On the symmetric $q$-analog on the bi-univalent functions with respect to symmetric points

Lemma 1.3. ([10, 17]). Let $\mathcal{P}$ be the class of all analytic functions h(z) being of the form $$h(z) = 1 + \sum_{n=1}^{\infty} p_n z^n, \quad (z \in \Delta)$$ satisfying $\Re h(z) > 0$ and h(0) = 1. Then the sharp estimates $|p_n| \le 2(n \in \mathbb{N})$ hold. Specially, the equality is true for all n when $$h(z) = \frac{1+z}{1-z} = 1 + \sum_{n=1}^{\infty} 2z^n.$$

Lemma 1.4. ([20]). If $h(z) \in \mathcal{P}$ , then the equalities $$2p_2 = p_1^2 + x(4 - p_1^2)$$ and $$4p_3 = p_1^3 + 2p_1(4 - p_1^2)x - p_1(4 - p_1^2)x^2 + 2(4 - p_1^2)(1 - |x|^2)s$$ hold for some x, s with $|x| \le 1$ and $|s| \le 1$ .

Theorem 2.1. If f(z) given by (1.1) belongs to the class $\widetilde{\mathcal{S}}_{\Sigma_a}^*(\lambda;\phi)$ , then $$|a_2a_4 - a_3^2| \le \begin{cases} \frac{E_1^2}{(3\lambda - 1)^2}, & \text{if } Q \le 0, \ P \le -\frac{Q}{4}, \\ 16P + 4Q + R, & \text{if } Q \ge 0, \ P \ge -\frac{Q}{8} \text{ or } Q \le 0, \ P \ge -\frac{Q}{4}, \\ \frac{4PQ - Q^2}{4P}, & \text{if } Q \ge 0, \ P \ge -\frac{Q}{8}. \end{cases}$$ where $$P = [6(|E_1 - E_2 + E_3| - 2E_1)\lambda^3 \widetilde{[2]}_q^3 + E_1^3 \{\lambda(\lambda - 1)|\lambda - 2|\widetilde{[2]}_q^3 + 3\lambda(\lambda - 1)\widetilde{[2]}_q\widetilde{[3]}_q + 6\lambda(\widetilde{[4]}_q - \widetilde{[2]}_q) + 15(\lambda - 1)\}] \frac{E_1}{96\lambda^5 \widetilde{[2]}_q^4 \widetilde{[4]}_q},$$ $$Q = \frac{E_1(5E_1 + 2|2E_1 - E_2|)}{8\lambda^2 \widetilde{[2]}_q \widetilde{[4]}_q} + \frac{5E_1^2}{4\lambda^3 \widetilde{[4]}_q \widetilde{[2]}_q^2 (\lambda \widetilde{[3]}_q - 1)} + \frac{E_1^3}{2\lambda^2 (3\widetilde{[3]}_q - 1)\widetilde{[2]}_q}$$ and $$R = \frac{E_1^2}{(\lambda \widetilde{[3]}_q - 1)^2}.$$ Proof. Assume that $f(z) \in \mathcal{S}^*_{\Sigma}(\lambda; \phi)$ . Then, according to Remark 1.2 and Lemma 1.3 there exist two analytic functions u(z) and $v(w) \in \mathcal{P}$ so that <span id="page-4-0"></span> $$\frac{2z[\widetilde{\mathcal{D}}_q f(z)]^{\lambda}}{f(z) - f(-z)} = \phi(u(z))$$ (2.6) and <span id="page-4-1"></span> $$\frac{2w[\widetilde{\mathcal{D}}_q g(w)]^{\lambda}}{g(w) - g(-w)} = \phi(v(w)). \tag{2.7}$$ Clearly, the left hand sides of (2.6) and (2.7) can be expanded into the following forms: $$\begin{split} \frac{2z[\widetilde{\mathcal{D}}_{q}f(z)]^{\lambda}}{f(z) - f(-z)} &= 1 + \lambda[\widetilde{2}]_{q}a_{2}z + \left[ (\lambda[\widetilde{3}]_{q} - 1)a_{3} + \frac{\lambda(\lambda - 1)}{2}[\widetilde{2}]_{q}^{2}a_{2}^{2} \right]z^{2} \\ &+ \left\{ \lambda[\widetilde{4}]_{q}a_{4} + \lambda\left(\frac{\lambda - 1}{2}[\widetilde{3}]_{q} - 1\right)[\widetilde{2}]_{q}a_{2}a_{3} + \frac{\lambda(\lambda - 1)(\lambda - 2)}{6}[\widetilde{2}]_{q}^{3}a_{2}^{3} \right\}z^{3} + \dots (2.8) \end{split}$$ and <span id="page-4-2"></span> $$\frac{2w[\widetilde{\mathcal{D}}_{q}g(w)]^{\lambda}}{g(w) - g(-w)} = 1 - \lambda[\widetilde{2}]_{q}a_{2}w + \left[ (\lambda[\widetilde{3}]_{q} - 1)(2a_{2}^{2} - a_{3}) + \frac{\lambda(\lambda - 1)}{2} \widetilde{[2]}_{q}^{2}a_{2}^{2} \right]w^{2} \\ - [\lambda(5a_{2}^{3} - 5a_{2}a_{3} + a_{4})\widetilde{[4]}_{q} + \lambda\left(\frac{\lambda - 1}{2}\widetilde{[3]}_{q} - 1\right)\widetilde{[2]}_{q}a_{2}(2a_{2}^{2} - a_{3}) \\ + \frac{\lambda(\lambda - 1)(\lambda - 2)}{6}\widetilde{[2]}_{q}^{3}a_{2}^{3}]w^{3} + \dots \tag{2.9}$$ Hence, from (2.4-2.5) and (2.6-2.9) we get that <span id="page-4-3"></span> $$\lambda[\widetilde{2}]_q a_2 = \frac{1}{2} E_1 c_1, \tag{2.10}$$ <span id="page-5-0"></span> $$(\lambda \widetilde{[3]}_q - 1)a_3 + \frac{\lambda(\lambda - 1)}{2} \widetilde{[2]}_q^2 a_2^2 = \frac{1}{2} E_1 \left( c_2 - \frac{c_1^2}{2} \right) + \frac{1}{4} E_2 c_1^2, \tag{2.11}$$ <span id="page-5-2"></span> $$\lambda \widetilde{[4]}_{q} a_{4} + \lambda \left(\frac{\lambda - 1}{2} \widetilde{[3]}_{q} - 1\right) \widetilde{[2]}_{q} a_{2} a_{3} + \frac{\lambda(\lambda - 1)(\lambda - 2)}{6} \widetilde{[2]}_{q}^{3} a_{2}^{3}$$ $$= \frac{1}{2} E_{1} \left(c_{3} - c_{1} c_{2} + \frac{c_{1}^{3}}{4}\right) + \frac{1}{4} E_{2} c_{1} \left(c_{2} - \frac{c_{1}^{2}}{2}\right) + \frac{1}{8} E_{3} c_{1}^{3}, \tag{2.12}$$ $$-\lambda [\widetilde{2}]_q a_2 = \frac{1}{2} E_1 d_1, \tag{2.13}$$ <span id="page-5-1"></span> $$(\lambda \widetilde{[3]}_q - 1)(2a_2^2 - a_3) + \frac{\lambda(\lambda - 1)}{2} \widetilde{[2]}_q^2 a_2^2 = \frac{1}{2} E_1 \left( d_2 - \frac{d_1^2}{2} \right) + \frac{1}{4} E_2 d_1^2$$ (2.14) and <span id="page-5-3"></span> $$-\lambda(5a_{2}^{3} - 5a_{2}a_{3} + a_{4})\widetilde{[4]}_{q} - \lambda\left(\frac{\lambda - 1}{2}\widetilde{[3]}_{q} - 1\right)\widetilde{[2]}_{q}a_{2}(2a_{2}^{2} - a_{3}) - \frac{\lambda(\lambda - 1)(\lambda - 2)}{6}\widetilde{[2]}_{q}^{3}a_{2}^{3}$$ $$= \frac{1}{2}E_{1}\left(d_{3} - d_{1}d_{2} + \frac{d_{1}^{3}}{4}\right) + \frac{1}{4}E_{2}d_{1}\left(d_{2} - \frac{d_{1}^{2}}{2}\right) + \frac{1}{8}E_{3}d_{1}^{3}. \tag{2.15}$$ Further, by (2.10) and (2.11) we derive that <span id="page-5-4"></span> $$a_2 = \frac{E_1 c_1}{2\lambda \widetilde{[2]}_q} = -\frac{E_1 d_1}{2\lambda \widetilde{[2]}_q} \tag{2.16}$$ such that <span id="page-5-7"></span> $$c_1 = -d_1 (2.17)$$ and $$E_1^2(c_1^2 + d_1^2) = 8\lambda^2 \widetilde{[2]}_a^2 a_2^2. \tag{2.18}$$ Making use of (2.11) and (2.14) we also obtain that $$2(\lambda[\widetilde{3}]_q - 1)(a_3 - a_2^2) = \frac{1}{2}E_1(c_2 - d_2). \tag{2.19}$$ Therefore <span id="page-5-5"></span> $$a_3 = \frac{E_1(c_2 - d_2)}{4(\lambda[\widetilde{3}]_q - 1)} + \frac{E_1^2 c_1^2}{4\lambda^2 [\widetilde{2}]_q^2}.$$ (2.20) In addition, by (2.12) and (2.15) we conclude that <span id="page-5-6"></span> $$a_{4} = -\frac{5}{2[\widetilde{4}]_{q}} a_{2}(a_{2}^{2} - a_{3}) - \frac{\widetilde{[2]}_{q}}{6[\widetilde{4}]_{q}} \{ (\lambda - 1)[3[\widetilde{3}]_{q} + (\lambda - 2)[\widetilde{2}]_{q}^{2}] - 6 \} a_{2}^{3}$$ $$+ \frac{E_{1}}{4\lambda \widetilde{[4]}_{q}} (c_{3} - d_{3}) + \frac{c_{1}}{8\lambda \widetilde{[4]}_{q}} (E_{2} - 2E_{1})(c_{2} + d_{2}) + \frac{c_{1}^{3}}{8\lambda \widetilde{[4]}_{q}} (E_{1} - E_{2} + E_{3}), \qquad (2.21)$$ then we show from (2.16) and (2.20-2.21) that <span id="page-6-2"></span> $$a_{2}a_{4} - a_{3}^{2} = -\frac{5}{2[\widetilde{4}]_{q}} a_{2}^{2} (a_{2}^{2} - a_{3}) - \frac{\widetilde{[2]}_{q}}{6[\widetilde{4}]_{q}} \{ (\lambda - 1)[\widetilde{3}[\widetilde{3}]_{q} + (\lambda - 2)[\widetilde{2}]_{q}^{2}] - 6 \} a_{2}^{4}$$ $$+ \frac{E_{1}a_{2}}{4\lambda[\widetilde{4}]_{q}} (c_{3} - d_{3}) + \frac{c_{1}a_{2}}{8\lambda[\widetilde{4}]_{q}} (E_{2} - 2E_{1})(c_{2} + d_{2}) + \frac{c_{1}^{3}a_{2}}{8\lambda[\widetilde{4}]_{q}} (E_{1} - E_{2} + E_{3}) - a_{3}^{2}$$ $$= \frac{E_{1}c_{1}^{4}}{96\lambda^{4}[\widetilde{2}]_{q}^{4}[\widetilde{4}]_{q}} [-E_{1}^{3} \{ (\lambda - 1)(\lambda - 2)[\widetilde{2}]_{q}^{3} + 3(\lambda - 1)[\widetilde{2}]_{q}[\widetilde{3}]_{q} - 6([\widetilde{2}]_{q} - [\widetilde{4}]_{q}) \}$$ $$+ 6(E_{1} - E_{2} + E_{3})\lambda^{2}[\widetilde{2}]_{q}^{3}] + \frac{E_{1}^{2}c_{1}(c_{3} - d_{3})}{8\lambda^{2}[\widetilde{2}]_{q}[\widetilde{4}]_{q}} + \frac{E_{1}(E_{2} - 2E_{1})c_{1}^{2}(c_{2} + d_{2})}{16\lambda^{2}[\widetilde{2}]_{q}[\widetilde{4}]_{q}}$$ $$- \frac{E_{1}^{2}(c_{2} - d_{2})^{2}}{16(\lambda[\widetilde{3}]_{q} - 1)^{2}} + \frac{(5 - 4[\widetilde{4}]_{q})E_{1}^{3}c_{1}^{2}(c_{2} - d_{2})}{32\lambda^{2}[\widetilde{2}]_{q}^{2}[\widetilde{4}]_{q}(\lambda[\widetilde{3}]_{q} - 1)}.$$ $$(2.22)$$ From Lemma 1.4 and (2.17), it derives that <span id="page-6-0"></span> $$c_2 - d_2 = \frac{1}{2}(4 - c_1^2)(x - y)$$ and $c_2 + d_2 = c_1^2 + \frac{1}{2}(4 - c_1^2)(x + y),$ (2.23) and <span id="page-6-1"></span> $$c_3 - d_3 = \frac{c_1^3}{2} + \frac{c_1}{2}(4 - c_1^2)(x + y) - \frac{c_1}{4}(4 - c_1^2)(x^2 + y^2) + \frac{1}{2}(4 - c_1^2)[(1 - |x|^2)s - (1 - |y|^2)t]$$ (2.24) for some $x, y, s, t \in [-1, 1]$ . Together with (2.23) and (2.24), (2.22) follows that $$a_{2}a_{4} - a_{3}^{2} = \frac{-E_{1}c_{1}^{4}}{96\lambda^{4}[\widetilde{2}]_{q}^{4}[\widetilde{4}]_{q}} [E_{1}^{3}\{(\lambda - 1)(\lambda - 2)[\widetilde{2}]_{q}^{3} + 3(\lambda - 1)[\widetilde{2}]_{q}[\widetilde{3}]_{q} - 6([\widetilde{2}]_{q} - [\widetilde{4}]_{q})\}$$ $$-6E_{3}\lambda^{2}[\widetilde{2}]_{q}^{3}] + \frac{E_{1}E_{2}c_{1}^{2}(4 - c_{1}^{2})(x + y)}{32\lambda^{2}[\widetilde{2}]_{q}[\widetilde{4}]_{q}}$$ $$-\frac{E_{1}^{2}c_{1}^{2}(4 - c_{1}^{2})}{32\lambda^{2}[\widetilde{2}]_{q}[\widetilde{4}]_{q}}(x^{2} + y^{2}) + \frac{E_{1}^{2}c_{1}(4 - c_{1}^{2})}{16\lambda^{2}[\widetilde{2}]_{q}[\widetilde{4}]_{q}}[(1 - x^{2})s - (1 - y^{2})t]$$ $$-\frac{E_{1}^{2}(4 - c_{1}^{2})^{2}(x - y)^{2}}{64(\lambda[\widetilde{3}]_{q} - 1)^{2}} + \frac{(5 - 4[\widetilde{4}]_{q})E_{1}^{3}c_{1}^{2}(4 - c_{1}^{2})(x - y)}{64\lambda^{2}[\widetilde{2}]_{q}^{2}[\widetilde{4}]_{q}}.$$ (2.25) Then $$\begin{split} |a_2a_4 - a_3^2| &\leq \frac{E_1c_1^4}{96\lambda^4 \widetilde{[2]}_q^4\widetilde{[4]}_q} [E_1^3\{(\lambda - 1)|\lambda - 2|\widetilde{[2]}_q^3 + 3(\lambda - 1)\widetilde{[2]}_q\widetilde{[3]}_q + 6(\widetilde{[4]}_q - \widetilde{[2]}_q)\} \\ &+ 6|E_3|\lambda^2 \widetilde{[2]}_q^3] + \frac{E_1|E_2|c_1^2(4 - c_1^2)(|x| + |y|)}{32\lambda^2 \widetilde{[2]}_q\widetilde{[4]}_q} \end{split}$$ $$\begin{split} &+\frac{E_{1}^{2}c_{1}^{2}(4-c_{1}^{2})(|x|^{2}+|y|^{2})}{32\lambda^{2}[\widetilde{2}]_{q}[\widetilde{4}]_{q}} + \frac{E_{1}^{2}c_{1}(4-c_{1}^{2})}{16\lambda^{2}[\widetilde{2}]_{q}[\widetilde{4}]_{q}}[(1-|x|^{2})+(1-|y|^{2})] \\ &+\frac{E_{1}^{2}(4-c_{1}^{2})^{2}(|x|+|y|)^{2}}{64(\lambda[\widetilde{3}]_{q}-1)^{2}} + \frac{|5-4[\widetilde{4}]_{q}|E_{1}^{3}c_{1}^{2}(4-c_{1}^{2})(|x|+|y|)}{64\lambda^{2}[\widetilde{2}]_{q}^{2}[\widetilde{4}]_{q}(\lambda[\widetilde{3}]_{q}-1)} \\ &=\frac{E_{1}c_{1}^{4}}{96\lambda^{4}[\widetilde{2}]_{q}^{4}[\widetilde{4}]_{q}}[E_{1}^{3}\{(\lambda-1)|\lambda-2|\widetilde{2}]_{q}^{3}+3(\lambda-1)\widetilde{[2}]_{q}^{2}[\widetilde{3}]_{q}+6(\widetilde{[4}]_{q}-\widetilde{[2}]_{q})\} \\ &+6|E_{3}|\lambda^{2}[\widetilde{2}]_{q}^{3}] + \frac{E_{1}^{2}c_{1}(4-c_{1}^{2})}{8\lambda^{2}[\widetilde{2}]_{q}\widetilde{[4}]_{q}} \\ &+\left\{\frac{E_{1}|E_{2}|}{32\lambda^{2}[\widetilde{2}]_{q}\widetilde{[4}]_{q}} + \frac{|5-4[\widetilde{4}]_{q}|E_{1}^{3}}{64\lambda^{2}[\widetilde{2}]_{q}^{2}\widetilde{[4}]_{q}(\lambda[\widetilde{3}]_{q}-1)}\right\}c_{1}^{2}(4-c_{1}^{2})(|x|+|y|) \\ &-\frac{E_{1}^{2}c_{1}(2-c_{1})(4-c_{1}^{2})}{32\lambda^{2}[\widetilde{2}]_{q}\widetilde{[4}]_{q}}(|x|^{2}+|y|^{2}) + \frac{E_{1}^{2}(4-c_{1}^{2})^{2}(|x|+|y|)^{2}}{64(\lambda[\widetilde{3}]_{q}-1)^{2}}. \end{split}$$ Since $|c_1| \le 2$ , we assume that $c = c_1 \in [0, 2]$ . Letting $\xi = |x|$ and $\zeta = |y|$ , we remark that $$|a_2 a_4 - a_3^2| \le \mathcal{F}_1 + (\xi + \zeta)\mathcal{F}_2 + (\xi^2 + \zeta^2)\mathcal{F}_3 + (\xi + \zeta)^2 \mathcal{F}_4 =: \Pi(\xi, \zeta)$$ (2.26) in the closed square $\mathcal{D} = \{(\xi, \zeta) : 0 \le \xi \le 1, 0 \le \zeta \le 1\}$ , where $$\begin{split} \mathcal{F}_{1} &= \mathcal{F}_{1}(c) := \frac{E_{1}c^{4}}{96\lambda^{4}\widetilde{[2]}_{q}^{4}\widetilde{[4]}_{q}} [E_{1}^{3}\{(\lambda-1)|\lambda-2|\widetilde{[2]}_{q}^{3}+3(\lambda-1)\widetilde{[2]}_{q}\widetilde{[3]}_{q}+6(\widetilde{[4]}_{q}-\widetilde{[2]}_{q})\} \\ &+ 6|E_{3}|\lambda^{2}\widetilde{[2]}_{q}^{3}] + \frac{E_{1}^{2}c(4-c^{2})}{8\lambda^{2}\widetilde{[2]}_{q}\widetilde{[4]}_{q}} \geq 0, \\ \mathcal{F}_{2} &= \mathcal{F}_{2}(c) := \left\{ \frac{E_{1}|E_{2}|}{32\lambda^{2}\widetilde{[2]}_{q}\widetilde{[4]}_{q}} + \frac{|5-4\widetilde{[4]}_{q}|E_{1}^{3}}{64\lambda^{2}\widetilde{[2]}_{q}^{2}\widetilde{[4]}_{q}(\lambda\widetilde{[3]}_{q}-1)} \right\} c^{2}(4-c^{2}) \geq 0, \\ \mathcal{F}_{3} &= \mathcal{F}_{3}(c) := -\frac{E_{1}^{2}c(2-c)(4-c^{2})}{32\lambda^{2}\widetilde{[2]}_{1}\widetilde{[4]}} \leq 0 \quad and \quad \mathcal{F}_{4} = \mathcal{F}_{4}(c) := \frac{E_{1}^{2}(4-c^{2})^{2}}{64(2\widetilde{[3]}_{1}-1)^{2}} \geq 0. \end{split}$$ For $c \in [0, 2]$ , we will maximize $\Pi(\xi, \zeta)$ by the sign of $\Pi_{\xi\xi}\Pi_{\zeta\zeta} - (\Pi_{\xi\zeta})^2$ in the closed square $\mathcal{D} = \{(\xi, \zeta) : 0 \le \xi \le 1, 0 \le \zeta \le 1\}$ . Consider $c \in (0,2)$ . By the simple calculation, we observe that $\mathcal{F}_3 < 0$ and $\mathcal{F}_3 + 2\mathcal{F}_4 > 0$ for $c \in (0,2)$ . Then, we know that $\Pi_{\xi\xi}\Pi_{\zeta\zeta} - (\Pi_{\xi\zeta})^2 < 0$ such that $\Pi$ doesn't take the local maximum in the interior the square $\mathcal{D}$ . For $\xi = 0$ and $0 \le \zeta \le 1$ (or, for $\zeta = 0$ and $0 \le \xi \le 1$ ), we see that $$\Pi(0,\zeta) = H(\zeta) = \mathcal{F}_1 + \mathcal{F}_2\zeta + (\mathcal{F}_3 + \mathcal{F}_4)\zeta^2.$$ Case (I) If $\mathcal{F}_3 + \mathcal{F}_4 \ge 0$ , then $H'(\zeta) = \mathcal{F}_2 + 2(\mathcal{F}_3 + \mathcal{F}_4)\zeta > 0$ for $0 < \zeta < 1$ and $c \in (0, 2)$ , and $H(\zeta)$ is an increasing function in [0, 1]. Hence, $H(\zeta)$ reaches the maximum via $$\max H(\zeta) = H(1) = \mathcal{F}_1 + \mathcal{F}_2 + \mathcal{F}_3 + \mathcal{F}_4.$$ Case (II) If $\mathcal{F}_3 + \mathcal{F}_4 < 0$ , we study two versions for the critical point $\widetilde{\zeta} = \frac{-\mathcal{F}_2}{2(\mathcal{F}_3 + \mathcal{F}_4)}$ . When $\widetilde{\zeta} > 1$ , we infer that $0 \le -(\mathcal{F}_3 + \mathcal{F}_4) \le \mathcal{F}_2 + \mathcal{F}_3 + \mathcal{F}_4$ so that $H(0) \le \mathcal{F}_1 + \mathcal{F}_2 + \mathcal{F}_3 + \mathcal{F}_4 = H(1)$ . Inversely, when $\widetilde{\zeta} \le 1$ , we have that $$H(1) = \mathcal{F}_1 + \frac{1}{2}\mathcal{F}_2 + \frac{1}{2}(\mathcal{F}_2 + 2\mathcal{F}_3 + 2\mathcal{F}_4) \le \mathcal{F}_1 + \frac{1}{2}\mathcal{F}_2$$ and $$H(0) \le H(\widetilde{\zeta}) = \mathcal{F}_1 + \frac{-\mathcal{F}_2^2}{4(\mathcal{F}_3 + \mathcal{F}_4)} \le \mathcal{F}_1 + \frac{1}{2}\mathcal{F}_2.$$ As for c = 0 or c = 2, we see that $$\Pi(\xi,\zeta) = (\xi+\zeta)^2 \mathcal{F}_4, \mathcal{F}_1 = \mathcal{F}_2 = \mathcal{F}_3 = 0 \text{ and } \mathcal{T}_4 > 0$$ or $$\Pi(\xi,\zeta) = \mathcal{F}_1 + (\xi + \zeta)^2 \mathcal{F}_4, \mathcal{F}_2 = \mathcal{F}_3 = 0 \text{ and } \mathcal{F}_1, \mathcal{F}_4 > 0.$$ When $\xi = \zeta = 1$ , we note that F reaches the corresponding maximum $4\mathcal{T}_4$ or $\mathcal{F}_1 + 4\mathcal{F}_4$ . For $\xi = 1$ and $0 \le \zeta \le 1$ (or, for $\zeta = 1$ and $0 \le \xi \le 1$ ), we derive that $$\Pi(1,\zeta) =: G(\zeta) = \mathcal{F}_1 + \mathcal{F}_2 + \mathcal{F}_3 + \mathcal{F}_4 + (\mathcal{F}_2 + 2\mathcal{F}_4)\zeta + (\mathcal{F}_3 + \mathcal{F}_4)\zeta^2.$$ Case (III) If $\mathcal{F}_3 + \mathcal{F}_4 \ge 0$ , then $G'(\zeta) = \mathcal{F}_2 + 2\mathcal{F}_4 + 2(\mathcal{F}_3 + \mathcal{F}_4)\zeta > 0$ for $0 < \zeta < 1$ and $c \in (0, 2)$ , and so $G(\zeta)$ is also an increasing function in [0, 1]. Therefore, the maximum of $G(\zeta)$ is $$\max G(\zeta) = G(1) = \mathcal{F}_1 + 2\mathcal{F}_2 + 2\mathcal{F}_3 + 4\mathcal{F}_4.$$ Case (IV) If $\mathcal{F}_3 + \mathcal{F}_4 < 0$ , then there have two versions with respect to the critical point $\widetilde{\zeta} = -\frac{\mathcal{F}_2 + 2\mathcal{F}_4}{2(\mathcal{F}_3 + \mathcal{F}_4)}$ . When $\widetilde{\zeta} > 1$ , we know that $0 \le -(\mathcal{F}_3 + \mathcal{F}_4) \le \mathcal{F}_2 + \mathcal{F}_3 + 3\mathcal{F}_4$ so that $G(0) \le \mathcal{F}_1 + 2\mathcal{F}_2 + 2\mathcal{F}_3 + 4\mathcal{F}_4 = G(1)$ . Oppositely, when $\widetilde{\zeta} \le 1$ , $$G(0) \leq G(\widetilde{\zeta}) = \mathcal{F}_1 + \mathcal{F}_2 + \mathcal{F}_3 + \mathcal{F}_4 - \frac{(\mathcal{F}_2 + 2\mathcal{F}_4)^2}{4(\mathcal{F}_3 + \mathcal{F}_4)} \leq \mathcal{F}_1 + \frac{3}{2}\mathcal{F}_2 + \mathcal{F}_3 + 2\mathcal{F}_4 \leq G(1).$$ All in all, by all the above cases, the maximum of $\Pi$ in the closed square $\mathcal{D} = \{(\xi, \zeta) : 0 \le \xi \le 1, 0 \le \zeta \le 1\}$ occurs at $\xi = 1$ and $\zeta = 1$ , that is to say, $$\max_{(\xi,\zeta)\in\mathcal{D}}\Pi(\xi,\zeta)=\Pi(1,1).$$ Consider the function $G: [0,2] \to \mathbb{R}$ by $$\mathcal{G}(c) = \max_{(\xi,\zeta)\in\mathcal{D}} \Pi(\xi,\zeta) = \Pi(1,1) = \mathcal{F}_1 + 2\mathcal{F}_2 + 2\mathcal{F}_3 + 4\mathcal{F}_4.$$ Then $$\mathcal{G}(c) = \frac{E_1}{96\lambda^4 \widetilde{[2]}_q^4 \widetilde{[4]}_q} [E_1^3 \{ (\lambda-1)|\lambda-2|\widetilde{[2]}_q^3 + 3(\lambda-1)\widetilde{[2]}_q \widetilde{[3]}_q + 6(\widetilde{[4]}_q - \widetilde{[2]}_q) \}$$ $$\begin{split} &+6(|E_{3}|-E_{1}-|E_{2}|)\lambda^{2}\widetilde{[2]}_{q}^{3}-\frac{3|5-4\widetilde{[4]}_{q}|E_{1}^{2}\lambda^{2}\widetilde{[2]}_{q}^{2}}{\lambda\widetilde{[3]}_{q}-1}+\frac{6E_{1}\lambda^{4}\widetilde{[2]}_{q}^{4}\widetilde{[4]}_{q}}{(\lambda\widetilde{[3]}_{q}-1)^{2}}]c^{4}\\ &+\left[\frac{E_{1}(E_{1}+|E_{2}|)}{4\lambda^{2}\widetilde{[2]}_{q}\widetilde{[4]}_{q}}+\frac{|5-4\widetilde{[4]}_{q}|E_{1}^{3}}{8\lambda^{2}\widetilde{[2]}_{q}^{2}\widetilde{[4]}_{q}(\lambda\widetilde{[3]}_{q}-1)}-\frac{E_{1}^{2}}{2(\lambda\widetilde{[3]}_{q}-1)^{2}}\right]c^{2}\\ &+\frac{E_{1}^{2}}{(\lambda\widetilde{[3]}_{q}-1)^{2}}. \end{split}$$ Let $t = c^2$ . Denote <span id="page-9-0"></span> $$P = \frac{E_{1}}{96\lambda^{4}[\widetilde{2}]_{q}^{4}[\widetilde{4}]_{q}} [E_{1}^{3}\{(\lambda - 1)|\lambda - 2|\widetilde{[2]}_{q}^{3} + 3(\lambda - 1)\widetilde{[2]}_{q}[\widetilde{3}]_{q} + 6(\widetilde{[4]}_{q} - \widetilde{[2]}_{q})\}$$ $$+ 6(|E_{3}| - E_{1} - |E_{2}|)\lambda^{2}[\widetilde{2}]_{q}^{3} - \frac{3|5 - 4\widetilde{[4]}_{q}|E_{1}^{2}\lambda^{2}\widetilde{[2]}_{q}^{2}}{\lambda^{2}\widetilde{[3]}_{q} - 1} + \frac{6E_{1}\lambda^{4}\widetilde{[2]}_{q}^{4}\widetilde{[4]}_{q}}{(\lambda^{2}\widetilde{[3]}_{q} - 1)^{2}}], \qquad (2.27)$$ $$Q = \frac{E_1(E_1 + |E_2|)}{4\lambda^2 \widetilde{[2]}_q \widetilde{[4]}_q} + \frac{|5 - 4\widetilde{[4]}_q |E_1^3}{8\lambda^2 \widetilde{[2]}_a^2 \widetilde{[4]}_q (\lambda \widetilde{[3]}_q - 1)} - \frac{E_1^2}{2(\lambda \widetilde{[3]}_q - 1)^2}$$ (2.28) and <span id="page-9-1"></span> $$R = \frac{E_1^2}{(\lambda \widetilde{[3]}_a - 1)^2}. (2.29)$$ By following the standard computations of the optimal value of quadratic as follows: $$\max_{0 \le t \le 4} (Pt^2 + Qt + R) = \begin{cases} R, & \text{if } Q \le 0, \ P \le -\frac{Q}{4}, \\ 16P + 4Q + R, & \text{if } Q \ge 0, \ P \ge -\frac{Q}{8} \text{ or } Q \le 0, \ P \ge -\frac{Q}{4}, \\ \frac{4PQ - Q^2}{4P}, & \text{if } Q \ge 0, \ P \ge -\frac{Q}{8}, \end{cases}$$ we imply that $$|a_2a_4 - a_3^2| \le \begin{cases} R, & \text{if } Q \le 0, \ P \le -\frac{Q}{4}, \\ 16P + 4Q + R, & \text{if } Q \ge 0, \ P \ge -\frac{Q}{8} \text{ or } Q \le 0, \ P \ge -\frac{Q}{4}, \\ \frac{4PQ - Q^2}{4P}, & \text{if } Q \ge 0, \ P \ge -\frac{Q}{8}, \end{cases}$$ where P, Q and R are given by (2.27-2.29). - Remark 2.2. By the same method as in Theorem 2.1, further we can study the second Hankel determinants for the classes $\widetilde{C}_{\Sigma_q}(\lambda;\phi)$ with respect to symmetric points, and leave them to the interested readers. In addition, when $q \to 1_-$ , there are some versions of the second Hankel determinants for some function classes under some appropriate assumption conditions: - (I) For $\phi(z) = z + \sqrt{1 + z^2}$ , refer to [50] for the classes $\widetilde{S}_{\Sigma_{1_-}}^*(1;\phi)$ and $\widetilde{C}_{\Sigma_{1_-}}(1;\phi)$ related to a shell shaped region; - (II) For $\phi(z) = \frac{1+Az}{1+Bz}$ , refer to [18] for the classes $\widetilde{\mathcal{S}}_{\Sigma_{1_{-}}}^{*}(1;\phi)$ and $\widetilde{\mathcal{C}}_{\Sigma_{1_{-}}}(1;\phi)$ with respect to a symmetric or conjugate point.

Theorem 3.1. If f(z) given by (1.1) is in the class $\widetilde{\mathcal{S}_{\Sigma_a}^*}^{\eta}(\mu,\lambda;\phi)$ , then <span id="page-10-2"></span> $$|a_{2}| \leq \min \left\{ \frac{E_{1}}{\lambda |[\mu - (\mu - 1)\widetilde{[2]}_{q}]L_{2}|\widetilde{[2]}_{q}}, \sqrt{\frac{2(|E_{2} - E_{1}| + E_{1})}{|\Omega|}}, \frac{E_{1}\sqrt{2E_{1}}}{\sqrt{|\Theta|}} \right\}$$ (3.1) and $$|a_{3}| \leq \frac{E_{1}}{(\lambda[\widetilde{3}]_{q} - 1)|[\mu - (\mu - 1)\widetilde{[3]}_{q}]L_{3}|} + \min \left\{ \frac{E_{1}^{2}}{\lambda^{2}[\mu - (\mu - 1)\widetilde{[2]}_{q}]^{2}\widetilde{[2]}_{q}^{2}|L_{2}|^{2}}, \frac{|2(E_{2} - E_{1}| + E_{1})}{|\Omega|} \right\},$$ (3.2) where $$\Omega := 2(\lambda[\widetilde{3}]_q - 1)[\mu - (\mu - 1)\widetilde{[3]}_q]L_3 + \lambda(\lambda[\mu - (\mu - 1)\widetilde{[2]}_q]^2 - [\mu - (\mu - 1)\widetilde{[2]}_q^2])\widetilde{[2]}_q^2 L_2^2$$ (3.3) and $$\begin{split} \Theta :&= 2(\lambda \widetilde{[3]}_q - 1)[\mu - (\mu - 1)\widetilde{[3]}_q]E_1^2L_3 \\ &+ \lambda \{\lambda [\mu - (\mu - 1)\widetilde{[2]}_q]^2(E_1^2 + 2E_1 - 2E_2) - [\mu - (\mu - 1)\widetilde{[2]}_q^2]E_1^2\}\widetilde{[2]}_q^2L_2^2. \end{split} \tag{3.4}$$ Proof. Suppose that $f(z) \in \widetilde{S}_{\sum q}^{*\eta}(\mu, \lambda; \phi)$ . Then, by Definition 1.1 and Lemma 1.3 there exist two analytic functions u(z) and $v(w) \in \mathcal{P}$ such that <span id="page-10-0"></span> $$\left\{ \frac{2z[\widetilde{\mathcal{D}}_q(\mathcal{J}_q^{\eta}f)(z)]^{\lambda}}{\mathcal{J}_q^{\eta}f(z) - \mathcal{J}_q^{\eta}f(-z)} \right\}^{\mu} \left\{ \frac{2\{\widetilde{\mathcal{D}}_q[z\widetilde{\mathcal{D}}_q(\mathcal{J}_q^{\eta}f)(z)]\}^{\lambda}}{\widetilde{\mathcal{D}}_q[\mathcal{J}_q^{\eta}f(z) - \mathcal{J}_q^{\eta}f(-z)]} \right\}^{1-\mu} = \phi(u(z)) \tag{3.5}$$ and <span id="page-10-1"></span> $$\left\{ \frac{2w[\widetilde{\mathcal{D}}_q(\mathcal{J}_q^{\eta}g)(w)]^{\lambda}}{\mathcal{J}_q^{\eta}g(w) - \mathcal{J}_q^{\eta}g(-w)} \right\}^{\mu} \left\{ \frac{2\{\widetilde{\mathcal{D}}_q[w\widetilde{\mathcal{D}}_q(\mathcal{J}_q^{\eta}g)(w)]\}^{\lambda}}{\widetilde{\mathcal{D}}_q[\mathcal{J}_q^{\eta}g(w) - \mathcal{J}_q^{\eta}g(-w)]} \right\}^{1-\mu} = \phi(v(w)) \tag{3.6}$$ By expanding the left half parts of (3.5) and (3.6), it leads to $$\left\{ \frac{2z[\widetilde{\mathcal{D}}_q(\mathcal{J}_q^{\eta}f)(z)]^{\lambda}}{\mathcal{J}_q^{\eta}f(z) - \mathcal{J}_q^{\eta}f(-z)} \right\}^{\mu} = 1 + \lambda \mu[\widetilde{2}]_q L_2 a_2 z + \left\{ \mu[(\lambda[\widetilde{3}]_q - 1)L_3 a_3 + \frac{\lambda(\lambda - 1)}{2} [\widetilde{2}]_q^2 L_2^2 a_2^2] + \frac{\mu(\mu - 1)}{2} \lambda^2 [\widetilde{2}]_q^2 L_2^2 a_2^2 \right\} z^2 + \dots,$$ $$\left\{ \frac{2\{\widetilde{\mathcal{D}}_{q}[z\widetilde{\mathcal{D}}_{q}(\mathcal{J}_{q}^{\eta}f)(z)]\}^{\lambda}}{\widetilde{\mathcal{D}}_{q}[\mathcal{J}_{q}^{\eta}f(z) - \mathcal{J}_{q}^{\eta}f(-z)]} \right\}^{1-\mu} = 1 - \lambda(\mu - 1)[\widetilde{2}]_{q}^{2}L_{2}a_{2}z \\ - \left\{ (\mu - 1)[(\lambda[\widetilde{3}]_{q} - 1)[\widetilde{3}]_{q}L_{3}a_{3} + \frac{\lambda(\lambda - 1)}{2}[\widetilde{2}]_{q}^{4}L_{2}^{2}a_{2}^{2}] - \frac{\mu(\mu - 1)}{2}\lambda^{2}[\widetilde{2}]_{q}^{4}L_{2}^{2}a_{2}^{2} \right\}z^{2} + \dots$$ so that $$\left\{ \frac{2z[\widetilde{\mathcal{D}}_{q}(\mathcal{J}_{q}^{\eta}f)(z)]^{\lambda}}{\mathcal{J}_{q}^{\eta}f(z) - \mathcal{J}_{q}^{\eta}f(-z)} \right\}^{\mu} \left\{ \frac{2\{\widetilde{\mathcal{D}}_{q}[z\widetilde{\mathcal{D}}_{q}(\mathcal{J}_{q}^{\eta}f)(z)]\}^{\lambda}}{\widetilde{\mathcal{D}}_{q}[\mathcal{J}_{q}^{\eta}f(z) - \mathcal{J}_{q}^{\eta}f(-z)]} \right\}^{1-\mu} = 1 + \lambda[\mu - (\mu - 1)[\widetilde{2}]_{q}][\widetilde{2}]_{q}L_{2}a_{2}z + \{(\lambda[\widetilde{3}]_{q} - 1)[\mu - (\mu - 1)[\widetilde{3}]_{q}]L_{3}a_{3} + \frac{\lambda}{2} \left(\lambda[\mu - (\mu - 1)[\widetilde{2}]_{q}]^{2} - [\mu - (\mu - 1)[\widetilde{2}]_{q}^{2}]\right)[\widetilde{2}]_{q}^{2}L_{2}^{2}a_{2}^{2}\}z^{2} + \dots, \tag{3.7}$$ and $$\begin{split} &\left\{ \frac{2w[\widetilde{\mathcal{D}}_{q}(\mathcal{J}_{q}^{\eta}g)(w)]^{\lambda}}{\mathcal{J}_{q}^{\eta}g(w) - \mathcal{J}_{q}^{\eta}g(-w)} \right\}^{\mu} = 1 - \lambda\mu[\widetilde{2}]_{q}L_{2}a_{2}w + \\ &\left\{ \mu[(\lambda[\widetilde{3}]_{q} - 1)L_{3}(2a_{2}^{2} - a_{3}) + \frac{\lambda(\lambda - 1)}{2}[\widetilde{2}]_{q}^{2}L_{2}^{2}a_{2}^{2}] + \frac{\mu(\mu - 1)}{2}\lambda^{2}[\widetilde{2}]_{q}^{2}L_{2}^{2}a_{2}^{2} \right\}w^{2} + \dots, \end{split}$$ $$\begin{split} &\left\{\frac{2\{\widetilde{\mathcal{D}}_{q}[w\widetilde{\mathcal{D}}_{q}(\mathcal{J}_{q}^{\eta}g)(w)]\}^{\lambda}}{\widetilde{\mathcal{D}}_{q}[\mathcal{J}_{q}^{\eta}g(w)-\mathcal{J}_{q}^{\eta}g(-w)]}\right\}^{1-\mu} = 1 + \lambda(\mu-1)\widetilde{[2]}_{q}^{2}L_{2}a_{2}w \\ &-\left\{(\mu-1)[(\lambda\widetilde{[3]}_{q}-1)\widetilde{[3]}_{q}L_{3}(2a_{2}^{2}-a_{3}) + \frac{\lambda(\lambda-1)}{2}\widetilde{[2]}_{q}^{4}L_{2}^{2}a_{2}^{2}\right] - \frac{\mu(\mu-1)}{2}\lambda^{2}\widetilde{[2]}_{q}^{4}L_{2}^{2}a_{2}^{2}\right\}w^{2} + \dots \end{split}$$ so that <span id="page-11-0"></span> $$\left\{ \frac{2w[\widetilde{\mathcal{D}}_{q}(\mathcal{J}_{q}^{\eta}g)(w)]^{\lambda}}{\mathcal{J}_{q}^{\eta}g(w) - \mathcal{J}_{q}^{\eta}g(-w)} \right\}^{\mu} \left\{ \frac{2\{\widetilde{\mathcal{D}}_{q}[w\widetilde{\mathcal{D}}_{q}(\mathcal{J}_{q}^{\eta}g)(w)]\}^{\lambda}}{\widetilde{\mathcal{D}}_{q}[\mathcal{J}_{q}^{\eta}g(w) - \mathcal{J}_{q}^{\eta}g(-w)]} \right\}^{1-\mu} = 1 - \lambda[\mu - (\mu - 1)[\widetilde{2}]_{q}][\widetilde{2}]_{q}L_{2}a_{2}w + \{(\lambda[\widetilde{3}]_{q} - 1)[\mu - (\mu - 1)[\widetilde{3}]_{q}]L_{3}(2a_{2}^{2} - a_{3}) + \frac{\lambda}{2}(\lambda[\mu - (\mu - 1)[\widetilde{2}]_{q}]^{2} - [\mu - (\mu - 1)[\widetilde{2}]_{q}^{2}])[\widetilde{2}]_{q}^{2}L_{2}^{2}a_{2}^{2}\}w^{2} + \dots$$ (3.8) Therefore, according to (2.4-2.5) and (3.5-3.8) we have that <span id="page-11-1"></span> $$\lambda[\mu - (\mu - 1)[2]_q][2]_q L_2 a_2 = \frac{1}{2} E_1 c_1, \tag{3.9}$$ <span id="page-11-2"></span> $$(\lambda \widetilde{[3]}_{q} - 1)[\mu - (\mu - 1)\widetilde{[3]}_{q}]L_{3}a_{3} + \frac{\lambda}{2} \left(\lambda [\mu - (\mu - 1)\widetilde{[2]}_{q}]^{2} - [\mu - (\mu - 1)\widetilde{[2]}_{q}^{2}]\right) \widetilde{[2]}_{q}^{2} L_{2}^{2} a_{2}^{2}$$ $$= \frac{1}{2} E_{1} \left(c_{2} - \frac{c_{1}^{2}}{2}\right) + \frac{1}{4} E_{2} c_{1}^{2}, \tag{3.10}$$ <span id="page-12-0"></span> $$-\lambda[\mu - (\mu - 1)\widetilde{[2]}_q]\widetilde{[2]}_q L_2 a_2 = \frac{1}{2} E_1 d_1$$ (3.11) and <span id="page-12-1"></span> $$\begin{split} (\lambda\widetilde{[3]}_q - 1)[\mu - (\mu - 1)\widetilde{[3]}_q]L_3(2a_2^2 - a_3) + \frac{\lambda}{2} \left(\lambda[\mu - (\mu - 1)\widetilde{[2]}_q]^2 - [\mu - (\mu - 1)\widetilde{[2]}_q^2]\right)\widetilde{[2]}_q^2 L_2^2 a_2^2 \\ = \frac{1}{2} E_1 \left(d_2 - \frac{d_1^2}{2}\right) + \frac{1}{4} E_2 d_1^2. \end{split} \tag{3.12}$$ From (3.9) and (3.11), it infers that $$a_{2} = \frac{E_{1}c_{1}}{2\lambda[\mu - (\mu - 1)\widetilde{[2]}_{q}]\widetilde{[2]}_{q}L_{2}} = -\frac{E_{1}d_{1}}{2\lambda[\mu - (\mu - 1)\widetilde{[2]}_{q}]\widetilde{[2]}_{q}L_{2}}$$ (3.13) such that <span id="page-12-5"></span> $$c_1 = -d_1 (3.14)$$ and <span id="page-12-2"></span> $$E_1^2(c_1^2 + d_1^2) = 8\lambda^2 [\mu - (\mu - 1)\widetilde{[2]}_q]^2 \widetilde{[2]}_q^2 L_2^2 a_2^2.$$ (3.15) By (3.10) and (3.12), we get that <span id="page-12-3"></span> $$\begin{aligned} \{2(\lambda[\widetilde{3}]_{q} - 1)[\mu - (\mu - 1)\widetilde{[3]}_{q}]L_{3} + \lambda \left(\lambda[\mu - (\mu - 1)\widetilde{[2]}_{q}]^{2} - [\mu - (\mu - 1)\widetilde{[2]}_{q}^{2}]\right)\widetilde{[2]}_{q}^{2}L_{2}^{2}\}a_{2}^{2} \\ &= \frac{1}{4}(E_{2} - E_{1})(c_{1}^{2} + d_{1}^{2}) + \frac{1}{2}E_{1}(c_{2} + d_{2}). \end{aligned} (3.16)$$ Therefore, by (3.15-3.16) we obtain that <span id="page-12-4"></span> $$a_2^2 = \frac{E_1^3(c_2 + d_2)}{2\Theta},\tag{3.17}$$ where $$\begin{split} \Theta :&= 2(\lambda \widetilde{[3]}_q - 1)[\mu - (\mu - 1)\widetilde{[3]}_q]E_1^2L_3 \\ &+ \lambda \{\lambda [\mu - (\mu - 1)\widetilde{[2]}_q]^2(E_1^2 + 2E_1 - 2E_2) - [\mu - (\mu - 1)\widetilde{[2]}_q^2]E_1^2\}\widetilde{[2]}_q^2L_2^2. \end{split}$$ Then, it enables us to follow from Lemma 1.3 and (3.15-3.17) that $$|a_2| \le \frac{E_1}{\lambda |[\mu - (\mu - 1)\widetilde{[2]}_q]L_2|\widetilde{[2]}_q},$$ $$|a_{2}| \leq \sqrt{\frac{2(|E_{2} - E_{1}| + E_{1})}{|2(\lambda[\widetilde{3}]_{q} - 1)[\mu - (\mu - 1)[\widetilde{3}]_{q}]L_{3} + \lambda(\lambda[\mu - (\mu - 1)[\widetilde{2}]_{q}]^{2} - [\mu - (\mu - 1)[\widetilde{2}]_{q}^{2}])[\widetilde{2}]_{q}^{2}L_{2}^{2}|}}$$ and $$|a_2| \leq \frac{E_1 \sqrt{2E_1}}{\sqrt{|\Theta|}}.$$ As a consequence, (3.1) holds true. Similarly, from (3.10), (3.12) and (3.14), it also implies that <span id="page-13-0"></span> $$2(\lambda[\widetilde{3}]_q - 1)[\mu - (\mu - 1)\widetilde{[3]}_q]L_3(a_3 - a_2^2) = \frac{1}{2}E_1(c_2 - d_2). \tag{3.18}$$ Hence, inserting (3.15) into (3.18) we obtain that $$a_{3} = \frac{E_{1}(c_{2} - d_{2})}{4(\lambda[\widetilde{3}]_{q} - 1)[\mu - (\mu - 1)[\widetilde{3}]_{q}]L_{3}} + \frac{E_{1}^{2}(c_{1}^{2} + d_{1}^{2})}{8\lambda^{2}[\mu - (\mu - 1)[\widetilde{2}]_{q}]^{2}[\widetilde{2}]_{q}^{2}L_{2}^{2}}.$$ (3.19) Therefore, from Lemma 1.3 it demonstrates that $$\mid a_{3} \mid \leq \frac{E_{1}}{(\lambda[\widetilde{3}]_{q}-1)|[\mu-(\mu-1)[\widetilde{3}]_{q}]L_{3}|} + \frac{E_{1}^{2}}{\lambda^{2}[\mu-(\mu-1)[\widetilde{2}]_{a}]^{2}[\widetilde{2}]_{a}^{2}|L_{2}|^{2}}.$$ On the other hand, by (3.16) and (3.18) we infer that $$\begin{split} a_3 &= \frac{E_1(c_2-d_2)}{4(\lambda\widetilde{[3]}_q-1)[\mu-(\mu-1)\widetilde{[3]}_q]L_3} \\ &+ \frac{\frac{1}{4}(E_2-E_1)(c_1^2+d_1^2)+\frac{1}{2}E_1(c_2+d_2)}{2(\lambda\widetilde{[3]}_q-1)[\mu-(\mu-1)\widetilde{[3]}_q]L_3+\lambda\left(\lambda[\mu-(\mu-1)\widetilde{[2]}_q]^2-[\mu-(\mu-1)\widetilde{[2]}_q^2]\right)\widetilde{[2]}_q^2L_2^2}. \end{split}$$ Thus, from Lemma 1.3 we see that $$|a_{3}| \leq \frac{E_{1}}{(\lambda[\widetilde{3}]_{q} - 1)|[\mu - (\mu - 1)\widetilde{[3}]_{q}]L_{3}|} + \frac{2(|E_{2} - E_{1}| + E_{1})}{|2(\lambda[\widetilde{3}]_{q} - 1)[\mu - (\mu - 1)\widetilde{[3}]_{q}]L_{3} + \lambda\left(\lambda[\mu - (\mu - 1)\widetilde{[2}]_{q}]^{2} - [\mu - (\mu - 1)\widetilde{[2}]_{q}^{2}]\right)\widetilde{[2]}_{q}^{2}L_{2}^{2}|}.$$

Corollary 3.2. If f(z) given by (1.1) is in the class $\widetilde{\mathcal{S}}_{\Sigma q}^{*\eta}(\lambda;\phi)$ , then $$\mid a_2 \mid \leq \min \left\{ \frac{E_1}{\lambda |L_2|\widetilde{[2]}_q}, \sqrt{\frac{2(|E_2 - E_1| + E_1)}{|\Gamma|}}, \frac{E_1 \sqrt{2E_1}}{\sqrt{|\Xi|}} \right\}$$ and $$|a_3| \le \frac{E_1}{(\lambda[\widetilde{3}]_q - 1)|L_3|} + \min \left\{ \frac{E_1^2}{\lambda^2[\widetilde{2}]_q^2|L_2|^2}, \frac{2(E_2 - E_1| + E_1)}{|\Gamma|} \right\},$$ where $$\Gamma := 2(\lambda \widetilde{[3]}_q - 1)L_3 + \lambda(\lambda - 1)\widetilde{[2]}_q^2 L_2^2$$ and $$\Xi := 2(\lambda \widetilde{[3]}_q - 1)E_1^2 L_3 + \lambda [(\lambda - 1)E_1^2 + 2\lambda (E_1 - E_2)]\widetilde{[2]}_q^2 L_2^2.$$

Corollary 3.3. If f(z) given by (1.1) is in the class $\widetilde{C}_{\Sigma q}^{\eta}(\mu, \lambda; \phi)$ , then $$|a_2| \le \min \left\{ \frac{E_1}{\lambda |L_2| \widetilde{[2]}_q^2}, \sqrt{\frac{2(|E_2 - E_1| + E_1)}{|\Psi|}}, \frac{E_1 \sqrt{2E_1}}{\sqrt{|\Upsilon|}} \right\}$$ and $$|a_3| \le \frac{E_1}{(\lambda[\widetilde{3}]_q - 1)[\widetilde{3}]_q |L_3|} + \min \left\{ \frac{E_1^2}{\lambda^2[\widetilde{2}]_q^4 |L_2|^2}, \frac{|2(E_2 - E_1| + E_1)}{|\Psi|} \right\},$$ where $$\Psi := 2(\lambda[\widetilde{3}]_q - 1)[\widetilde{3}]_q L_3 + \lambda(\lambda - 1)[\widetilde{2}]_q^4 L_2^2$$ and $$\Upsilon:=2(\lambda\widetilde{[3]}_q-1)\widetilde{[3]}_qE_1^2L_3+\lambda\left[(\lambda-1)E_1^2+2\lambda(E_1-E_2)\right]\widetilde{[2]}_q^4L_2^2.$$ From now on, we pay attention to Fekete-Szegö problems for the class $\widetilde{\mathcal{S}_{\Sigma q}^*}^{\eta}(\mu, \lambda; \phi)$ .

Theorem 3.4. If f(z) given by (1.1) is in the class $\widetilde{\mathcal{S}_{\Sigma_q}^*}^{\eta}(\mu, \lambda; \phi)$ and $\varrho \in \mathbb{R}$ , then $$\mid a_{3}-\varrho a_{2}^{2}\mid \leq \begin{cases} \frac{E_{1}}{(\widetilde{\lambda[3]_{q}}-1)|[\mu-(\mu-1)\widetilde{[3]_{q}}]L_{3}|}, & if \ 2|(1-\varrho)L_{3}[\mu-(\mu-1)\widetilde{[3]_{q}}]|E_{1}^{2}(\lambda\widetilde{[3]_{q}}-1)\leq |\Theta|, \\ \frac{2|1-\varrho|E_{1}^{3}}{|\Theta|}, & if \ 2|(1-\varrho)L_{3}[\mu-(\mu-1)\widetilde{[3]_{q}}]|E_{1}^{2}(\lambda\widetilde{[3]_{q}}-1)\leq |\Theta|. \end{cases}$$ where $\Theta$ is the same as in Theorem 3.1. Proof. From (3.18), it infers that $$a_3 - a_2^2 = \frac{E_1(c_2 - d_2)}{4(\lambda[\widetilde{3}]_q - 1)[\mu - (\mu - 1)[\widetilde{3}]_q]L_3}.$$ By (3.17) we obtain that $$\begin{split} a_3 - \varrho a_2^2 &= \frac{E_1\{2(1-\rho)E_1^2(\lambda\widetilde{[3]}_q - 1)[\mu - (\mu-1)\widetilde{[3]}_q]L_3 + \Theta\}c_2}{4(\lambda\widetilde{[3]}_q - 1)[\mu - (\mu-1)\widetilde{[3]}_q]L_3\Theta} \\ &+ \frac{E_1\{2(1-\varrho)E_1^2(\lambda\widetilde{[3]}_q - 1)[\mu - (\mu-1)\widetilde{[3]}_q]L_3 - \Theta\}d_2}{4(\lambda\widetilde{[3]}_q - 1)[\mu - (\mu-1)\widetilde{[3]}_q]L_3\Theta}. \end{split}$$ Hence, we know from Lemma 1.3 that $$|a_3 - \varrho a_2^2| \le \frac{E_1}{(\lambda[\widetilde{3}]_a - 1)|[\mu - (\mu - 1)[\widetilde{3}]_a]L_3|}$$ when $2|(1-\varrho)L_3[\mu-(\mu-1)\widetilde{[3]}_q]|E_1^2(\lambda\widetilde{[3]}_q-1) \le |\Theta|$ , or $$|a_3 - \varrho a_2^2| \le \frac{2|1 - \varrho|E_1^2}{|\Theta|}$$ when $2|(1-\varrho)L_3[\mu-(\mu-1)\widetilde{[3]}_q]|E_1^2(\lambda\widetilde{[3]}_q-1) \ge |\Theta|$ . Then, Theorem 3.4 is completely proved. $\square$

Corollary 3.5. If f(z) given by (1.1) is in the class $\widetilde{S}_{\sum q}^{*\eta}(\lambda;\phi)$ and $\varrho \in \mathbb{R}$ , then $$\mid a_3 - \varrho a_2^2 \mid \leq \begin{cases} \frac{E_1}{(\lambda[\widetilde{3}]_q - 1)|L_3|}, & \text{if } 2 \mid (1 - \varrho) L_3 \mid E_1^2(\lambda[\widetilde{3}]_q - 1) \leq \mid \Xi \mid, \\ \frac{2 \mid 1 - \varrho \mid E_1^3}{\mid \Xi \mid}, & \text{if } 2 \mid (1 - \varrho) L_3 \mid E_1^2(\lambda[\widetilde{3}]_q - 1) \geq \mid \Xi \mid, \end{cases}$$ where $$\Xi := 2(\lambda \widetilde{[3]}_q - 1)E_1^2L_3 + \lambda [(\lambda - 1)E_1^2 + 2\lambda (E_1 - E_2)]\widetilde{[2]}_q^2L_2^2$$

Corollary 3.6. If f(z) given by (1.1) is in the class $\widetilde{C_{\Sigma}}_{q}^{\eta}(\lambda;\phi)$ and $\varrho \in \mathbb{R}$ , then $$\mid a_3-\varrho a_2^2\mid \leq \begin{cases} \frac{E_1}{(\widetilde{\lambda[\widetilde{3}]_q-1)[\widetilde{3}]_q}|L_3|}, & if \ 2|(1-\varrho)L_3|E_1^2(\widetilde{\lambda[\widetilde{3}]_q-1)[\widetilde{3}]_q}\leq |\Upsilon|, \\ \frac{2|1-\varrho|E_1^3}{|\Upsilon|}, & if \ 2|(1-\varrho)L_3|E_1^2(\widetilde{\lambda[\widetilde{3}]_q-1)[\widetilde{3}]_q}\geq |\Upsilon|, \end{cases}$$ where $$\Upsilon := 2(\lambda[3]_q - 1)[3]_q E_1^2 L_3 + \lambda[(\lambda - 1)E_1^2 + \lambda(E_1 - E_2)][2]_q^4 L_2^2.$$

Corollary 3.7. If f(z) given by (1.1) is in the class $\widetilde{\mathcal{S}_{\Sigma_a}^*}^{\eta}(\mu,\lambda;\phi)$ , then $$|a_3 - a_2^2| \le \frac{E_1}{(\lambda[\widetilde{3}]_q - 1)|[\mu - (\mu - 1)\widetilde{[3]}_q]L_3|}$$

Definition 1.1. A function $f(z) \in \Sigma$ given by (1.1), is in the class $\widetilde{S}_{\Sigma q}^{*}(\mu, \lambda; \phi)$ if the following subordinations are satisfied: $$\left\{ \frac{2z[\widetilde{\mathcal{D}}_q(\mathcal{J}_q^{\eta}f)(z)]^{\lambda}}{\mathcal{J}_q^{\eta}f(z) - \mathcal{J}_q^{\eta}f(-z)} \right\}^{\mu} \left\{ \frac{2\{\widetilde{\mathcal{D}}_q[z\widetilde{\mathcal{D}}_q(\mathcal{J}_q^{\eta}f)(z)]\}^{\lambda}}{\widetilde{\mathcal{D}}_q[\mathcal{J}_q^{\eta}f(z) - \mathcal{J}_q^{\eta}f(-z)]} \right\}^{1-\mu} < \phi(z) \tag{1.7}$$ and $$\left\{ \frac{2w[\widetilde{\mathcal{D}}_q(\mathcal{J}_q^{\eta}g)(w)]^{\lambda}}{\mathcal{J}_q^{\eta}g(w) - \mathcal{J}_q^{\eta}g(-w)} \right\}^{\mu} \left\{ \frac{2\{\widetilde{\mathcal{D}}_q[w\widetilde{\mathcal{D}}_q(\mathcal{J}_q^{\eta}g)(w)]\}^{\lambda}}{\widetilde{\mathcal{D}}_q[\mathcal{J}_q^{\eta}g(w) - \mathcal{J}_q^{\eta}g(-w)]} \right\}^{1-\mu} < \phi(w) \tag{1.8}$$ <span id="page-2-1"></span>for $z, w \in \Delta$ , where $\mu \ge 0$ and $\lambda \ge 1$ , and the function g is the inverse of f and given by (1.2). Remark 1.2. If we remove the generalized Bernardi integral operator, then the class $\widetilde{\mathcal{S}_{\Sigma q}}^{\eta}(\mu,\lambda;\phi)$ is exactly $\widetilde{\mathcal{S}_{\Sigma q}}(\mu,\lambda;\phi)$ . Furthermore, the classes $\widetilde{\mathcal{S}_{\Sigma q}}^{\eta}(1,\lambda;\phi)$ and $\mathcal{S}_{\Sigma q}^{b}(0,\lambda;\phi)$ reduce to $\widetilde{\mathcal{S}_{\Sigma q}}(\lambda;\phi)$ and $\widetilde{\mathcal{C}_{\Sigma q}}(\lambda;\phi)$ , respectively; refer to [13] for $\mathcal{S}_{\Sigma}^(\lambda;\phi)$ and $C_{\Sigma}(\lambda;\phi)$ departed from the symmetric q-derivative operator,.