k=2}^{\infty} \varrho_{k} \eta^{k} \right\} = \sum_{k=2}^{\infty} [k]_{q} \varrho_{k} \eta^{k-1}, \qquad (k \in \mathbb{N}, \eta \in \nabla), \tag{6}$$ and $$\partial_{q}^{\kappa} \left\{ \sum_{k=2}^{\infty} \varrho_{k} \eta^{k} \right\} = \partial_{q} \left( \partial_{q}^{\kappa - 1} \left\{ \sum_{k=2}^{\infty} \varrho_{k} \eta^{k} \right\} \right) = \varrho_{k}[k]_{q}!, \qquad (k \in \mathbb{N}). \tag{7}$$ In 2019, Alsoboh and Darus [48] introduced the q-derivative operator $Y_{q,\mu,\beta,\gamma,\delta}^{(\ell)}: \mathbf{A} \to \mathbf{A}$ , as: $$Y_{q,\mu,\beta,\gamma,\delta}^{(\ell)}\Phi(\eta) = z + \sum_{k=2}^{\infty} \Omega_k^{\ell} \varrho_k \eta^k, \tag{8}$$ where $$\Omega_{\mathbb{k}}^{\ell} = \left[ (\gamma - \delta)(\beta - \mu)([\mathbb{k}]_q - 1) + 1 \right]^{\ell}, \tag{9}$$ and $(\mu, \beta, \gamma, \delta \ge 0, \gamma > \delta, \beta > \mu, \ell \in \mathbb{N}_0, \eta \in \nabla)$ .

Definition 3. For $\mu, \beta, \gamma, \delta \geq 0$ , $\gamma > \delta, \beta > \mu$ and $k \in \mathbb{N}_0$ , a bi-univalent function $\Phi$ of the form (1) is in the class $D_{\Sigma_a}^{\ell}(\chi, \delta, \gamma, \mu; \varphi)$ if it satisfies the following subordination conditions: $$(1-\chi)\frac{Y_{q,\mu,\beta,\gamma,\delta}^{(\ell)}\Phi(\eta)}{\eta} + \chi \partial_q Y_{q,\mu,\beta,\gamma,\delta}^{(\ell)}\Phi(\eta) \prec \varphi(\eta), \quad (\eta \in \nabla; \chi \ge 1), \tag{11}$$ and $$(1-\chi)\frac{Y_{q,\mu,\beta,\gamma,\delta}^{(\ell)}h(\rho)}{\rho} + \chi \partial_q Y_{q,\mu,\beta,\gamma,\delta}^{(\ell)}h(\rho) \prec \varphi(\rho), \quad (\rho \in \nabla; \chi \ge 1), \tag{12}$$ where $h(\rho)$ and $Y_{q,\mu,\beta,\gamma,\delta}^{(\ell)}\Phi(\eta)$ are defined by (3) and (8), respectively. Example 1. A bi-univalent function f of the form (1) is referred to as being in the class $D_{\Sigma_q}^0(\chi, \delta, \gamma, \mu; \varphi) = D_{\Sigma}(q; \chi, \varphi)$ , if the following conditions of subordination are met: $$(1-\chi)\frac{\Phi(\eta)}{\eta} + \chi \partial_q \Phi(\eta) \prec \varphi(\eta), \quad (\eta \in \nabla),$$ and $$(1-\chi)\frac{h(\rho)}{\rho} + \chi \partial_q h(\rho) \prec \varphi(\rho), \quad (\rho \in \nabla),$$ where $h(\rho)$ is defined by (3). This class was introduced by Altınkaya and Yalçın [37]. Example 2. A bi-univalent function f of the form (1) is referred to as being in the class $\lim_{q\to 1^-} D^0_{\Sigma_q}(\chi, \delta, \gamma, \mu; \varphi) = R_{\sigma}(\chi, \varphi)$ , if the following conditions of subordination are met: $$(1-\chi)\frac{\Phi(\eta)}{\eta} + \chi(\Phi(\eta))' \prec \varphi(\eta), \quad (\eta \in \nabla),$$ and $$(1-\chi)\frac{h(\rho)}{\rho} + \chi(h(\rho))' \prec \varphi(\rho), \quad (\rho \in \nabla),$$ where $h(\rho)$ is defined by (3). This class was introduced by Kumar et al. [53].

Definition 1. The q-shifted factorial $(\varkappa;q)_l$ is presented as $$(\varkappa;q)_l = \prod_{j=0}^{l-1} \left(1 - \varkappa q^j\right) \qquad (l \in \mathbb{N}; \ \varkappa, q \in \mathbb{C}),\tag{3}$$ where, as usual, $\mathbb{C}$ is the set of complex numbers. If $\varkappa \neq q^{-m} \ (m \in \mathbb{N}_0 := \{0,1,2,3,\cdots\})$ , then $$(\varkappa;q)_{\infty} = \prod_{j=0}^{\infty} \left(1 - \varkappa q^{j}\right) \qquad (\varkappa \in \mathbb{C}; |q| < 1).$$ (4) In the case when $\varkappa \neq 0$ and $q \geq 1$ , $(\varkappa;q)_{\infty}$ diverges. Therefore, when we take $(\varkappa;q)_{\infty}$ , then we will assume that |q| < 1. Remark 1. For $q \to 1-in(\varkappa;q)_1$ , we have $$(\varkappa;q)_l=(\varkappa)_l=\prod_{i=0}^{l-1}(\varkappa+j)\qquad (l\in\mathbb{N}).$$ The q-factorial $[l]_q!$ is defined by $$[n]_q! = \prod_{l=1}^n [l]_q \qquad (l \in \mathbb{N}), \tag{5}$$ where the q-number $[l]_q$ is given below: $$[l]_q = \frac{1 - q^l}{1 - q} \qquad (l \in \mathbb{N}).$$ If l = 0, then $[l]_q! = 1$ .

Definition 2. The $(\varkappa;q)_l$ in (3) can be given more precisely in the form of the q-gamma function as follows: $$\Gamma_q(\varkappa) = \frac{(1-q)^{1-\alpha}(q;q)_{\infty}}{(q^a;q)_{\infty}} \qquad (0 < q < 1),$$ or $$(q^{\varkappa};q)_{l} = \frac{\left(1-q^{l}\right)\Gamma_{q}(\varkappa+l)}{\Gamma_{q}(\varkappa)} \qquad (l \in \mathbb{N}).$$

Definition 3. (Jackson [10]). For $h \in A$ , the q-difference operator is defined by $$D_q h(z) = \frac{h(z) - h(qz)}{z(1-q)} \qquad (z \in \mathbb{E}).$$ We recall that for $l \in \mathbb{N}$ and $z \in \mathbb{E}$ , we have $$D_q(z^l) = [l]_q z^{l-1}$$ and $D_q\left(\sum_{l=1}^{\infty} a_l z^l\right) = \sum_{l=1}^{\infty} [l]_q a_l z^{l-1}$ , where the q-number $[l]_q$ is already given along with (5). The q-generalized Pochhammer symbol is expressed as follows: $$[\varkappa]_{q,l} = \frac{\Gamma_q(\varkappa+l)}{\Gamma_q(\varkappa)} \qquad (l \in \mathbb{N}; \ \varkappa \in \mathbb{C}).$$ Remark 2. If $q \rightarrow 1-$ , then $$[\varkappa]_{q,l} = (\varkappa)_l = \frac{\Gamma(\varkappa + l)}{\Gamma(\varkappa)}.$$

Definition 4. (see [52]). For $\varrho > 0$ , the fractional q-integral operator is defined by $$I_{q,z}^{\varrho}h(z) = \frac{1}{\Gamma_{q}(\varrho)} \int_{0}^{z} (z - tq)_{\varrho - 1} h(t) d_{q}(t), \tag{6}$$ where $(z - tq)_{\rho-1}$ is given by $$(z-tq)_{\varrho-1} = z^{\varrho-1}{}_1\Phi_0\bigg(q^{-\varrho+1}; \dots; q, \frac{tq^{\varrho}}{z}\bigg).$$ The representation of the q-binomial series $_1\Phi_0$ is given by $$_{1}\Phi_{0}(a; --; q, z) = 1 + \sum_{l=1}^{\infty} \frac{(a, q)_{l}}{(q, q)_{l}} z^{l}$$ $(|q| < 1; |z| < 1).$

Definition 5. (see, for example, [53,54]). For an analytic function h, the fractional q-derivative operator $\mathfrak{D}_q$ of order $\varrho$ is described by $$\begin{split} \mathfrak{D}_{q}h(z) &= D_{q}I_{q,z}^{1-\varrho}h(z) \\ &= \frac{1}{\Gamma_{q}(1-\varrho)}D_{q}\int_{0}^{z}(z-tq)_{-\varrho}h(t)d_{q}(t) \qquad (0 \leq \varrho < 1). \end{split}$$ In Geometric Function Theory, linear operators (both derivative and integral operators) are extensively utilized. The most important aspect of this study is that we are simultaneously examining the characteristics of many classes of analytic functions under a certain linear operator. Taking the aforementioned importance of linear operators into consideration, we now define the operator below.

Definition 6. The extended fractional q-derivative $\mathfrak{D}_q^\varrho$ of order $\varrho$ is specified as follows: $$\mathfrak{D}_{q}^{\varrho}h(z) = D_{q}^{m}I_{q,z}^{m-\varrho}h(z),\tag{7}$$ where m is assumed to be the smallest integer. We find from (7) that $$\mathfrak{D}_{q}^{\varrho}z^{l} = \frac{\Gamma_{q}(l+1)}{\Gamma_{q}(l+1-\varrho)}z^{l-\varrho} \qquad (0 \le \varrho; \ l > -1).$$ Remark 3. For $-\infty < \varrho < 0$ , $\mathfrak{D}_q^{\varrho}$ denotes a fractional q-integral of h of order $\varrho$ . Additionally, for $0 \le \varrho < 2$ , $\mathfrak{D}_q^{\varrho}$ denotes a q-derivative of h of order $\varrho$ .

Definition 7. Following the work of Selvakumaran et al. [55], we introduce the $(\varrho, q)$ -differintegral operator $\Omega_q^{\varrho}: \mathcal{A} \to \mathcal{A}$ , which they defined as follows: $$\Omega_{q}^{\varrho}h(z) = \frac{\Gamma_{q}(2-\varrho)}{\Gamma_{q}(2)} z^{\varrho} \mathfrak{D}_{q}^{\varrho}h(z) = z + \sum_{l=2}^{\infty} \frac{\Gamma_{q}(2-\varrho)\Gamma_{q}(l+1)}{\Gamma_{q}(2)\Gamma_{q}(l+1-\varrho)} a_{l}z^{l} \qquad (z \in \mathbb{E}),$$ (8) where $0 \le \varrho < 2$ and 0 < q < 1. Each of the following properties of the $(\varrho, q)$ -differintegral operator $\Omega_q^{\varrho} h$ are worthy of note. Property 1. $$\lim_{\varrho \to 1} \mathbf{\Omega}_q^{\varrho} h(z) = \mathbf{\Omega}_q^{\varrho} h(z) = z D_q h(z).$$ Property 2. $$\mathbf{\Omega}_q^\varrho\Big(\Omega_q^\delta h(z)\Big) = \Omega_q^\delta\Big(\mathbf{\Omega}_q^\varrho h(z)\Big) = z + \sum_{l=2}^\infty \frac{\Gamma_q(2-\varrho)\Gamma_q(2-\delta)\big(\Gamma_q(l+1)\big)^2}{\Gamma_q(2)\Gamma_q(l+1-\varrho)\Gamma_q(l+1-\delta)} a_l z^l.$$ Property 3. $$\frac{D_q\left(\Omega_q^{\varrho}h(z)\right)}{\Omega_q^{\varrho}h(z)} = \begin{cases} \frac{zD_qh(z)}{h(z)} & (\varrho = 0)\\ \\ 1 + z\frac{D_q\left(D_qh(z)\right)}{D_qh(z)} & (\varrho = 1). \end{cases}$$ Considering the operator $\Omega_q^\varrho$ defined in Definition 7 and inspired by the work given in [27], a new subclass of the class $\Sigma$ is introduced by means of this operator. The next section will provide proofs of the original findings by using the Faber polynomial method and one lemma.

Definition 8. Let the function h be of the form (1). Then, h is referred to as $\varrho$ -fractional bi-close-to-convex function in $\mathbb{E}$ if a suitable function $g \in \mathcal{S}^*$ exists such that $$\Re\left(\frac{D_q\left(\Omega_q^\varrho h(z)\right)}{g(z)}\right) > \alpha$$ and $$\Re\left(\frac{D_q\left(\mathbf{\Omega}_q^\varrho\alpha(\vartheta)\right)}{\delta(\vartheta)}\right)>\alpha,$$ where $0 \le \alpha < 1$ , $0 \le \varrho < 2$ and $z, \vartheta \in \mathbb{E}$ . All such functions are symbolized by $\mathcal{K}_{\Sigma}(q, \alpha, \varrho)$ . Remark 4. If we let $q \to 1-$ and $\varrho=0$ , then $\mathcal{K}_{\Sigma}(q,\alpha,\varrho)$ reduces to the class introduced by Hamidi and Jahangiri in [27]. Remark 5. If $q \to 1-$ and $\alpha = 0$ , then $\mathcal{K}_{\Sigma}(q, \alpha, \varrho)$ reduces to the class introduced by Sakar and Güney in [56].

Definition 1. Let $F \in A$ . Then, $F \in S_{BS}^*$ if and only if $$\frac{\xi F'(\xi)}{F(\xi)} \prec \left(\frac{\xi}{e^{\xi}-1}\right)^2 = \varphi_{BS}(\xi), \quad \xi \in \mathbf{E}.$$ In other words, a function $F \in \mathbf{S}_{BS}^*$ can be written as $$F(\xi) = \xi \exp\left(\int_0^{\xi} \frac{\varphi(s) - 1}{s} ds\right),\tag{3}$$ where $\varphi$ is analytic and satisfies $\varphi(\xi) \prec \varphi_{BS}(\xi) = (\frac{\xi}{e^{\xi}-1})^2$ $(\xi \in \mathbf{E})$ . To give some examples of functions in the class $\mathbf{S}_{BS}^*$ , consider $$\varphi_1(\xi) = 1 + \frac{\xi}{4}, \quad \varphi_2(\xi) = \frac{5 + 2\xi}{5 + \xi}, \quad \varphi_3(\xi) = \frac{3 + \xi e^{\xi}}{3}, \quad \varphi_4(\xi) = 1 + \frac{\xi \cos(\xi)}{4}.$$ The function $\varphi_0(\xi) = (\frac{\xi}{\rho \xi - 1})^2$ is univalent in E, $\varphi_i(0) = \varphi_0(0)$ (i = 1, 2, 3, 4) and $\varphi_i(E) \subset$ $\varphi_0(\mathbf{E})$ ; it is easy to conclude that $\varphi_i(\xi) \prec \varphi_0(\xi)$ . The functions $F_i \in \mathbf{S}_{BS}^*$ corresponding to every $\varphi_i$ respectively, are given as follows: $$\begin{split} F_1(\xi) &= \xi e^{\xi/4}, \quad F_2(\xi) = \xi \left( 1 + \frac{\xi}{5} \right), \\ F_3(\xi) &= \xi \exp\left( \frac{e^{\xi} - 1}{3} \right), \quad F_4(\xi) = \xi \exp\left( \frac{\sin(\xi)}{4} \right). \end{split}$$ In particular, if $\varphi_0(\xi) = \left(\frac{\xi}{e^{\xi}-1}\right)^2$ , then (3) takes the form $$F_0(\xi) = \xi \exp\left(\int_0^{\xi} \frac{\varphi_0(s) - 1}{s} ds\right) = \xi - \xi^2 + \frac{17\xi^3}{24} - \frac{29\xi^4}{72} + \frac{377\xi^5}{1920} - \frac{11\xi^6}{120} + \cdots$$ (4) The above function acts as an extremal function for $S_{RS}^*$ . The following theorem gives the sharp estimates for $\varphi_{BS}$ :

Definition 1. The f function belonging to the class $\Sigma$ in type (1) is said to belong to the class $\mathcal{B}^{\sigma,m,\wp}_{i,\delta,\Sigma}(\lambda,z)$ if f satisfies the following inequalities: $$\left(\frac{z^{1-\lambda} (\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1} f(z))'}{[\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1} f(z)]^{1-\lambda}}\right) \prec \Xi(z) \tag{9}$$ and $$\left(\frac{w^{1-\lambda}(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}g(w))'}{[\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}g(w)]^{1-\lambda}}\right) \prec \Xi(w), \tag{10}$$ here $0 \le \lambda \le 1$ ; $z, w \in \mathbb{U}$ and it is assumed that the g function is as in (8). The new subclasses of the $\Sigma$ class created by the special selection of the parameters in this definition can be defined as in the following two examples. Example 1. For $\lambda = 0$ , the f function belonging to the class $\Sigma$ in type (1) is said to belong to the class $\mathcal{S}^{\sigma,m;\wp}_{i,\delta\Sigma}$ if f satisfies the following inequalities: $$\left(\frac{z(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}f(z))'}{\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}f(z)}\right) \prec \Xi(z) \tag{11}$$ and $$\left(\frac{w(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}g(w))'}{\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}g(w)}\right) \prec \Xi(w)$$ (12) in here $z, w \in \mathbb{U}$ and it is assumed that the g function is as in (8). Example 2. For $\lambda = 1$ , the f function belonging to the class $\Sigma$ in type (1) is said to belong to the class $\mathcal{R}_{i,\delta,\Sigma}^{\sigma,m;\wp}$ if f satisfies the following inequalities: $$\left(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}f(z))'\right) \prec \Xi(z) \tag{13}$$ and $$\left( \left( \mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1} g(w) \right)' \right) \prec \Xi(w) \tag{14}$$ in here $z, w \in \mathbb{U}$ and it is assumed that the g function is as in (8). In [40], Obradovic et al. gave some criteria for univalence expressing by $\Re(f'(z)) > 0$ , for the linear combinations $$\tau \left(1 + \frac{zf''(z)}{f'(z)}\right) + (1 - \tau)\frac{1}{f'(z)} > 0, \qquad (\tau \ge 1, z \in \mathbb{U}).$$ According to the above definitions, Lashin [41] defined the new subclasses of bi-univalent function.

Definition 2. A function f belonging to the class $\Sigma$ in type (1) is considered to be in the class $\mathcal{M}_{i,\delta,\Sigma}^{\sigma,m;\wp}(\tau,\Xi)$ if f satisfies the following inequalities: $$\tau \left( 1 + \frac{z(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1} f(z))''}{(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1} f(z))'} \right) + (1-\tau) \frac{1}{(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1} f(z))'} \prec \Xi(z)$$ $$(15)$$ and $$\tau \left( 1 + \frac{w(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}g(w))''}{(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}g(w))'} \right) + (1-\tau) \frac{1}{(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}g(w))'} \prec \Xi(w)$$ $$(16)$$ where $z, w \in \mathbb{U}$ , $\tau \geq 1$ , and it is assumed that the function g is as defined in (8). Example 3. A function f belonging to the class $\Sigma$ in type (1) is considered to be in the class $\mathcal{M}_{j,\delta,\Sigma}^{\sigma,m;\wp}(1,\Xi) \equiv \mathcal{K}_{j,\delta,\Sigma}^{\sigma,m;\wp}(\Xi)$ if f satisfies the following inequalities: $$\left(1 + \frac{z(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}f(z))''}{(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}f(z))'}\right) \prec \Xi(z) \text{ and } \left(1 + \frac{w(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}g(w))''}{(\mathcal{D}_{j,\delta,\kappa}^{\sigma,m-1}g(w))'}\right) \prec \Xi(w)$$ where $z, w \in \mathbb{U}$ , and it is assumed that the function g is as defined in (8).

Definition 1. A function $s \in \sigma_{\kappa}$ of the form (3) is said to be in the class $\mathfrak{P}_{\sigma_{\kappa}}^{\tau}(\eta, \nu, \varphi)$ if $$\frac{1}{\eta} \left( \nu \frac{\{(\varsigma s'(\varsigma))'\}^{\tau}}{s'(\varsigma)} + (1 - \nu)(s'(\varsigma))^{\tau} - 1 \right) + 1 \prec \varphi(\varsigma)$$ and $$\frac{1}{\eta} \left( \nu \frac{\{(\varkappa g'(\varkappa))'\}^{\tau}}{g'(\varkappa)} + (1 - \nu)(g'(\varkappa))^{\tau} - 1 \right) + 1 \prec \varphi(\varkappa),$$ where $g = s^{-1}$ , $\tau \ge 1$ , $\eta \in \mathbb{C}^*$ , and $0 \le \nu < 1$ . Remark 1. (i) The subclass $\mathfrak{P}_{\sigma_{\kappa}}^{\tau}(\eta, 0, \varphi) \equiv \mathscr{H}_{\sigma_{\kappa}}^{\tau}(\eta, \varphi)$ , and was explored in [24]. (ii) $\mathfrak{P}_{\sigma_{\kappa}}^{1}(\eta, \nu, \varphi) \equiv \mathscr{I}_{\sigma_{\kappa}}(\eta, \nu, \varphi)$ is the subclass of functions $s \in \sigma_{\kappa}$ satisfying $$\frac{1}{\eta} \left( \nu \frac{(\varsigma s'(\varsigma))'}{s'(\varsigma)} + (1 - \nu) s'(\varsigma) - 1 \right) + 1 \prec \varphi(\varsigma)$$ and its inverse $g = s^{-1}$ satisfies $$\frac{1}{\eta} \left( \nu \frac{(\varkappa g'(\varkappa))'}{g'(\varkappa)} + (1 - \nu)g'(\varkappa) - 1 \right) + 1 \prec \varphi(\varkappa),$$ where $\eta \in \mathbb{C}^*$ and $0 \le \nu < 1$ .

Definition 1. A function $f \in \Sigma$ is said to be in the class $\mathcal{N}_{\Sigma}(\lambda, \delta, h)$ for $\lambda \geq 0$ and $\delta \geq 0$ if $$I_{\lambda,\delta}[f](z) := (1-\lambda)\frac{f(z)}{z} + \lambda f'(z) + \delta z f''(z) \prec h(z), \quad and$$ $$I_{\lambda,\delta}[g](w) := (1-\lambda)\frac{g(w)}{w} + \lambda g'(w) + \delta w g''(w) \prec h(w), \quad g = f^{-1}.$$ Here, we present an example that helps prove that this class is nonempty and contains functions other than the identity one. Remark 1. (i) We emphasize that the class $\mathcal{N}_{\Sigma}(\lambda, \delta, h)$ is not empty for appropriate choices of the parameters. Thus, letting $$h_*(z) = 1 + 0.35z + 0.1z^2$$ like we may see in Figure 1a made using the MAPLETM computer software, we have Re $$h_*(z) > 0, z \in \mathbb{D}$$ , $B_1 = 0.35 = h'_*(0) \neq 0, B_2 = 0.1$ , and $$B_n = 0$$ for $n \ge 3$ . It is easy to show that $$\operatorname{Re} \frac{zh_'(z)}{h_(z) - 1} = \operatorname{Re} \frac{2z + 3.5}{z + 3.5} > 0.6 > 0, \ z \in \mathbb{D}.$$ Hence, $h_*$ is a starlike (univalent) function in $\mathbb D$ with respect to the point $z_0 = 1$ . The function $$f_*(z) = \frac{z}{1 + 0.2z} \in \mathcal{S}$$ and its inverse $$g_(w) = f_^{-1}(w) = \frac{w}{1 - 0.2w}$$ are analytic in $\mathbb{D}$ ; Hence, $f_* \in \Sigma$ . Also, for $\lambda = 0.2$ and $\delta = 0.1$ , a simple computation shows that $$I_{0.2,0.1}[f_*](z) = \frac{1}{0.2z+1} - \frac{0.04z}{(0.2z+1)^2} + 0.1z \left( -\frac{0.4}{(0.2z+1)^2} + \frac{0.08z}{(0.2z+1)^3} \right) \quad and$$ $$I_{0.2,0.1}[g_*](w) = \frac{1}{-0.2w+1} + \frac{0.04w}{(-0.2w+1)^2} + 0.1w \left( \frac{0.4}{(-0.2w+1)^2} + \frac{0.08w}{(-0.2w+1)^3} \right).$$ Since $h_*$ is univalent in $\mathbb{D}$ , using the inclusions $$I_{0.2,0.1}[f_*](\mathbb{D}) \subset h(\mathbb{D})$$ and $I_{0.2,0.1}[g_*](\mathbb{D}) \subset h(\mathbb{D})$ that follow from Figure 1b and Figure 1c, respectively, also made using MAPLE™, we conclude that $$I_{0.2,0.1}[f_](z) \prec h_(z)$$ and $I_{0.2,0.1}[g_](w) \prec h_(w)$ . Therefore, $f_ \in \mathcal{N}_{\Sigma}(0.2, 0.1, h_)$ . Therefore, there exists values of the parameters $\lambda$ , $\delta$ , and functions h, such that $$\mathcal{N}_{\Sigma}(\lambda, \delta, h) \setminus \{\mathrm{Id}\} \neq \emptyset$$ , where Id denotes the identity function. To not lengthen the paper unnecessarily, we omit the $MAPLE^{TM}$ codes for the figures we used throughout the article.  Figure 1. Figures for Remark 1. (ii) If, in the above example, the values of $|\lambda|$ and $|\delta|$ decrease to 0, then the behavior of the functions $I_{\lambda,\delta}[f_]$ and $I_{\lambda,\delta}[g_]$ becomes very similar to that of the functions $\frac{f_(z)}{z}$ and $\frac{g_(z)}{z}$ . In some examples we made using MAPLE^TM software, we saw that the above set inclusions hold. Hence, these new functions belong to the classes of Definition 1. These indicate a consequence of the general fact that $$\lim_{(\lambda,\delta)\to(0,0)}\mathrm{I}_{\lambda,\delta}[f_](z)=\frac{f_(z)}{z}\quad and \lim_{(\lambda,\delta)\to(0,0)}\mathrm{I}_{\lambda,\delta}[g_](z)=\frac{f_(z)}{z},\ z\in\mathbb{D};$$ that is. $$\mathcal{N}_{\Sigma}(0,0,h) = \mathcal{N}_{\Sigma}\bigg(0,0,\frac{f(z)}{z}\bigg) \text{ for all } f \in \Sigma.$$ (iii) If, in similar examples, the values of $|\lambda|$ and $|\delta|$ increase, then there are some cases when the subordinations of Definition 1 do or do not hold, as follows (to not lengthen the paper, we omit the corresponding graphical representations): (a) $$\widetilde{f}(z) = \frac{z}{1+0.1z} \in \mathcal{N}_{\Sigma}(1.1, 0.3, h_*);$$ (b) $f_(z) = \frac{z}{1+0.2z} \notin \mathcal{N}_{\Sigma}(1.1, 0.3, h_), \text{ if } h_*(z) = 1+0.35z+0.1z^2.$ In a similar way, the authors of [26] defined the following family of analytic functions: $$\mathscr{S}(\nu,\rho;h) = \left\{ f \in \mathcal{A} : 1 + \frac{1}{\rho} \left( \frac{zf'(z) + \nu z^2 f''(z)}{(1-\nu)f(z) + \nu z f'(z)} - 1 \right) \prec h(z), 0 \leq \nu \leq 1, \rho \in \mathbb{C} \setminus \{0\} \right\}$$ and obtained a bound for the general coefficients of the bi-univalent functions of this class by using the Faber polynomials subject to a series of assumptions. In our paper, we replace the assumptions for the function h from [26] with some weaker ones as stated above (i.e., omitting the conditions that $h(\mathbb{D})$ is symmetric with respect of the real axis and $B_1 > 0$ ). Here, we present an example that helps to better understand the above explanation for the function h and proves that this family is nonempty, containing other functions than the identity one. Remark 2. In the below example, we consider a case when $h(\mathbb{D})$ is not symmetric with respect of the real axis and $B_1 \neq 0$ , as we assumed in Definition 1. We show that for some values of the parameters, the class $\mathcal{S}(\nu, \rho; h)$ is not empty. Taking $$\hat{h}(z) = 1 + 0.35(1+i)z + 0.1z^2$$ since $\widehat{h}(\overline{z}) \neq \overline{\widehat{h}(z)}$ for all $z \in \mathbb{D}$ , it follows that the domain $\widehat{h}(\mathbb{D})$ is not symmetric with respect of the real axis and $$B_1 = 0.35(1+i) = \hat{h}'(0) \neq 0.$$ Like we may see in Figure 2a, we have $\operatorname{Re} \widehat{h}(z) > 0$ , $z \in \mathbb{D}$ , and Figure 2b, also made with MAPLE^TM software, shows that $$\operatorname{Re} J(z) := \operatorname{Re} \frac{z \hat{h}'(z)}{\hat{h}(z) - 1} = \operatorname{Re} \frac{3.5(1+i) + 2z}{3.5(1+i) + z} > 0.7 > 0, \ z \in \mathbb{D}.$$ Hence, $\hat{h}$ is a starlike (univalent) function with respect to the point $z_0 = 1$ . Denoting $$L_{\nu,\rho}[f](z) := 1 + \frac{1}{\rho} \left( \frac{zf'(z) + \nu z^2 f''(z)}{(1-\nu)f(z) + \nu z f'(z)} - 1 \right),$$ with the same notation as in Remark 1, we have (see Figure 2c) $$L_{0.5,4}[f_*](\mathbb{D}) \subset \widehat{h}(\mathbb{D}).$$ Using the fact that $\tilde{h}$ is univalent in $\mathbb{D}$ , the above inclusion shows that $$L_{4,0.5}[f_*](z) \prec \widehat{h}(z)$$ , i.e. $f_* \in \mathscr{S}(0.5,4;\widetilde{h})$ . In conclusion, for the above choices of functions and the corresponding parameters, we have $$\mathscr{S}(\nu,\rho;h)\setminus\{\mathrm{Id}\}\neq\emptyset.$$  Figure 2. Figures for Remark 2. In [18], the researchers proved the following result for analytic functions of the family $\mathcal{S}(\nu, \rho; h)$ : Theorem ([18] Theorem 4). Let $f(z) = z + \sum_{k=n}^{\infty} a_k z^k$ $(n \ge 2)$ and its inverse map $g = f^{-1}$ be in $\mathscr{S}(\nu, \rho; h)$ with $|B_2| \le B_1$ . Then, (i) $$|a_n| \leq \min \left\{ \frac{|\rho|B_1}{(n-1)[1+\nu(n-1)]}; \sqrt{\frac{2|\rho|B_1}{n(2n-2)[1+\nu(2n-2)]}} \right\},$$ (ii) $$\left| na_n^2 - a_{2n-1} \right| \le \frac{|\rho|B_1}{(2n-2)[1+\nu(2n-2)]}.$$ The goal of the current study is to estimate upper bounds for the coefficients $|a_n|$ for those functions that belong to the set of bi-univalent functions with missing coefficients and defined by the $\mathcal{N}_{\Sigma}(\lambda, \delta, h)$ . This paper aims to improve some of the results from [18,27]. Additionally, connections to some previously obtained results are made. The below lemmas are required to prove our results.

Definition 1. Suppose $\Xi: \mathbb{C}^4 \times \mathcal{U} \times \overline{\mathcal{U}} \longrightarrow \mathbb{C}$ and h is univalent in $\mathcal{U}$ . If $p \in \mathcal{H}$ and satisfies the third-order strong differential subordination $$\Xi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z, \zeta) \prec \prec h(z), \tag{7}$$ then p is said to be a solution of the strong differential subordination. Moreover, if $p \prec q$ for all p satisfying (7), then the univalent function q is a dominant of the solutions for the strong differential subordination. A dominant $\tilde{q}$ that satisfies $\tilde{q} \prec q$ for all dominants q of (7) is the best dominant of (7). For $\Pi \subset \mathbb{C}$ , with $\Xi$ and p given in Definition 1, relation (7) can be written as follows: $$\left\{\Xi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z, \zeta)\right\} \subset \Pi. \tag{8}$$ Condition (8) will also be referred to as strong differential subordination, and can be further extended to the definitions of the solution, dominant and best dominant.

Definition 2. ([1]). Let Q denote the collection of all injective and analytic functions q on $\overline{U}\setminus E(q)$ , where $$E(q) = \left\{ \xi : \xi \in \partial \mathcal{U} \ and \ \lim_{z \to \xi} q(z) = \infty \right\},$$ and $\min |q'(\xi)| = \rho > 0$ $(\xi \in \partial U \setminus E(q))$ . Also, Q(b) is the class of functions q with q(0) = b. We will use the following lemmas from the third-order differential subordinations to find dominants of strong differential subordinations.

Definition 3. Suppose $\Pi \in \mathbb{C}$ and $q \in \mathcal{Q}$ . The class of admissible functions $\Xi_n[\Pi, q]$ consists of those functions $$\Xi: \mathbb{C}^4 \times \mathcal{U} \times \overline{\mathcal{U}} \longrightarrow \mathbb{C}$$ that fulfill the following admissibility condition: $$\Xi(r_1, s_1, t_1, u_1; z, \zeta) \notin \Pi \tag{16}$$ whenever $r_1 = q(\xi)$ , $s_1 = n\xi q'(\xi)$ , $$\Re\left(\frac{t_1}{s_1} + 1\right) \ge n \left[\Re\frac{\xi \, q''(\xi)}{q'(\xi)} + 1\right]$$ and $$\Re\left(\frac{u_1}{s_1}\right) \ge n^2 \left[\Re\frac{\xi^2 q'''(\xi)}{q'(\xi)}\right],$$ for $\xi \in \partial \mathbb{U} \backslash E(q)$ . Here, $\Xi_1[\Pi, \mathbf{q}]$ is denoted as $\Xi[\Pi, \mathbf{q}]$ . We refer to two special subcases of this definition. If $\Xi: \mathbb{C}^3 \times \mathcal{U} \times \overline{\mathcal{U}} \longrightarrow \mathbb{C}$ , then (16) becomes $\Xi(r_1, s_1, t_1; z, \zeta) \notin \Pi$ when $r_1 = \mathbf{q}(\xi), s_1 = n\xi \mathbf{q}'(\xi)$ and $$\Re\left(\frac{t_1}{s_1}+1\right) \geq n\left[\Re\frac{\xi \mathsf{q}''(\xi)}{\mathsf{q}'(\xi)}+1\right], \quad \text{for } \xi \in \partial \mathbb{U} \backslash E(\mathsf{q}).$$ If $\Xi: \mathbb{C}^2 \times \mathcal{U} \times \overline{\mathcal{U}} \longrightarrow \mathbb{C}$ , then (16) becomes $\Xi(q(\xi), n\xi q'(\xi); z, \zeta) \notin \Pi$ when $\xi \in \partial \mathcal{U} \setminus E(q)$ . We also deduce from Definition 3 the inclusion relations $\Xi_n[\Pi', q] \subset \Xi_n[\Pi, q]$ if $\Pi' \subset \Pi$ . The following theorem is a key result in the notion of third-order strong differential subordination.

Definition 4. Consider q to be given by (14), $n \ge 2$ , and $\Pi$ is a set in $\mathbb{C}$ . For $\theta \in [0, 2\pi]$ , the class $\Xi_n[\Pi, M, b]$ which consists of those functions $$\Xi:\mathbb{C}^4\times\mathcal{U}\times\overline{\mathcal{U}}\longrightarrow\mathbb{C}$$ that fulfill the following admissibility condition $$\Xi(r_1, s_1, t_1, u_1; z, \zeta) \notin \Pi$$ whenever $$r_1=Me^{i\theta}$$ , $s_1=nM\frac{|M-\bar{b}e^{i\theta}|^2}{M^2-|b|^2}e^{i\theta}$ $$\Re \frac{t_1}{s_1} + 1 \ge n \frac{|M - \bar{b}e^{i\theta}|^2}{M^2 - |b|^2} \quad and$$ $$\Re \frac{u_1}{s_1} \ge 6n^2 \Re \frac{\left[\bar{b}M - |b|^2\right]^2}{\left[M^2 - |b|^2\right]^2},$$ $$for z \in \mathcal{U}, \zeta \in \overline{\mathcal{U}}.$$ (21) When b = 0, 0 ≤ θ ≤ 2π, we see from (21) that Ξn[Π, M, 0] consists of those functions $$\Xi: \mathbb{C}^4 \times \mathcal{U} \times \overline{\mathcal{U}} \longrightarrow \mathbb{C}$$ that fulfill $$\Xi(Me^{i\theta}, nMe^{i\theta}, L, N; z, \zeta) \notin \Pi$$ when $$\Re(Le^{-i\theta}) \ge (n^2 - n)M \text{ and } \Re(Ne^{-i\theta}) \ge 0.$$ (22) The following result is the immediate consequence.

Definition 5. Suppose $\Pi$ is a set in $\mathbb{C}$ and $q \in \mathcal{Q}$ . The family of admissible functions $\Theta_I[\Pi, q]$ consists of functions $$\Theta: \mathbb{C}^4 \times \mathcal{U} \times \overline{\mathcal{U}} \longrightarrow \mathbb{C}$$ fulfilling the admissibility $$\Theta(\alpha, \beta, \gamma, \eta; z, \zeta) \notin \Pi$$ when $\alpha=q(\xi)$ , $\beta=\frac{k\xi\,q'(\xi)+\delta\,q(\xi)}{p+\delta}$ $$\Re\left(\frac{(p+\delta)^2\gamma-\delta^2\alpha}{(p+\delta)\beta-\delta\alpha}-2\delta\right)\geq k\left[\Re\frac{\xi\,q''(\xi)}{q'(\xi)}+1\right]$$ and $$\Re\left(\frac{(p+\delta)^2(\eta(p+\delta)-3\gamma(1+\delta))+(3+2\delta)\delta^2\alpha}{(p+\delta)\beta-\delta\alpha}+2+3(2+\delta)\delta\right)\geq k^2\left[\Re\frac{\xi^2q'''(\xi)}{q'(\xi)}\right],$$ for $\xi \in \partial \mathcal{U} \setminus E(q)$ and $k \geq p$ .

Definition 6. If $\Pi$ is a set in $\mathbb{C}$ , M > 0 and $p \geq 2$ . The admissible functions class $\Theta_I[\Pi, M]$ consists of those functions $\Theta:\mathbb{C}^4\times\mathcal{U}\times\overline{\mathcal{U}}\longrightarrow\mathbb{C}$ such that $$\Theta\left(Me^{i\theta}, \frac{k+\delta}{p+\delta}Me^{i\theta}, \frac{L+((1+2\delta)k+\delta^2)Me^{i\theta}}{(p+\delta)^2}, \frac{N+3(1+\delta)L+((1+3\delta+3\delta^2)k+\delta^3)Me^{i\theta}}{(p+\delta)^2}; z, \zeta\right)$$ whenever $$\Re Le^{-i\theta} \ge (k^2 - k)M$$ , $\Re Ne^{-i\theta} \ge 0$ for $0 \le \theta \le 2\pi$ and $k \ge p$ .