Hankel determinant for a general subclass of m-fold symmetric bi-univalent functions defined by Ruscheweyh operator

Lemma 1.1. [12] If the function $h \in \mathcal{P}$ is given by the series (1.2), then $$|h_k| \le 2 \quad (k \in \mathbb{N}),\tag{1.8}$$ and $$\left| h_2 - \frac{h_1^2}{2} \right| \le 2 - \frac{\left| h_2 \right|^2}{2}. \tag{1.9}$$

Lemma 1.2. [14] If the function $h \in \mathcal{P}$ is given by the series (1.2), then $$2h_2 = h_1^2 + x\left(4 - h_1^2\right),\tag{1.10}$$ and $$4h_3 = h_1^3 + 2(4 - h_1^2)h_1x - h_1(4 - h_1^2)x^2 + 2(4 - h_1^2)(1 - |x|^2)z,$$ (1.11) for some x, z with $|x| \le 1$ and $|z| \le 1$ .

Theorem 2.1. Let $f \in \Xi_{\Sigma_m}(\lambda, \gamma; \beta)$ be given by (1.5). Then $$\left|a_{m+1}a_{3m+1} - a_{2m+1}^{2}\right| \leq \begin{cases} \frac{4(1-\beta)^{2}}{(\gamma+1)^{2}(m\lambda+1)} \left[\frac{(m+1)^{2}(1-\beta)^{2}}{(\gamma+1)^{2}(m\lambda+1)^{3}} + \frac{6}{(\gamma+2)(\gamma+3)(3m\lambda+1)}\right], & \beta \in [0,\tau] \\ \frac{4(1-\beta)^{2}}{(\gamma+1)^{2}(\gamma+2)^{2}(2m\lambda+1)^{2}} \left[4 - \frac{[\omega_{2}(1-\beta) + 9\omega_{3} - 4\omega_{4}]^{2}}{\omega_{4}[\omega_{1}(1-\beta)^{2} - 2\omega_{2}(1-\beta) - 12\omega_{3} + 4\omega_{4}]}\right], \beta \in [\tau, 1) \end{cases}$$ where $$\omega_1 := (m+1)^2 (\gamma+2)^2 (\gamma+3)(2m\lambda+1)^2 (3m\lambda+1), \tag{2.3}$$ $$\omega_2 := m(\gamma + 1)(\gamma + 2)(\gamma + 3)(m\lambda + 1)^2(2m\lambda + 1)(3m\lambda + 1),\tag{2.4}$$ $$\omega_3 := (\gamma + 1)^2 (\gamma + 2)(m\lambda + 1)^3 (2m\lambda + 1)^2, \tag{2.5}$$ $$\omega_4 := (\gamma + 1)^2 (\gamma + 3)(m\lambda + 1)^4 (3m\lambda + 1), \tag{2.6}$$ and $$\tau := 1 - \frac{\omega_2 + \sqrt{\omega_2^2 + 12\omega_1\omega_3}}{2\omega_1}.$$ Proof. It follows from (2.1) and (2.2) that there exist p and q in the class P such that $$(1 - \lambda)\frac{\mathcal{R}^{\gamma} f(z)}{z} + \lambda \left(\mathcal{R}^{\gamma} f(z)\right)' = \beta + (1 - \beta)p(z), \tag{2.7}$$ and $$(1 - \lambda)\frac{\mathcal{R}^{\gamma} f(w)}{w} + \lambda \left(\mathcal{R}^{\gamma} f(w)\right)' = \beta + (1 - \beta)q(z)$$ (2.8) where p and q are given by the series (1.6). We also find that $$(1-\lambda)\frac{\mathcal{R}^{\gamma}f(z)}{z} + \lambda \left(\mathcal{R}^{\gamma}f(z)\right)'$$ $$= 1 + (\gamma+1)(m\lambda+1)a_{m+1}z^{m} + \frac{1}{2}(\gamma+1)(\gamma+2)(2\lambda m+1)a_{2m+1}z^{2m}$$ $$+ \frac{1}{6}(\gamma+1)(\gamma+2)(\gamma+3)(3\lambda m+1)a_{3m+1}z^{3m} + \cdots,$$ (2.9) and $$(1-\lambda)\frac{\mathcal{R}^{\gamma}g(w)}{w} + \lambda \left(\mathcal{R}^{\gamma}g(w)\right)'$$ $$= 1 - (\gamma+1)(m\lambda+1)a_{m+1}w^{m} + \frac{1}{2}(\gamma+1)(\gamma+2)(2m\lambda+1) \times$$ $$\left[ (m+1)a_{m+1}^{2} - a_{2m+1} \right]w^{2m} - \frac{1}{6}(\gamma+1)(\gamma+2)(\gamma+3)(3m\lambda+1) \times$$ $$\left[ \frac{1}{2}(m+1)(3m+2)a_{m+1}^{3} - (3m+2)a_{m+1}a_{2m+1} + a_{3m+1} \right]w^{3m} + \cdots$$ (2.10) Equating coefficients in (2.7) and (2.8) we have $$(\gamma + 1)(m\lambda + 1)a_{m+1} = (1 - \beta)p_m, \tag{2.11}$$ $$\frac{1}{2}(\gamma+1)(\gamma+2)(2\lambda m+1)a_{2m+1} = (1-\beta)p_{2m},$$ (2.12) $$\frac{1}{6}(\gamma+1)(\gamma+2)(\gamma+3)(3m\lambda+1)a_{3m+1} = (1-\beta)p_{3m},$$ (2.13) and $$-(\gamma + 1)(m\lambda + 1)a_{m+1} = (1 - \beta)q_m, \tag{2.14}$$ $$\frac{1}{2}(\gamma+1)(\gamma+2)(2m\lambda+1)\left[(m+1)a_{m+1}^2 - a_{2m+1}\right] = (1-\beta)q_{2m},\tag{2.15}$$ $$-\frac{1}{6}(\gamma+1)(\gamma+2)(\gamma+3)(3m\lambda+1)\times$$ $$\left[\frac{1}{2}(m+1)(3m+2)a_{m+1}^{3} - (3m+2)a_{m+1}a_{2m+1} + a_{3m+1}\right] = (1-\beta)q_{3m}.$$ (2.16) From (2.11) and (2.14), we get $$p_m = -q_m, (2.17)$$ and $$a_{m+1} = \frac{1-\beta}{(\gamma+1)(m\lambda+1)} p_m. \tag{2.18}$$ Now, from (2.12) and (2.15) and (2.18), we obtain $$a_{2m+1} = \frac{(m+1)(1-\beta)^2}{2(\gamma+1)^2(m\lambda+1)^2}p_m^2 + \frac{(1-\beta)}{(\gamma+1)(\gamma+2)(2m\lambda+1)}(p_{2m} - q_{2m}).$$ (2.19) Also, from (2.13), (2.16), (2.18) and (2.19), we find that $$a_{3m+1} = \frac{(3m+2)(1-\beta)^2}{2(\gamma+1)^2(\gamma+2)(m\lambda+1)(2m\lambda+1)} p_m (p_{2m} - q_{2m}) + \frac{3(1-\beta)}{(\gamma+1)(\gamma+2)(\gamma+3)(3m\lambda+1)} (p_{3m} - q_{3m}).$$ (2.20) Then, from (2.18), (2.19) and (2.20) we have that $$a_{m+1}a_{3m+1} - a_{2m+1}^2 = -\frac{(m+1)^2(1-\beta)^4}{4(\gamma+1)^4(m\lambda+1)^4}p_m^4$$ $$+ \frac{m(1-\beta)^3}{2(\gamma+1)^3(\gamma+2)(m\lambda+1)^2(2m\lambda+1)}p_m^2(p_{2m} - q_{2m})$$ $$+ \frac{3(1-\beta)^2}{(\gamma+1)^2(\gamma+2)(\gamma+3)(m\lambda+1)(3m\lambda+1)}p_m(p_{3m} - q_{3m})$$ $$- \frac{(1-\beta)^2}{(\gamma+1)^2(\gamma+2)^2(2m\lambda+1)^2}(p_{2m} - q_{2m})^2.$$ (2.21) According to Lemma 1.2 and (2.17), we can write $$p_{2m} - q_{2m} = \frac{4 - p_m^2}{2}(x - y), \tag{2.22}$$ and $$p_{3m} - q_{3m} = \frac{p_m^3}{2} + \frac{p_m \left(4 - p_m^2\right)}{2} (x+y) - \frac{p_m \left(4 - p_m^2\right)}{4} \left(x^2 + y^2\right) + \frac{4 - p_m^2}{2} \left[ \left(1 - |x|^2\right) z - \left(1 - |y|^2\right) w \right],$$ (2.23) $$p_{2m} + q_{2m} = p_m^2 + \frac{4 - p_m^2}{2}(x + y), \tag{2.24}$$ for some x, y, z and w with $|x| \le 1, |y| \le 1, |z| \le 1$ and $|w| \le 1$ . Using (2.22) and (2.23) in (2.21) we obtain $$\begin{split} &|a_{m+1}a_{3m+1}-a_{2m+1}^2|\\ &=|-\frac{(m+1)^2(1-\beta)^4}{4(\gamma+1)^4(m\lambda+1)^4}p_m^4+\frac{m(1-\beta)^3}{4(\gamma+1)^3(\gamma+2)(m\lambda+1)^2(2m\lambda+1)}p_m^2\left(4-p_m^2\right)\left(x-y\right)\\ &+\frac{3(1-\beta)^2}{(\gamma+1)^2(\gamma+2)(\gamma+3)(m\lambda+1)(3m\lambda+1)}p_m\left[\frac{p_m^3}{2}+\frac{p_m\left(4-p_m^2\right)}{2}(x+y)\right.\\ &-\frac{p_m\left(4-p_m^2\right)}{4}\left(x^2+y^2\right)+\frac{4-p_m^2}{2}\left[\left(1-|x|^2\right)z-\left(1-|y|^2\right)w\right]\\ &-\frac{(1-\beta)^2}{4(\gamma+1)^2(\gamma+2)^2(2m\lambda+1)^2}\left(4-p_m^2\right)^2\left(x-y\right)^2|\\ &\leq\frac{(m+1)^2(1-\beta)^4}{4(\gamma+1)^4(m\lambda+1)^4}p_m^4+\frac{3(1-\beta)^2}{2(\gamma+1)^2(\gamma+2)(\gamma+3)(m\lambda+1)(3m\lambda+1)}p_m^4\\ &+\frac{3(1-\beta)^2}{(\gamma+1)^2(\gamma+2)(\gamma+3)(m\lambda+1)(3m\lambda+1)}p_m\left(4-p_m^2\right)\\ &+\left[\frac{m(1-\beta)^3}{4(\gamma+1)^3(\gamma+2)(m\lambda+1)^2(2m\lambda+1)}p_m^2\left(4-p_m^2\right)\right]\left(|x|+|y|\right)\\ &+\frac{3(1-\beta)^2}{2(\gamma+1)^2(\gamma+2)(\gamma+3)(m\lambda+1)(3m\lambda+1)}p_m^2\left(4-p_m^2\right)\\ &-\frac{3(1-\beta)^2}{2(\gamma+1)^2(\gamma+2)(\gamma+3)(m\lambda+1)(3m\lambda+1)}p_m^2\left(4-p_m^2\right)\\ &-\frac{3(1-\beta)^2}{2(\gamma+1)^2(\gamma+2)(\gamma+3)(m\lambda+1)(3m\lambda+1)}p_m\left(4-p_m^2\right)\right]\left(|x|^2+|y|^2\right)\\ &+\frac{(1-\beta)^2}{4(\gamma+1)^2(\gamma+2)^2(2m\lambda+1)^2}\left(4-p_m^2\right)^2\left(|x|+|y|\right)^2. \end{split}$$ Since p in the class $\mathcal{P}$ , we have (Lemma 1.1) $|p_m| \leq 2$ . Letting $p_m = \rho$ , we may assume without loss of generality that $\rho \in [0,2]$ . Thus, for $\mu_1 = |x| \leq 1$ and $\mu_2 = |y| \leq 1$ , we get $$\left|a_{m+1}a_{3m+1}-a_{2m+1}^{2}\right| \leq F_{1}+F_{2}\left(\mu_{1}+\mu_{2}\right)+F_{3}\left(\mu_{1}^{2}+\mu_{2}^{2}\right)+F_{4}\left(\mu_{1}+\mu_{2}\right)^{2},$$ where $$\begin{split} F_1 &= F_1(\rho) = \frac{(m+1)^2(1-\beta)^4\rho^4}{4(\gamma+1)^4(m\lambda+1)^4} + \frac{3(1-\beta)^2\rho^4}{2(\gamma+1)^2(\gamma+2)(\gamma+3)(m\lambda+1)(3m\lambda+1)} \\ &\quad + \frac{3(1-\beta)^2\rho\left(4-\rho^2\right)}{(\gamma+1)^2(\gamma+2)(\gamma+3)(m\lambda+1)(3m\lambda+1)} \geq 0, \\ F_2 &= F_2(\rho) = \frac{m(1-\beta)^3\rho^2\left(4-\rho^2\right)}{4(\gamma+1)^3(\gamma+2)(m\lambda+1)^2(2m\lambda+1)} \\ &\quad + \frac{3(1-\beta)^2\rho^2\left(4-\rho^2\right)}{2(\gamma+1)^2(\gamma+2)(\gamma+3)(m\lambda+1)(3m\lambda+1)} \geq 0, \\ F_3 &= F_3(\rho) = \frac{3(1-\beta)^2\rho^2\left(4-\rho^2\right)}{4(\gamma+1)^2(\gamma+2)(\gamma+3)(m\lambda+1)(3m\lambda+1)} \\ &\quad - \frac{3(1-\beta)^2\rho\left(4-\rho^2\right)}{2(\gamma+1)^2(\gamma+2)(\gamma+3)(m\lambda+1)(3m\lambda+1)} \leq 0, \\ F_4 &= F_4(\rho) = \frac{(1-\beta)^2\left(4-\rho^2\right)^2}{4(\gamma+1)^2(\gamma+2)^2(2m\lambda+1)^2} \geq 0. \end{split}$$  Figure 1: Graph of F1, F2, F³ and F⁴ for γ = 0 and λ = m = 1. Now, we need to maximize $$F(\mu_1, \mu_2) = F_1 + F_2(\mu_1 + \mu_2) + F_3(\mu_1^2 + \mu_2^2) + F_4(\mu_1 + \mu_2)^2$$ in the closed square ^S = [0, 1] ^× [0, 1] for ^ρ ^∈ [0, 2]. We investigate the maximum of ^F (µ1, µ2) when ρ ∈ (0, 2), ρ = 0 and ρ = 2, keeping in mind the sign of $$F_{\mu_1\mu_1}F_{\mu_2\mu_2} - (F_{\mu_1\mu_2})^2$$ (according to the Second Derivative Test for functions of the two dependent variables µ¹ and µ2) First, let ρ ∈ (0, 2). Since F³ < 0 and F³ + 2F⁴ > 0 for ρ ∈ (0, 2), we see that $$F_{\mu_1\mu_1}F_{\mu_2\mu_2} - (F_{\mu_1\mu_2})^2 = 4F_3(F_3 + 2F_4) < 0.$$  Figure 2: Graph of F³ + 2F⁴ for γ = 0 and λ = m = 1. Thus, the function F cannot have a local maximum in the interior of the square S. Now, we investigate the maximum of F on the boundary of the square S. Case 1. For $\mu_1 = 0$ and $\mu_2 \in [0, 1]$ (a similar argument can be applied for $\mu_2 = 0$ and $\mu_1 \in [0, 1]$ , so we omit the details in that case), we obtain $$F(0, \mu_2) = G(\mu_2) \equiv F_1 + F_2 \mu_2 + (F_3 + F_4) \mu_2^2$$ Subcase 1. Let $F_3 + F_4 \ge 0$ . In this case, for $0 < \mu_2 < 1$ we have that $$G'(\mu_2) = F_2 + 2(F_3 + F_4)\mu_2 > 0,$$ that is, $G(\mu_2)$ is an increasing function. Hence the maximum of $G(\mu_2)$ occurs at $\mu_2 = 1$ and $$\max \left\{ F\left(0,\mu_{2}\right) : \, \mu_{2} \in [0,1] \right\} = \max \left\{ G\left(\mu_{2}\right) : \, \mu_{2} \in [0,1] \right\} = G(1) = F_{1} + F_{2} + F_{3} + F_{4}.$$ Subcase 2. Let $F_3 + F_4 < 0$ . Note $$F_2 + 2(F_3 + F_4) \ge 0.$$ For $\mu_2 \in (0,1)$ since $F_3 + F_4 < 0$ we have that $$F_2 + 2(F_3 + F_4)\mu_2 > F_2 + 2(F_3 + F_4) \ge 0$$ so $G'(\mu_2) > 0$ . Thus $\max \{G(\mu_2) : \mu_2 \in [0,1]\} = G(1)$ .  Figure 3: Graph of $F_2 + 2(F_3 + F_4)$ for $\gamma = 0$ and $\lambda = m = 1$ . Case 2. For $\mu_1 = 1$ and $\mu_2 \in [0, 1]$ (a similar argument can be applied for $\mu_2 = 1$ and $\mu_1 \in [0, 1]$ , so we omit the details in that case), we obtain $$F(1, \mu_2) = H(\mu_2) = F_1 + F_2 + F_3 + F_4 + (F_2 + 2F_4)\mu_2 + (F_2 + F_4)\mu_2^2$$ Thus an argument like in Subcases 1 and 2 yields $$\max\{F(1,\mu_2): \mu_2 \in [0,1]\} = \max\{H(\mu_2): \mu_2 \in [0,1]\} = H(1) = F_1 + 2(F_2 + F_3) + 4F_4.$$ Next let $\rho = 2$ . Now let $(\mu_1, \mu_2) \in \mathbb{S}$ and note $$F(\mu_1, \mu_2) = \frac{4(\gamma+2)(\gamma+3)(3m\lambda+1)(m+1)^2(1-\beta)^4 + 24(\gamma+1)^2(m\lambda+1)^3(1-\beta)^2}{(\gamma+1)^4(\gamma+2)(\gamma+3)(m\lambda+1)^4(3m\lambda+1)}.$$ (2.25) Keeping in mind the constant value in (2.25) we have $$\max\{F(\mu_1,\mu_2): \mu_1 \in [0,1], \mu_2 \in [0,1]\} = F(1,1) = F_1 + 2(F_2 + F_3) + 4F_4.$$ Finally, let $\rho = 0$ . Now let $(\mu_1, \mu_2) \in \mathbb{S}$ and note $$F(\mu_1, \mu_2) = \frac{4(1-\beta)^2(\mu_1 + \mu_2)^2}{(\gamma+1)^2(\gamma+2)^2(2m\lambda+1)^2}.$$ We see that, the maximum of $F(\mu_1, \mu_2)$ occurs at $\mu_1 = \mu_2 = 1$ and $$\max \{F(\mu_1, \mu_2): \mu_1 \in [0, 1], \mu_2 \in [0, 1]\} = F(1, 1) = F_1 + 2(F_2 + F_3) + 4F_4.$$ Combining all cases note since $F_1 + 2(F_2 + F_3) + 4F_4 \ge 0$ when $\rho \in [0, 2]$ we have $$\max \{F(\mu_1, \mu_2): \ \mu_1 \in [0, 1], \ \mu_2 \in [0, 1]\} = F(1, 1).$$  Figure 4: Graph of $F_1 + 2(F_2 + F_3) + 4F_4$ for $\gamma = 0$ and $\lambda = m = 1$ . Let $K:[0,2]\to\mathbb{R}$ be given by $$K(\rho) = F(1,1) = F_1 + 2(F_2 + F_3) + 4F_4.$$ (2.26) Substituting the values of $F_1, F_2, F_3$ and $F_4$ in the function K defined by (2.26), yields $$K(\rho) = \frac{(1-\beta)^2}{4(\gamma+1)^4(\gamma+2)^2(\gamma+3)(1+m\lambda)^4(1+2m\lambda)^2(1+3m\lambda)} \times \left[ \left[ (m+1)^2(\gamma+2)^2(\gamma+3)(2m\lambda+1)^2(3m\lambda+1)(1-\beta)^2 - 2m(\gamma+1)(\gamma+2)(\gamma+3)(m\lambda+1)^2(2m\lambda+1)(3m\lambda+1)(1-\beta) - 12(\gamma+1)^2(\gamma+2)(m\lambda+1)^3(2m\lambda+1)^2 + 4(\gamma+1)^2(\gamma+3)(m\lambda+1)^4(3m\lambda+1) \right] \rho^4 + \left[ 8m(\gamma+1)(\gamma+2)(\gamma+3)(1+m\lambda)^2(1+2m\lambda)(1+3m\lambda)(1-\beta) + 72(\gamma+1)^2(\gamma+2)(m\lambda+1)^3(2m\lambda+1)^2 - 32(\gamma+1)^2(\gamma+3)(m\lambda+1)^4(3m\lambda+1) \right] \rho^2 + 64(\gamma+1)^2(\gamma+3)(1+m\lambda)^4(1+3m\lambda) \right].$$ Now the maximum of $K(\rho)$ occurs either at $\rho = 0$ , $\rho \in (0,2)$ or $\rho = 2$ . Suppose first the maximum of $K(\rho)$ occurs at some $\rho \in (0,2)$ . Note for any $\rho \in (0,2)$ we have $$K'(\rho) = \frac{(1-\beta)^2}{(\gamma+1)^4(\gamma+2)^2(\gamma+3)(1+m\lambda)^4(1+2m\lambda)^2(1+3m\lambda)} \times \left[ \left[ (m+1)^2(\gamma+2)^2(\gamma+3)(2m\lambda+1)^2(3m\lambda+1)(1-\beta)^2 - 2m(\gamma+1)(\gamma+2)(\gamma+3)(m\lambda+1)^2(2m\lambda+1)(3m\lambda+1)(1-\beta) - 12(\gamma+1)^2(\gamma+2)(m\lambda+1)^3(2m\lambda+1)^2 + 4(\gamma+1)^2(\gamma+3)(m\lambda+1)^4(3m\lambda+1) \right] \rho^3 + \left[ 4m(\gamma+1)(\gamma+2)(\gamma+3)(\lambda m+1)^2(1+2m\lambda)(1+3m\lambda)(1-\beta) + 36(\gamma+1)^2(\gamma+2)(m\lambda+1)^3(2m\lambda+1)^2 - 16(\gamma+1)^2(\gamma+3)(m\lambda+1)^4(3m\lambda+1) \right] \rho^3 \right].$$ Next, we conclude the following results Result 1. Let $$\omega_1(1-\beta)^2 - 2\omega_2(1-\beta) - 12\omega_3 + 4\omega_4 \ge 0,$$ that is, $$\beta \in \left[0, 1 - \frac{\omega_2 + \sqrt{\omega_2^2 + \omega_1 \left[12\omega_3 - 4\omega_4\right]}}{\omega_1}\right],$$ where $\omega_1, \omega_2, \omega_3$ and $\omega_4$ are given by (2.3), (2.4), (2.5) and (2.6), respectively. Note, $K'(\rho) > 0$ for every $\rho \in (0, 2)$ . Thus, $$\max\{K(\rho): 0 < \rho < 2\} = K(2^{-}) = \frac{4(1-\beta)^{2}}{(\gamma+1)^{2}(m\lambda+1)} \left[ \frac{(m+1)^{2}(1-\beta)^{2}}{(\gamma+1)^{2}(m\lambda+1)^{3}} + \frac{6}{(\gamma+2)(\gamma+3)(3m\lambda+1)} \right].$$ Result 2. Let $$\omega_1(1-\beta)^2 - 2\omega_2(1-\beta) - 12\omega_3 + 4\omega_4 < 0,$$ that is, $$\beta \in \left(1 - \frac{\omega_2 + \sqrt{\omega_2^2 + \omega_1 \left[12\omega_3 - 4\omega_4\right]}}{\omega_1}, 1\right).$$ Then $K'(\rho) = 0$ gives the critical point $\rho_1 = 0$ or $$\rho_2 = \sqrt{\frac{16\omega_4 - 4\omega_2(1-\beta) - 36\omega_3}{\omega_1(1-\beta)^2 - 2\omega_2(1-\beta) - 12\omega_3 + 4\omega_4}}.$$ When $$\beta \in \left(1 - \frac{\omega_2 + \sqrt{\omega_2^2 + \omega_1 \left[12\omega_3 - 4\omega_4\right]}}{\omega_1}, 1 - \frac{\omega_2 + \sqrt{\omega_2^2 + 12\omega_1\omega_3}}{2\omega_1}\right],$$ we observe that $\rho_2 \geq 2$ . Then the maximum value of $K(\rho)$ occurs at $0^+$ or $2^-$ . This is a contradiction since we assumed the maximum of $K(\rho)$ occurs at some $\rho \in (0,2)$ . When $$\beta \in \left(1 - \frac{\omega_2 + \sqrt{\omega_2^2 + 12\omega_1\omega_3}}{2\omega_1}, 1\right),\,$$ we observe that $\rho_2 \in (0,2)$ . Since $K''(\rho_2) < 0$ , the maximum value of $K(\rho)$ occurs at $\rho = \rho_2$ . Thus, we have $$\max\{K(\rho): \rho \in (0,2)\} = K(\rho_2) = \frac{4(1-\beta)^2}{(\gamma+1)^2(\gamma+2)^2(2m\lambda+1)^2} \times \left[4 - \frac{[\omega_2(1-\beta) + 9\omega_3 - 4\omega_4]^2}{\omega_4 [\omega_1(1-\beta)^2 - 2\omega_2(1-\beta) - 12\omega_3 + 4\omega_4]}\right].$$ (2.27) Next suppose if $\beta \in [0, \tau]$ and the maximum of $K(\rho)$ occurs at $\rho = 2$ . Then $$\max\{K(\rho): \rho \in [0,2]\} = K(2) = \frac{4(1-\beta)^2}{(\gamma+1)^2(m\lambda+1)} \left[ \frac{(m+1)^2(1-\beta)^2}{(\gamma+1)^2(m\lambda+1)^3} + \frac{6}{(\gamma+2)(\gamma+3)(3m\lambda+1)} \right].$$ We only now need to note (see the idea in the second part of Result 2) if $\beta \in (\tau, 1)$ then the maximum of $K(\rho)$ cannot occur at $\rho = 2$ since $$K(2) \le \frac{4(1-\beta)^2}{(\gamma+1)^2(\gamma+2)^2(2m\lambda+1)^2} \times \left[4 - \frac{\left[\omega_2(1-\beta) + 9\omega_3 - 4\omega_4\right]^2}{\omega_4\left[\omega_1(1-\beta)^2 - 2\omega_2(1-\beta) - 12\omega_3 + 4\omega_4\right]}\right] (= K(\rho_2)).$$ Finally let us consider $\beta \in [0,1)$ and the maximum of $K(\rho)$ occurring at $\rho = 0$ . Then $$\max\{K(\rho): \rho \in [0,2]\} = K(0) = \frac{16(1-\beta)^2}{(\gamma+1)^2(\gamma+2)^2(2m\lambda+1)^2}.$$ We note (see the ideas in the second part of Result 2) if $\beta \in (\tau, 1)$ then the maximum of $K(\rho)$ cannot occur at $\rho = 0$ since $$K(0) \le \frac{4(1-\beta)^2}{(\gamma+1)^2(\gamma+2)^2(2m\lambda+1)^2} \times \left[4 - \frac{\left[\omega_2(1-\beta) + 9\omega_3 - 4\omega_4\right]^2}{\omega_4 \left[\omega_1(1-\beta)^2 - 2\omega_2(1-\beta) - 12\omega_3 + 4\omega_4\right]}\right] (= K(\rho_2)).$$ Finally note (see the ideas in Result 1 and the details in Result 2) if $$\beta \in \left[0, 1 - \frac{\omega_2 + \sqrt{\omega_2^2 + \omega_1 \left[12\omega_3 - 4\omega_4\right]}}{\omega_1}\right],$$ or $$\beta \in \left(1 - \frac{\omega_2 + \sqrt{\omega_2^2 + \omega_1 \left[12\omega_3 - 4\omega_4\right]}}{\omega_1}, 1\right),$$ then the maximum of $K(\rho)$ cannot occur at $\rho = 0$ since $K(0) \leq K(2)$ . This completes the proof. By setting $\lambda = 1$ and $\gamma = 0$ in Theorem 2.1, we obtain the following consequence. Corollary 2.1. [3] Let $f \in \Xi_{\Sigma_m}(\beta)$ $(0 \le \beta < 1)$ be given by (1.5). Then $$\left|a_{m+1}a_{3m+1} - a_{2m+1}^2\right| \leq \begin{cases} &\frac{4(1-\beta)^2}{m+1} \left[\frac{(1-\beta)^2}{m+1} + \frac{1}{3m+1}\right], & \beta \in [0,v] \\ \\ &\frac{(1-\beta)^2}{(2m+1)^2} \left[4 - \frac{[m(1-\beta)\psi_1 + 3\psi_2 - 2\psi_3]^2}{\psi_3[(2m+1)(1-\beta)^2\psi_1 - m(1-\beta)\psi_1 + \psi_3 - 2\psi_2]}\right], \beta \in [v,1) \end{cases}$$ where $$\psi_1 := (2m+1)(3m+1),$$ $$\psi_2 := (m+1)(2m+1)^2,$$ $$\psi_3 := (m+1)^2(3m+1),$$ and $$v := \frac{(3m+1)(7m+4) - \sqrt{m^2(3m+1)^2 + 8\psi_2(3m+1)}}{4\psi_1}.$$ By taking m=1 in Theorem 2.1, we conclude the following result. Corollary 2.2. Let $f \in \Xi_{\Sigma}(\lambda, \gamma; \beta)$ $(\lambda \geq 1, \gamma \in \mathbb{N}_0, 0 \leq \beta < 1)$ be given by (1.1). Then $$|a_2 a_4 - a_3^2| \le \begin{cases} \frac{8(1-\beta)^2}{(\gamma+1)^2(\lambda+1)} \left[ \frac{2(1-\beta)^2}{(\gamma+1)^2(\lambda+1)^3} + \frac{3}{(\gamma+2)(\gamma+3)(3\lambda+1)} \right], & \beta \in [0,\xi] \\ \frac{4(1-\beta)^2}{(\gamma+1)^2(\gamma+2)^2(2\lambda+1)^2} \left[ 4 - \frac{[\vartheta_2(1-\beta) + 9\vartheta_3 - 4\vartheta_4]^2}{\vartheta_4[4\vartheta_1(1-\beta)^2 - 2\vartheta_2(1-\beta) - 12\vartheta_3 + 4\vartheta_4]} \right], \beta \in [\xi, 1) \end{cases}$$ where $$\vartheta_1 := (\gamma + 2)^2 (\gamma + 3)(2\lambda + 1)^2 (3\lambda + 1),$$ $$\vartheta_2 := (\gamma + 1)(\gamma + 2)(\gamma + 3)(\lambda + 1)^2 (2\lambda + 1)(3\lambda + 1),$$ $$\vartheta_3 := (\gamma + 1)^2 (\gamma + 2)(\lambda + 1)^3 (2\lambda + 1)^2,$$ $$\vartheta_4 := (\gamma + 1)^2 (\gamma + 3)(\lambda + 1)^4 (3\lambda + 1),$$ and $$\xi := 1 - \frac{\vartheta_2 + \sqrt{\vartheta_2^2 + 48\vartheta_1\vartheta_3}}{8\vartheta_1}$$ Remark 2.1. Corollary 2.2 improves a result in Altinkaya and Yalçin [2, Theorem 3]. By putting $\gamma = 0$ in Corollary 2.2, we obtain the following result. Corollary 2.3. Let $f \in \Xi_{\Sigma}(\lambda; \beta)$ $(\lambda \geq 1, 0 \leq \beta < 1)$ be given by (1.1). Then $$|a_2 a_4 - a_3^2| \le \begin{cases} \frac{8(1-\beta)^2}{\lambda+1} \left[ \frac{2(1-\beta)^2}{(\lambda+1)^3} + \frac{1}{2(3\lambda+1)} \right], & \beta \in [0,\epsilon] \\ \frac{2(1-\beta)^2}{(2\lambda+1)^2} \left[ 4 - \frac{[\eta_2(1-\beta) + 3\eta_3 - 2\eta_4]^2}{\eta_4[8\eta_1(1-\beta)^2 - 2\eta_2(1-\beta) - 4\eta_3 + 2\eta_4]} \right], \beta \in [\epsilon, 1) \end{cases}$$ where $$\eta_1 := (2\lambda + 1)^2 (3\lambda + 1), \eta_2 := (\lambda + 1)^2 (2\lambda + 1)(3\lambda + 1), \eta_3 := (\lambda + 1)^3 (2\lambda + 1)^2, \eta_4 := (\lambda + 1)^4 (3\lambda + 1),$$ and $$\epsilon := 1 - \frac{(\lambda+1)^2(3\lambda+1) + \sqrt{(\lambda+1)^4(3\lambda+1)^2 + 32(\lambda+1)^3(2\lambda+1)^2(3\lambda+1)}}{16(2\lambda+1)(3\lambda+1)}$$ Remark 2.2. Corollary 2.3 improves a result in Altinkaya and Yalçin [2, Corollary 5]. By setting $\lambda = 1$ in Corollary 2.3, we get the following consequence. Corollary 2.4. [10] Let $f \in \Xi_{\Sigma}(\beta)$ $(0 \le \beta < 1)$ be given by (1.1). Then $$|a_2 a_4 - a_3^2| \le \begin{cases} (1-\beta)^2 \left[ (1-\beta)^2 + \frac{1}{2} \right], & \beta \in \left[ 0, \frac{11-\sqrt{37}}{12} \right] \\ \frac{(1-\beta)^2}{16} \left[ \frac{60\beta^2 - 84\beta - 25}{9\beta^2 - 15\beta + 1} \right], & \beta \in \left[ \frac{11-\sqrt{37}}{12}, 1 \right) \end{cases}.$$ Remark 2.3. Corollary 2.4 recovers a result in Altinkaya and Yalçin [2, Corollary 4].

Definition 2.1. A function $f \in \Sigma_m$ given by (1.5) is said to be in the class $\Xi_{\Sigma_m}(\lambda, \gamma; \beta)$ ( $\lambda \geq 1, \gamma \in \mathbb{N}_0, 0 \leq \beta < 1$ and $m \in \mathbb{N}$ ) if it satisfies the conditions $$\operatorname{Re}\left\{ (1-\lambda)\frac{\mathcal{R}^{\gamma}f(z)}{z} + \lambda \left(\mathcal{R}^{\gamma}f(z)\right)' \right\} > \beta, \tag{2.1}$$ and $$\operatorname{Re}\left\{ (1-\lambda)\frac{\mathcal{R}^{\gamma}f(w)}{w} + \lambda \left(\mathcal{R}^{\gamma}f(w)\right)' \right\} > \beta, \tag{2.2}$$ where $z, w \in \mathbb{U}$ and the function $g = f^{-1}$ is given by (1.7).