Fourth order Hankel determinants for certain subclasses of modified sigmoid-activated analytic functions involving the trigonometric sine function

Lemma 2.1. Let the function p be of the form (1.2), then $$|c_n| \le 2, \quad n \ge 1,\tag{2.1}$$ $$|c_{n+k} - \mu c_n c_k| < 2$$ , for $0 \le \mu \le 1$ (2.2) $$|c_m c_n - c_k c_l| \le 4$$ for $n + m = k + l$ , (2.3) $$|c_{n+2k} - \mu c_n c_k^2| \le 2(1+2\mu) \quad \text{for } \mu \in \mathbb{R}.$$ (2.4) $$\left|c_2 - \frac{c_l^2}{2}\right| \le 2 - \frac{|c_l^2|}{2},\tag{2.5}$$ for complex number $\mu$ , we have $$\left|c_2 - \mu c_2^2\right| \le 2 \max\left\{1, |2\mu - 1|\right\}.$$ (2.6) For the results in (2.1), (2.5), (2.2), (2.4), (2.3), see [45]. Also, see [46] for inequality (2.6)

Lemma 2.2. [47]. Let the function $p \in \mathcal{P}$ be given by (1.2)), then $$\left|c_3 - 2Bc_1c_2 + Dc_1^3\right| \le 2$$ , if $$0 \le B \le 1$$ , and $B(2B-1) \le D \le B$ .

Lemma 2.3. [48, 49]. Let the function $p \in \mathcal{P}$ be given by (1.2)), then $$2c_2 = c_1^2 + x(4 - c_1^2)$$ and $$4c_3 = c_1^3 + 2(4 - c_1^2)c_1x - (4 - c_1^2)c_1x^2 + 2(4 - c_1^2)(1 - |x^2|)z$$ where $x, z \in \mathbb{C}$ with $|z| \leq 1$ and $|x| \leq 1$ .

Lemma 2.4. [50]. Consider the function $p \in \mathcal{P}$ of the form (1.2), $0 < \gamma < 1$ , $0 < \alpha < 1$ , and $$8\gamma(1-\gamma)\left\{(\alpha\beta-2\lambda)^{2}+\left(\alpha(\gamma+\alpha)-\beta\right)^{2}\right\}$$ $$+\alpha(1-\alpha)(\beta-2\gamma\alpha)^{2}$$ $$\leq 4\alpha^{2}\gamma(1-\alpha)^{2}(1-\gamma). \tag{2.7}$$ Then $$\left| \lambda b_1^4 + \gamma b_2^2 + 2\alpha b_1 b_3 - \frac{3}{2} \beta b_1^2 b_2 - b_4 \right| \le 2. \tag{2.8}$$ We divided our paper into four parts. In Sect. 1, we give some basic definitions including some subclasses of analytic functions, such as starlike, bounded turning, and convex functions, also we give the definitions of the Hankel determinant, the modified sigmoid function, and the sine function. Inspired by the above mentioned work, we create new classes of analytic functions related to modified sigmoid function and sine functions. Also we consider the Salagean type of differential operator and define a new class of analytic functions. In Sect. 2, we use known lemmas to prove our article's main results. For functions $\chi$ in the classes $\mathcal{R}^m_{\sin}(\mathfrak{T})$ and $\mathcal{R}_{\sin}(\mathfrak{T})$ , we first calculate the initial coefficient bounds, Fekete–Szego problem, and the second, third, and fourth Hankel determinants in Sect. 3 and highlight some known results.

Theorem 3.1. If $\chi_{\Im} \in \mathcal{R}^m_{\sin}(\Im)$ where $\chi_{\Im}$ has the form (1.5), then $$|d_2| \le \frac{1}{2(2^m \Im(\nu))},\tag{3.1}$$ $$|d_3| \le \frac{1}{3(3^m \Im(\nu))},\tag{3.2}$$ $$|d_4| \le \frac{1}{4(4^m \Im(\nu))},\tag{3.3}$$ $$|d_5| \le \frac{1}{5(5^m \Im(\nu))},\tag{3.4}$$ where $\Im(v)$ is given by (1.6). The results are sharp. Proof Let $\chi_{\Im} \in \mathcal{R}_{\sin}^m(\Im)$ , then from relation (1.8) we have $$\left(S^{m}\chi_{\Im}(z)\right)' = 1 + \sin(w(z)),\tag{3.5}$$ where w is a Schwarz function. Now consider a function p such that $$p(z) = \frac{1 + w(z)}{1 - w(z)} = 1 + c_1 z + c_2 z^2 + c_3 z^3 + \cdots,$$ (3.6) then $p \in \mathcal{P}$ . From (3.6), a simple computation yields $$w(z) = \frac{c_1 z + c_2 z^2 + c_3 z^3 + \cdots}{2 + c_1 z + c_2 z^2 + c_3 z^3 + \cdots}.$$ From (1.8), we can write $$(S^{m}\chi_{\Im}(z))' = 1 + 2(2^{m}\Im(v))d_{2}z + 3(3^{m}\Im(v))d_{3}z^{2} + 4(4^{m}\Im(v))d_{4}z^{3} + 5(5^{m}\Im(v))d_{5}z^{4} + \cdots$$ (3.7) By using the above values and after simplification, we get $$1 + \sin(w(z)) = 1 + \frac{1}{2}c_1z + \left(\frac{c_2}{2} - \frac{c_1^2}{4}\right)z^2 + \left(\frac{5c_1^3}{48} - \frac{c_1c_2}{2} + \frac{c_3}{2}\right)z^3 + \left(-\frac{c_1^4}{32} + \frac{5c_1^2c_2}{16} - \frac{c_1c_3}{2} - \frac{c_2^2}{2} + \frac{c_4}{2}\right)z^4 + \cdots$$ (3.8) From (3.5), (3.7), and (3.8), it follows that $$d_2 = \frac{c_1}{4(2^m \Im(\nu))},\tag{3.9}$$ $$d_3 = \frac{1}{3(3^m \Im(\nu))} \left(\frac{c_2}{2} - \frac{c_1^2}{4}\right),\tag{3.10}$$ $$d_4 = \frac{1}{4(4^m \Im(\nu))} \left( \frac{5c_1^3}{48} - \frac{c_1 c_2}{2} + \frac{c_3}{2} \right), \tag{3.11}$$ and $$d_5 = \frac{1}{5(5^m \Im(\nu))} \left( \frac{-1}{32} c_1^4 - \frac{c_2^2}{4} - \frac{c_1 c_3}{2} + \frac{5c_1^2 c_2}{16} + \frac{c_4}{2} \right). \tag{3.12}$$ Applying relation (2.1) in (3.9), we obtain $$|d_2| \leq \frac{1}{2(2^m\Im(\nu))}.$$ Using relations (2.1) and (2.5) on (3.10), we obtain $$|d_3| = \frac{1}{6(3^m \Im(\nu))} \left| c_2 - \frac{c_1^2}{2} \right|$$ $$\leq \frac{1}{6(3^m \Im(\nu))} \left( 2 - \frac{|c_1|^2}{2} \right)$$ $$= \frac{1}{3(3^m \Im(\nu))}.$$ Rearranging (3.11) gives $$|d_4| = \frac{1}{4(4^m \Im(\nu))} \left| \frac{c_3}{2} - \frac{c_1 c_2}{2} + \frac{5}{24} c_1^3 \right|.$$ $$= \frac{1}{8(4^m \Im(\nu))} \left| c_3 - 2Bc_1 c_2 + Dc_1^3 \right|.$$ Let $$B = \frac{1}{2}$$ and $D = \frac{5}{24}$ . We can observe that 0 < B and B(2B - 1) < D < B. Therefore, using Lemma 2.2 leads us to $$|d_4| \leq \frac{1}{4(4^m\Im(\nu))}.$$ From (3.12) it follows that $$|d_5| = \frac{1}{5(5^m \Im(v))} \left| \frac{-1}{32} c_1^4 - \frac{c_2^2}{4} - \frac{c_1 c_3}{2} + \frac{5c_1^2 c_2}{16} + \frac{c_4}{2} \right|$$ or $$\begin{aligned} |d_5| &= \frac{1}{10(5^m \Im(\nu))} \left| \frac{1}{16} c_1^4 + \frac{c_2^2}{2} + 2\left(\frac{c_1 c_3}{2}\right) - \frac{3}{2} \left(\frac{5c_1^2 c_2}{24}\right) - c_4 \right| \\ &= \frac{1}{10(5^m \Im(\nu))} \left| \lambda c_1^4 + \gamma c_2^2 + 2\alpha c_1 c_3 - \frac{3}{2} \beta c_1^2 c_2 - c_4 \right|, \end{aligned}$$ where $$\lambda = \frac{1}{16}$$ , $\gamma = \frac{1}{2}$ , $\alpha = \frac{1}{2}$ , $\beta = \frac{5}{24}$ . We see that $0 < \gamma < 1$ and $0 < \alpha < 1$ . Now we calculate inequality (3.1), we see that $$8\gamma(1-\gamma)\left\{(\alpha\beta-2\lambda)^2+\left(\alpha(\gamma+\alpha)-\beta\right)^2\right\}+\alpha(1-\alpha)(\beta-2\gamma\alpha)^2$$ $$\leq 4\alpha^2\gamma(1-\alpha)^2(1-\gamma),$$ where $$8\gamma(1-\gamma) = 2, \qquad (\alpha\beta - 2\lambda)^2 = \frac{1}{2304}, \qquad (\alpha(\gamma + \alpha) - \beta)^2 = \frac{49}{576},$$ $$\alpha(1-\alpha)(\beta - 2\gamma\alpha)^2 = \frac{49}{2304}, \qquad 4\alpha^2\gamma(1-\alpha)^2(1-\gamma) = \frac{1}{16}.$$ By using Lemma 2.4, we get $$|d_5| \leq \frac{1}{5(5^m \Im(\nu))}.$$ For n = 2, 3, 4, 5, we take the function $\chi_n(z) = z + \cdots$ , such that $$(S^m \chi_{n,\Im}(z))' = 1 + \sin(z^{n-1}), \quad z \in E,$$ then $$(S^m \chi_{\Im}(z))' \prec 1 + \sin(z)$$ and $\chi_{n,\Im} \in \mathcal{R}^m_{\sin}(\Im)$ , where $$\chi_{2,\Im}(z) = z + \frac{1}{2(2^m\Im(\nu))}z^2 + \cdots, \quad z \in E,$$ (3.13) $$\chi_{3,\Im}(z) = z + \frac{1}{3(3^m\Im(\nu))}z^3 + \cdots, \quad z \in E,$$ (3.14) $$\chi_{4,\Im}(z) = z + \frac{1}{4(4^m\Im(\nu))}z^4 + \cdots, \quad z \in E,$$ (3.15) $$\chi_{5,\Im}(z) = z + \frac{1}{5(5^m\Im(\nu))}z^5 + \cdots, \quad z \in E,$$ (3.16) which shows that the bounds are sharp. For m = 0 and v = 0, we get the known sharp result proved in [44]. Conjecture 3.2 If a function $\chi_{\Im} \in \mathcal{R}^m_{\sin}(\Im)$ is of the form (1.8), then $$|d_n| \le \frac{1}{n(n^m \Im(\nu))} \quad \text{for } n \ge 6, \tag{3.17}$$ where $\Im(v)$ is given by (1.6).

Theorem 3.3. If a function $\chi_{\Im}$ given in (1.5) belongs to the class $\mathcal{R}_{sin}(\Im)$ , then $$|d_2| \leq \frac{1}{2\Im(\nu)}, \qquad |d_3| \leq \frac{1}{3\Im(\nu)}, \qquad |d_4| \leq \frac{1}{4\Im(\nu)}, \qquad |d_5| \leq \frac{1}{5\Im(\nu)},$$ where $\Im(v)$ is given by (1.6). The results are sharp for functions (3.13) to (3.16). Proof Using the same procedure as we adopted in Theorem 3.2, we obtain the required result of Theorem 3.3. Conjecture 3.4 If a function $\chi_{\Im} \in \mathcal{R}_{sin}(\Im)$ is of the form (1.9), then $$|d_n| \leq \frac{1}{n\Im(\nu)}, \quad n \geq 6,$$ where $\Im(v)$ is given by (1.6).

Theorem 3.5. Let $\chi_{\Im} \in \mathcal{R}^m_{\sin}(\Im)$ . Then, for a complex number $\rho$ , $$\left| d_3 - \rho d_2^2 \right| \le \frac{1}{3(3^m \Im(\nu))} \max \left\{ 1, \left| \frac{3\rho(3^m \Im(\nu))}{4(2^m \Im(\nu))^2} \right| \right\},$$ (3.18) where $\Im(v)$ is given by (1.6). The result is sharp. Proof Using (3.9) and (3.10), one may write $$\left| d_3 - \rho d_2^2 \right| = \frac{1}{6(3^m \Im(\nu))} \left| c_2 - \left( \frac{4(2^m \Im(\nu))^2 + 3\rho(3^m \Im(\nu))}{8(2^m \Im(\nu))^2} \right) c_1^2 \right|.$$ Application of relation (2.6) gives $$\left| d_3 - \rho d_2^2 \right| \leq \frac{1}{3(3^m \Im(\nu))} \max \left\{ 1, \left| \frac{3\rho(3^m \Im(\nu))}{4(2^m \Im(\nu))^2} \right| \right\}.$$ For the sharpness of (3.18), we consider (3.13) with n = 2 $$\chi_{2,\Im}(z) = z + \frac{1}{2(2^m\Im(v))}z^2 + \cdots, \quad z \in E,$$ which gives equality in (3.18) when $|\rho| \ge \frac{4(2^m\Im(\nu))^2}{3(3^m\Im(\nu))}$ , namely $$|d_3 - \rho d_2^2| = |\rho d_2^2| = \frac{|\rho|}{4(2^m \Im(\nu))^2}.$$ For $|\rho| \leq \frac{4(2^m \Im(\nu))^2}{3(3^m \Im(\nu))}$ , consider $$\chi_3(z) = z + \frac{1}{3(3^m \Im(\nu))} z^3 + \cdots, \quad z \in E,$$ which gives $$\left| d_3 - \rho d_2^2 \right| = |d_3| = \frac{1}{3(3^m \Im(\nu))} = \frac{1}{3(3^m \Im(\nu))} \max \left\{ 1, \left| \frac{3\rho(3^m \Im(\nu))}{4(2^m \Im(\nu))^2} \right| \right\}.$$

Corollary 3.6. If $\chi_{\Im} \in \mathcal{R}^m_{sin}(\Im)$ , then for a complex number $|\rho| \leq \frac{4(2^m\Im(v))^2}{\Im(3^m\Im(v))}$ , we have $$\left| d_3 - \rho d_2^2 \right| \le \frac{1}{3(3^m \Im(\nu))},$$ (3.19) where $\Im(v)$ is given by (1.6). This inequality is sharp.

Corollary 3.7. [44]. Let $\chi \in \mathcal{R}_{sin}$ . Then, for a complex number $\rho$ , $$\left|d_3 - \rho d_2^2\right| \le \frac{1}{3} \max \left\{1, \left|\frac{3\rho}{4}\right|\right\}.$$

Theorem 3.8. Let $\chi_{\Im} \in \mathcal{R}_{sin}(\Im)$ . Then, for a complex number $\rho$ , $$|d_3 - \rho d_2^2| \le \frac{1}{3\Im(\nu)} \max \left\{ 1, \frac{\rho(3\Im(\nu))}{(2\Im(\nu))^2} \right\},$$ where $\Im(v)$ is given by (1.6). The result is sharp. Proof Using the same procedure as we adopted in Theorem 3.5, we get the required result. $\Box$

Theorem 3.9. Let $\chi_{\Im} \in \mathcal{R}^m_{\sin}(\Im)$ . Then $$\left| d_3 - d_2^2 \right| \le \frac{1}{3(3^m \Im(\nu))},\tag{3.20}$$ where $\Im(v)$ is given by (1.6). This inequality is sharp for the function $$\chi_3(z) = z + \frac{1}{3(3^m \Im(v))} z^3 + \cdots, \quad z \in E.$$ Proof From (3.9) and (3.10), we have $$\begin{aligned} \left| d_3 - d_2^2 \right| &= \frac{1}{6(3^m \Im(\nu))} \left| c_2 - \frac{1}{2} \left( 1 + \frac{3(3^m \Im(\nu))}{(2^m \Im(\nu))^2} \right) c_1^2 \right|, \\ &= \frac{1}{6(3^m \Im(\nu))} \left| c_2 - \phi c_1^2 \right|, \end{aligned}$$ where $\phi = \frac{1}{2}(1 + \frac{3(3^m\Im(\nu))}{(2^m\Im(\nu))^2})$ since $\nu \ge 0$ and $0 < \phi < 1$ . Now, by using (2.2) for n = 2 and k = 1, we obtain (3.20).

Theorem 3.10. Let $\chi_{\Im} \in \mathcal{R}_{sin}(\Im)$ . Then $$\left|d_3-d_2^2\right|\leq \frac{1}{3\Im(\nu)},$$ where $\Im(v)$ is given by (1.6). This inequality is sharp for $$\chi_3(z) = z + \frac{1}{3\Im(v)}z^3 + \cdots, \quad z \in E.$$ Proof Using the same procedure as we adopted in Theorem 3.9, we obtain the required result of Theorem 3.10.

Theorem 3.11. Let $\chi_{\Im} \in \mathcal{R}^m_{\sin}(\Im)$ . Then $$|d_2d_3 - d_4| \le \frac{1}{4(4^m\Im(\nu))}. (3.21)$$ This inequality is sharp for $$\chi_{4,\Im}(z) = z + \frac{1}{4(4^m\Im(v))}z^4 + \cdots, \quad z \in E,$$ where $\Im(v)$ is given by (1.6). Proof From (3.9), (3.10), and (3.11), we have $$|d_{2}d_{3}-d_{4}| = \begin{vmatrix} \left(\frac{4(4^{m}\Im(\nu))+5(2^{m}\Im(\nu))(3^{m}\Im(\nu))}{192(2^{m}\Im(\nu))(3^{m}\Im(\nu))(4^{m}\Im(\nu))}\right)c_{1}^{3} \\ -\left(\frac{(4^{m}\Im(\nu))+3(2^{m}\Im(\nu))(3^{m}\Im(\nu))}{24(2^{m}\Im(\nu))(3^{m}\Im(\nu))(4^{m}\Im(\nu))}\right)c_{1}c_{2} + \left(\frac{1}{8(4^{m}4^{m}\Im(\nu))}\right)c_{3} \end{vmatrix}$$ By using the applications of Lemma 2.3, and after some simple calculations, we get $$|d_2d_3-d_4| \leq \frac{1}{4(4^m\Im(\nu))}.$$ This completes the

Corollary 3.12. [44]. Let $\chi \in \mathcal{R}_{sin}$ $(m = 0 \text{ and } \nu = 0)$ . Then $$|d_2d_3 - d_4| \le \frac{1}{4}.$$ This inequality is sharp for $$\chi_4(z)=z+\frac{1}{4}z^4+\cdots.$$

Theorem 3.13. Let $\chi_{\Im} \in \mathcal{R}_{sin}(\Im)$ . Then $$|d_2d_3-d_4|\leq \frac{1}{4\Im(\nu)},$$ where $\Im(v)$ is given by (1.6). This inequality is sharp for $$\chi_{4,\Im}(z)=z+\frac{1}{4\Im(\nu)}z^4+\cdots, \quad z\in E.$$ Proof Using the same procedure of Theorem 3.11, we obtain the result of Theorem 3.13. $\square$ Remark 3.14 For $\nu = 0$ , in Theorem 3.13, we get known result proved in [44].

Theorem 3.15. Let $\chi_{\Im} \in \mathcal{R}^m_{\sin}(\Im)$ . Then $$\left| d_2 d_4 - d_3^2 \right| \le \frac{1}{9(3^m \Im(\nu))^2}. \tag{3.22}$$ This inequality is sharp for $$\chi_{3,\Im}(z) = z + \frac{1}{3(3^m\Im(v))}z^3 + \cdots, \quad z \in E,$$ where $\Im(v)$ is given by (1.6). Proof From (3.9), (3.10), and (3.11), we have $$\left|d_{2}d_{4}-d_{3}^{2}\right|=\left|C(m,v)c_{1}c_{3}-B(m,v)c_{1}^{2}c_{2}-D(m,v)c_{2}^{2}-A(m,v)c_{1}^{4}\right|,$$ where $$A(m, \nu) = \frac{5}{(2^m \Im(\nu))(4^m \Im(\nu))}, \qquad D(m, \nu) = \frac{1}{32(3^m \Im(\nu))^2}$$ $$C(m, \nu) = \frac{1}{32(2^m \Im(\nu))(4^m \Im(\nu))},$$ $$B(m,\nu) = \frac{9(3^m \Im(\nu))^2 - 8(2^m \Im(\nu))(4^m \Im(\nu))}{288(2^m \Im(\nu))(4^m \Im(\nu))(3^m \Im(\nu))^2}.$$ Using Lemma 2.3, we obtain $$\left|d_2d_4-d_3^2\right| = \left| \begin{array}{l} T_1(m,v)c_1^4+T_2(m,v)c_1^2\big(4-c_1^2\big)x-\big(3^m\Im(v)\big)T_3(m,v)c_1^2x \\ -\frac{(4-c_1^2)c_1^2x^2}{128(2^m\Im(v))(4^m\Im(v))} -\frac{(4-c_1^2)^2x^2}{144(3^m\Im(v))^2} + \frac{c(4-c_1^2)(1-|x|^2)}{144(2^m\Im(v))(4^m\Im(v))}z \right|,$$ where $$T_{1}(m,\nu) = \frac{1}{128(2^{m}\Im(\nu))(4^{m}\Im(\nu))} - T_{3}(q)$$ $$+ \left(\frac{15(3^{m}\Im(\nu))^{2} - 16(2^{m}\Im(\nu))(4^{m}\Im(\nu))}{2304(2^{m}\Im(\nu))(4^{m}\Im(\nu))(3^{m}\Im(\nu))^{2}}\right) - \frac{1}{144(3^{m}\Im(\nu))^{2}},$$ $$T_{2}(m,\nu) = \left(\left(3^{m}\Im(\nu)\right)^{2} - \left(2^{m}\Im(\nu)\right)\left(4^{m}\Im(\nu)\right)\right),$$ $$T_{3}(m,\nu) = \frac{(9(3^{m}\Im(\nu))^{2} - 8(2^{m}\Im(\nu))(4^{m}\Im(\nu)))}{576(2^{m}\Im(\nu))(4^{m}\Im(\nu))(3^{m}\Im(\nu))^{2}}.$$ Let |z| = 1, |x| = t, $t \in [0; 1]$ , $|c_1| = c \in [0; 2]$ . Then, using the triangle inequality, we get $$\begin{split} \left| d_2 d_4 - d_3^2 \right| \\ & \leq \left| T_1(m, \nu) \right| c^4 + \left| T_2(m, \nu) \right| c^2 \left( 4 - c^2 \right) t + \left( 3^m \Im(\nu) \right) T_3(m, \nu) c_1^2 t \\ & + \frac{(4 - c_1^2) c_1^2 t^2}{128 (2^m \Im(\nu)) (4^m \Im(\nu))} + \frac{(4 - c_1^2)^2 t^2}{144 (3^m \Im(\nu))^2} + \frac{c(4 - c^2) (1 - |t|^2)}{144 (2^m \Im(\nu)) (4^m \Im(\nu))} \\ & = H(c, t). \end{split}$$ Taking the derivative of H(c, t) w.r.t., t we get $$\begin{split} \frac{\partial H(c,t)}{\partial t} &= \left| T_2(m,v) \middle| c^2 \big( 4 - c^2 \big) + \big( 3^m \Im(v) \big) \middle| T_3(m,v) \middle| c_1^2 \right. \\ &+ \frac{(4 - c_1^2)^2 t}{72 (3^m \Im(v))^2} \\ &> 0, \end{split}$$ which shows that H(c,t) increases on [0,1] with respect to t. That is, H(c,t) has a maximum value at t=1, which is $$\begin{aligned} \max H(c,t) &= H(c,1) \\ &= \left| T_1(m,v) \right| c^4 + \left| T_2(m,v) \right| c^2 \left( 4 - c^2 \right) + \left( 3^m \Im(v) \right) T_3(m,v) c^2 \\ &+ \frac{(4-c^2)c^2}{128(2^m \Im(v))(4^m \Im(v))} + \frac{(4-c^2)^2}{144(3^m \Im(v))^2} \\ &= G(c). \end{aligned}$$ Differentiation gives $$G'(c) = 4|T_1(m,\nu)|c^3 + |T_2(m,\nu)|8c + 2(3^m\Im(\nu))|T_3(m,\nu)|c$$ $$\begin{split} &+ \frac{8c - 2c^3 - 2c^3}{128(2^m \Im(\nu))(4^m \Im(\nu))} - \frac{4c(4 - c^2)}{144(3^m \Im(\nu))^2} \\ &= \left( \left| T_2(m, \nu) \right| 8 + 2 \left( 3^m \Im(\nu) \right) \right| T_3(m, \nu) \right| \\ &- \frac{16}{144(3^m \Im(\nu))^2} + \frac{8}{128(2^m \Im(\nu))(4^m \Im(\nu))} \right) c \\ &+ \left( \left| T_1(m, \nu) \right| - \frac{1}{128(2^m \Im(\nu))(4^m \Im(\nu))} + \frac{1}{144(3^m \Im(\nu))^2} \right) c^3. \end{split}$$ If G'(c) = 0, then the root is c = 0 and $$c^2 = \frac{-(|T_2(m,\nu)|8 + 2(3^m\Im(\nu))|T_3(m,\nu)| - \frac{16}{144(3^m\Im(\nu))^2} + \frac{8}{128(2^m\Im(\nu))(4^m\Im(\nu))})}{4(|T_1(m,\nu)| - \frac{1}{128(2^m\Im(\nu))(4^m\Im(\nu))} + \frac{1}{144(3^m\Im(\nu))^2})}.$$ Again, taking the derivative of G'(c), we have $$\begin{split} G''(0) &= \left| T_2(m,\nu) \middle| 8 + 2 \bigl( 3^m \Im(\nu) \bigr) \middle| T_3(m,\nu) \middle| \\ &- \frac{16}{144 (3^m \Im(\nu))^2} + \frac{8}{128 (2^m \Im(\nu)) (4^m \Im(\nu))} \\ &+ 3 \biggl( \left| T_1(m,\nu) \middle| - \frac{1}{128 (2^m \Im(\nu)) (4^m \Im(\nu))} + \frac{1}{144 (3^m \Im(\nu))^2} \biggr) c^2. \end{split}$$ We see that $$G''(0) < 0$$ , so the function G(c) can attain the maximum value at c = 0; which is $$|d_2d_4-d_3^2|\leq \frac{1}{9(3^m\Im(\nu))^2}$$ This completes the result.

Theorem 3.16. Let $\chi_{\Im} \in \mathcal{R}_{sin}(\Im)$ . Then $$|d_2d_4-d_3^2|\leq \frac{1}{9(\Im(\nu))^2},$$ where $\Im(v)$ is given by (1.6). This inequality is sharp for $$\chi_{3,\Im}(z) = z + \frac{1}{3\Im(\nu)}z^3 + \cdots, \quad z \in E.$$ Proof Using the same procedure as we adopted in Theorem 3.15, we obtain the result of Theorem 3.16. $\Box$

Theorem 3.17. Let $\chi_{\Im} \in \mathcal{R}^m_{\sin}(\Im)$ . Then $$|H_{3,1}(\chi)| \le \frac{1}{27(3^m\Im(\nu))(3^m\Im(\nu))^2} + \frac{1}{16(4^m\Im(\nu))(4^m\Im(\nu))}$$ $$+\frac{1}{15(5^m\Im(\nu))(3^m\Im(\nu))},$$ where $\Im(v)$ is given by (1.6). Proof From (1.4), it is easy to see that $$H_{3,1}(\chi) = d_3(d_2d_4 - d_3^2) + d_4(d_2d_3 - d_4) + d_5(d_3 - d_2^2),$$ where $d_1 = 1$ . This implies that $$|H_{3,1}(\chi)| \le |d_3||d_2d_4 - d_3^2| + |d_4||d_2d_3 - d_4| + |d_5||d_3 - d_2^2|$$ By using (3.1), (3.2), (3.3), (3.4), (3.19), (3.21), and (3.22), we have $$\begin{split} \left| H_{3,1}(\chi) \right| & \leq \frac{1}{27(3^m \Im(\nu))(3^m \Im(\nu))^2} + \frac{1}{16(4^m \Im(\nu))(4^m \Im(\nu))} \\ & + \frac{1}{15(5^m \Im(\nu))(3^m \Im(\nu))}, \end{split}$$ which is our required result.

Theorem 3.18. Let $\chi_{\Im} \in \mathcal{R}_{\sin}(\Im)$ . Then $$|H_{3,1}(\chi)| \le \frac{1}{27(\Im(\nu))^3} + \frac{1}{16(\Im(\nu))^2} + \frac{1}{15(\Im(\nu))^2},$$ where $\Im(v)$ is given by (1.6). Proof Using the same procedure as we adopted in Theorem 3.17, we obtain the result of Theorem 3.18. $\Box$ Remark 3.19 For m = 1, v = 0, in Theorem 3.18, we get known result proved in [44].

Theorem 4.1. Let $\chi_{\varphi} \in \mathcal{R}^m_{\sin}(\Im)$ . Then $$\begin{split} \left| H_{4,1}(\chi) \right| &\leq \frac{1}{7(7^m \Im(\nu))} \left( \frac{1}{27(3^m \Im(\nu))(3^m \Im(\nu))^2} \right. \\ &\quad + \frac{1}{16(4^m \Im(\nu))(4^m \Im(\nu))} + \frac{1}{15(5^m \Im(\nu))(3^m \Im(\nu))} \right) \\ &\quad + \frac{1}{108(4^m \Im(\nu))(6^m \Im(\nu))(3^m \Im(\nu))^2} \\ &\quad + \frac{1}{120(5^m \Im(\nu))(6^m \Im(\nu))(4^m \Im(\nu))} \\ &\quad + \frac{1}{108(3^m \Im(\nu))(6^m \Im(\nu))^2} + \frac{1}{225(3^m \Im(\nu))^2(5^m \Im(\nu))^2} \\ &\quad + \frac{1}{54(3^m \Im(\nu))^2(5^m \Im(\nu))^2} \\ &\quad - \frac{1}{125(5^m \Im(\nu))^3} + \frac{1}{256(4^m \Im(\nu))^4} \\ &\quad - \frac{1}{80(3^m \Im(\nu))(4^m \Im(\nu))^2(5^m \Im(\nu))}. \end{split}$$ where $\Im(v)$ is given by (1.6). Proof Taking modulus on both sides of (4.1) and then applying the triangle inequality, we obtain $$\begin{aligned} \left| H_{4,1}(\chi) \right| &\leq |d_7| \left| H_{3,1}(\chi) \right| + 2|d_4| |d_6| \left| d_2 d_4 - d_3^2 \right| + 2|d_5| |d_6| |d_2 d_3 - d_4| \\ &+ |d_6|^2 \left| d_3 - d_2^2 \right| + |d_5|^2 \left| d_2 d_4 - d_3^2 \right| + |d_5|^2 \left| d_2 d_4 + 2d_3^2 \right| \\ &+ |d_5|^3 + |d_4|^4 + 3|d_3| |d_4|^2 |d_5|. \end{aligned}$$ Now, by using (3.1), (3.2),(3.3),(3.4),(3.17), (3.20), (3.21), (3.22), and (4.2), we get the required result.

Theorem 4.2. Let $\chi_{\varphi} \in \mathcal{R}_{\sin}(\Im)$ . Then $$\begin{split} \left| H_{4,1}(\chi) \right| & \leq \frac{1}{7\Im(\nu)} \left( \frac{1}{27(\Im(\nu))^3} + \frac{1}{16(\Im(\nu))^2} + \frac{1}{15(\Im(\nu))^2} \right) \\ & + \frac{1}{108(\Im(\nu))^4} + \frac{1}{120(\Im(\nu))^3} \\ & + \frac{1}{108(\Im(\nu))^3} + \frac{1}{225(\Im(\nu))^4} + \frac{1}{54(\Im(\nu))^4} \\ & - \frac{1}{125(\Im(\nu))^3} + \frac{1}{256(\Im(\nu))^4} - \frac{1}{80(\Im(\nu))^4}, \end{split}$$ where $\Im(v)$ is given by (1.6). | | Proof By using a similar method as we adopted in the above theorem, we get the required | | |---------|-----------------------------------------------------------------------------------------|--| | result. | | |

Definition 1.1. [42]. The Sălăgean type differential operator $S^m: A_{\Im} \to A_{\Im}$ is defined by $$S_q^0 \chi_{\mathfrak{I}}(z) = \chi_{\mathfrak{I}}(z), \qquad S^1 \chi(z) = z \chi_{\mathfrak{I}}'(z), \qquad \dots,$$ $$S^m \chi_{\mathfrak{I}}(z) = z S(S^{m-1} \chi_{\mathfrak{I}}(z)), \qquad (1.7)$$ where $\chi_{\Im}(z) \in \mathcal{A}_{\Im}$ , $m \in \mathbb{N} \cup \{0\}$ . It is easy to prove that if $$\chi_{\Im}(z) = z + \sum_{n=2}^{\infty} \Im(v) d_n z^n \in \mathcal{A}_{\Im},$$ then $$S^{m}\chi_{\Im}(z)=z+\sum_{n=2}^{\infty}n^{m}\Im(v)d_{n}z^{n}.$$ Remark 1.2 When $\Im(v) = 1$ , we have the Sălăgean differential operator [43]. Here we define a new class of bounded turning functions connected with modified sigmoid function and sine functions as follows.

Definition 1.3. A function $\chi_{\Im} \in \mathcal{R}^m_{sin}(\Im)$ , where $\chi_{\Im}$ is of the form (1.5), if $$\left(S^{m}\chi_{\mathfrak{I}}(z)\right)' < 1 + \sin(z),\tag{1.8}$$ or it can be defined as $$\frac{S^{m+1}\chi_{\Im}(z)}{z} \prec 1 + \sin(z).$$ When $\nu = 0$ and m = 0 in Definition 1.3, we get a known class proved in [44]. When m = 0 in Definition 1.3, we get the following function class.

Definition 1.4. A function $\chi_{\Im} \in \mathcal{R}_{\sin}(\Im)$ , where $\chi_{\Im}$ is of the form (1.5), if $$\chi_{\gamma}'(z) < 1 + \sin(z). \tag{1.9}$$