Bounds for the Second Hankel Determinant of Certain Univalent Functions

Lemma 1.2. [11] If the function $p \in \mathcal{P}$ is given by the series (1.2), then <span id="page-1-2"></span>(1.4) $$2c_2 = c_1^2 + x(4 - c_1^2),$$ <span id="page-1-3"></span>(1.5) $$4c_3 = c_1^3 + 2(4 - c_1^2)c_1x - c_1(4 - c_1^2)x^2 + 2(4 - c_1^2)(1 - |x|^2)z,$$ for some x, z with |x| < 1 and |z| < 1.

Definition 2.2. Let $\varphi : \mathbb{D} \to \mathbb{C}$ be analytic and $\varphi(z)$ is given as in (2.1). The class $\mathscr{C}(\varphi)$ of Ma-Minda convex functions with respect to $\varphi$ consists of functions f satisfying the subordination $$1 + \frac{zf''(z)}{f'(z)} \prec \varphi(z).$$ <span id="page-5-1"></span>THEOREM 2.2. Let the function $f \in \mathcal{C}(\varphi)$ be given by (1.1). (1) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$B_1^2+4|B_2|-2B_1\leq 0,\quad B_1^4-B_1^2|B_2|-6B_1|B_3|+4B_2^2+4B_1^2\geq 0,$$ then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \le \frac{B_1^2}{36}.$$ (2) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$B_1^2 + 4|B_2| - 2B_1 \ge 0$$ , $2B_1^4 - 2B_1^2|B_2| - 12B_1|B_3| + 8B_2^2 + 4B_1|B_2| + B_1^3 + 6B_1^2 \le 0$ , or the conditions $$B_1^2 + 4|B_2| - 2B_1 \le 0$$ , $B_1^4 - B_1^2|B_2| - 6B_1|B_3| + 4B_2^2 + 4B_1^2 \le 0$ , then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \le \frac{1}{144}(-B_1^4 + B_1^2|B_2| + 6B_1|B_3| - 4B_2^2).$$ П (3) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$B_1^2 + 4|B_2| - 2B_1 > 0$$ , $2B_1^4 - 2B_1^2|B_2| - 12B_1|B_3| + 8B_2^2 + 4B_1|B_2| + B_1^3 + 6B_1^2 \ge 0$ , <span id="page-6-0"></span>then the second Hankel determinant satisfies $$|a_2a_4-a_3^2| \leq \frac{B_1^2}{576} \left( \frac{17B_1^4 - 8B_1^2|B_2| - 96B_1|B_3| + 80B_2^2 + 12B_1^3 + 48B_1|B_2| + 36B_1^2}{B_1^4 - B_1^2|B_2| - 6B_1|B_3| + 4B_2^2 + B_1^3 + 4B_1|B_2| + 2B_1^2} \right).$$ PROOF. Since $f \in \mathcal{C}(\varphi)$ , there exists an analytic function w with w(0) = 0 and |w(z)| < 1 in $\mathbb{D}$ such that (2.12) $$1 + \frac{zf''(z)}{f'(z)} = \varphi(w(z)).$$ Since <span id="page-6-1"></span> $$(2.13) 1 + \frac{zf''(z)}{f'(z)} = 1 + 2a_2z + (-4a_2^2 + 6a_3)z^2 + (8a_2^3 - 18a_2a_3 + 12a_4)z^3 + \cdots,$$ equations (2.4), (2.12) and (2.13) yield $$\begin{aligned} a_2 &= \frac{B_1 c_1}{4}, \\ a_3 &= \frac{1}{24} \left[ (B_1^2 - B_1 + B_2) c_1^2 + 2B_1 c_2 \right], \\ a_4 &= \frac{1}{192} \left[ (-4B_2 + 2B_1 + B_1^3 - 3B_1^2 + 3B_1 B_2 + 2B_3) c_1^3 + 2(3B_1^2 - 4B_1 + 4B_2) c_1 c_2 + 8B_1 c_3 \right]. \end{aligned}$$ Therefore $$a_{2}a_{4} - a_{3}^{2} = \frac{B_{1}}{768} \left[ c_{1}^{4} \left( -\frac{4}{3}B_{2} + \frac{2}{3}B_{1} - \frac{1}{3}B_{1}^{3} - \frac{1}{3}B_{1}^{2} + \frac{1}{3}B_{1}B_{2} + 2B_{3} - \frac{4}{3}\frac{B_{2}^{2}}{B_{1}} \right) + \frac{2}{3}c_{2}c_{1}^{2}(B_{1}^{2} - 4B_{1} + 4B_{2}) + 8B_{1}c_{1}c_{3} - \frac{16}{3}B_{1}c_{2}^{2} \right].$$ By writing <span id="page-6-2"></span> $$d_1 = 8B_1$$ , $d_2 = \frac{2}{3}(B_1^2 - 4B_1 + 4B_2)$ , (2.14) $$d_{3} = -\frac{16}{3}B_{1}, \quad d_{4} = -\frac{4}{3}B_{2} + \frac{2}{3}B_{1} - \frac{1}{3}B_{1}^{3} - \frac{1}{3}B_{1}^{2} + \frac{1}{3}B_{1}B_{2} + 2B_{3} - \frac{4}{3}\frac{B_{2}^{2}}{B_{1}},$$ $$T = \frac{B_{1}}{768},$$ we have $$(2.15) |a_2a_4 - a_3^2| = T|d_1c_1c_3 + d_2c_1^2c_2 + d_3c_2^2 + d_4c_1^4|.$$ Similar as in Theorems 2.1, it follows from (1.4) and (1.5) that $$|a_2a_4 - a_3^2| = \frac{T}{4} |c^4(d_1 + 2d_2 + d_3 + 4d_4) + 2xc^2(4 - c^2)(d_1 + d_2 + d_3) + (4 - c^2)x^2(-d_1c^2 + d_3(4 - c^2)) + 2d_1c(4 - c^2)(1 - |x|^2)z|$$ Replacing |x| by $\mu$ and then substituting the values of $d_1$ , $d_2$ , $d_3$ and $d_4$ from (2.14) yield <span id="page-7-0"></span> $$|a_{2}a_{4} - a_{3}^{2}| \leq \frac{T}{4} \left[ c^{4} \left( -\frac{4}{3}B_{1}^{3} + \frac{4}{3}B_{1}B_{2} + 8B_{3} - \frac{16B_{2}^{2}}{3B_{1}} \right) + 2\mu c^{2}(4 - c^{2}) \left( \frac{2}{3}B_{1}^{2} + \frac{8}{3}B_{2} \right) \right.$$ $$\left. + \mu^{2}(4 - c^{2}) \left( \frac{8}{3}B_{1}c^{2} + \frac{64}{3}B_{1} \right) + 16B_{1}c(4 - c^{2})(1 - \mu^{2}) \right]$$ $$(2.16) \qquad = T \left[ \frac{c^{4}}{3} \left( -B_{1}^{3} + B_{1}|B_{2}| + 6|B_{3}| - 4\frac{B_{2}^{2}}{B_{1}} \right) + 4B_{1}c(4 - c^{2}) + \frac{1}{3}\mu c^{2}(4 - c^{2})(B_{1}^{2} + 4|B_{2}|) \right.$$ $$\left. + \frac{2B_{1}}{3}\mu^{2}(4 - c^{2})(c - 4)(c - 2) \right]$$ $$\equiv F(c, \mu).$$ Again, differentiating $F(c, \mu)$ in (2.16) partially with respect to $\mu$ yield <span id="page-7-1"></span>(2.17) $$\frac{\partial F}{\partial \mu} = T \left[ \frac{c^2}{3} (4 - c^2) (B_1^2 + 4|B_2|) + \frac{4B_1}{3} \mu (4 - c^2) (c - 4) (c - 2) \right].$$ It is clear from (2.17) that $\frac{\partial F}{\partial \mu} > 0$ . Thus $F(c, \mu)$ is an increasing function of $\mu$ for $0 < \mu < 1$ and for any fixed c with 0 < c < 2. So the maximum of $F(c, \mu)$ occurs at $\mu = 1$ and $$\max F(c,\mu) = F(c,1) \equiv G(c).$$ Note that $$G(c) = T \left[ \frac{c^4}{3} \left( -B_1^3 + B_1 |B_2| + 6|B_3| - 4\frac{B_2^2}{B_1} - B_1^2 - 4|B_2| - 2B_1 \right) + \frac{4}{3}c^2 (B_1^2 + 4|B_2| - 2B_1) + \frac{64}{3}B_1 \right].$$ Let <span id="page-7-2"></span> $$P = \frac{1}{3} \left( -B_1^3 + B_1 |B_2| + 6|B_3| - 4\frac{B_2^2}{B_1} - B_1^2 - 4|B_2| - 2B_1 \right),$$ $$Q = \frac{4}{3} (B_1^2 + 4|B_2| - 2B_1),$$ $$R = \frac{64}{3} B_1,$$ By using (2.11), we have $$|a_2a_4 - a_3^2| \le \frac{B_1}{768} \begin{cases} R, & Q \le 0, P \le -\frac{Q}{4}; \\ 16P + 4Q + R, & Q \ge 0, P \ge -\frac{Q}{8} \text{ or } Q \le 0, P \ge -\frac{Q}{4}; \\ \frac{4PR - Q^2}{4P}, & Q > 0, P \le -\frac{Q}{8} \end{cases}$$ where P, Q, R are given in (2.18). REMARK 2.2. For the choice of $\varphi(z) = (1+z)/(1-z)$ , Theorem 2.2 reduces to [16, Theorem 3.2].

Definition 3.1. Let $\varphi : \mathbb{D} \to \mathbb{C}$ be analytic and $\varphi(z)$ as given in (2.1). Let $0 \le \gamma \le 1$ and $\tau \in \mathbb{C} \setminus \{0\}$ . A function $f \in \mathscr{A}$ is in the class $\mathscr{R}^{\tau}_{\gamma}(\varphi)$ if it satisfies the following subordination: $$1 + \frac{1}{\tau}(f'(z) + \gamma z f''(z) - 1) \prec \varphi(z).$$ <span id="page-8-2"></span>THEOREM 3.1. Let $0 \le \gamma \le 1$ , $\tau \in \mathbb{C} \setminus \{0\}$ and the function f as in (1.1) is in the class $\mathscr{R}^{\tau}_{\nu}(\varphi)$ . Also, let $$p = \frac{8}{9} \frac{(1+\gamma)(1+3\gamma)}{(1+2\gamma)^2}.$$ (1) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$2|B_2|(1-p) + B_1(1-2p) \le 0$$ , $|B_1B_3 - pB_2^2| - pB_1^2 \le 0$ , then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \le \frac{|\tau|^2 B_1^2}{9(1+2\gamma)^2}.$$ (2) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$2|B_2|(1-p)+B_1(1-2p) \ge 0$$ , $2|B_1B_3-pB_2^2|-2(1-p)B_1|B_2|-B_1 \ge 0$ , or the conditions $$2|B_2|(1-p)+B_1(1-2p) \le 0$$ , $|B_1B_3-pB_2^2|-B_1^2 \ge 0$ , then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \le \frac{|\tau|^2}{8(1+\gamma)(1+3\gamma)} |B_3B_1 - pB_2^2|.$$ (3) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$2|B_2|(1-p)+B_1(1-2p)>0, \quad 2|B_1B_3-pB_2^2|-2(1-p)B_1|B_2|-B_1^2\leq 0,$$ then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \leq \frac{|\tau|^2 B_1^2}{32(1+\gamma)(1+3\gamma)} \left( \frac{4p|B_3B_1 - pB_2^2| - 4(1-p)B_1[|B_2|(3-2p) + B_1]}{-4B_2^2(1-p)^2 - B_1^2(1-2p)^2}}{|B_3B_1 - pB_2^2| - (1-p)B_1(2|B_2| + B_1)} \right).$$ PROOF. For $f \in \mathcal{R}^{\tau}_{\gamma}(\varphi)$ , there exists an analytic function w with w(0) = 0 and |w(z)| < 1 in $\mathbb{D}$ such that <span id="page-8-0"></span>(3.1) $$1 + \frac{1}{\tau}(f'(z) + \gamma z f''(z) - 1) = \varphi(w(z)).$$ Since f has the Maclaurin series given by (1.1), a computation shows that <span id="page-8-1"></span> $$(3.2) \quad 1 + \frac{1}{\tau}(f'(z) + \gamma z f''(z) - 1) = 1 + \frac{2a_2(1+\gamma)}{\tau}z + \frac{3a_3(1+2\gamma)}{\tau}z^2 + \frac{4a_4(1+3\gamma)}{\tau}z^3 + \cdots$$ It follows from (3.1), (2.4) and (3.2) that $$\begin{split} a_2 &= \frac{\tau B_1 c_1}{4(1+\gamma)}, \\ a_3 &= \frac{\tau B_1}{12(1+2\gamma)} \left[ 2c_2 + c_1^2 \left( \frac{B_2}{B_1} - 1 \right) \right], \\ a_4 &= \frac{\tau}{32(1+3\gamma)} [B_1 (4c_3 - 4c_1c_2 + c_1^3) + 2B_2 c_1 (2c_2 - c_1^2) + B_3 c_1^3]. \end{split}$$ Therefore $$a_{2}a_{4} - a_{3}^{2} = \frac{\tau^{2}B_{1}c_{1}}{128(1+\gamma)(1+3\gamma)} \left[ B_{1}(4c_{3} - 4c_{1}c_{2} + c_{1}^{3}) + 2B_{2}c_{1}(2c_{2} - c_{1}^{2}) + B_{3}c_{1}^{3} \right]$$ $$- \frac{\tau^{2}B_{1}^{2}}{144(1+2\gamma)^{2}} \left[ 4c_{2}^{2} + c_{1}^{4} \left( \frac{B_{2}}{B_{1}} - 1 \right)^{2} + 4c_{2}c_{1}^{2} \left( \frac{B_{2}}{B_{1}} - 1 \right) \right]$$ $$= \frac{\tau^{2}B_{1}^{2}}{128(1+\gamma)(1+3\gamma)} \left\{ \left[ (4c_{1}c_{3} - 4c_{1}^{2}c_{2} + c_{1}^{4}) + \frac{2B_{2}c_{1}^{2}}{B_{1}}(2c_{2} - c_{1}^{2}) + \frac{B_{3}}{B_{1}}c_{1}^{4} \right]$$ $$- \frac{8}{9} \frac{(1+\gamma)(1+3\gamma)}{(1+2\gamma)^{2}} \left[ 4c_{2}^{2} + c_{1}^{4} \left( \frac{B_{2}}{B_{1}} - 1 \right)^{2} + 4c_{2}c_{1}^{2} \left( \frac{B_{2}}{B_{1}} - 1 \right) \right] \right\},$$ which yields $$|a_{2}a_{4} - a_{3}^{2}| = T \left| 4c_{1}c_{3} + c_{1}^{4} \left[ 1 - 2\frac{B_{2}}{B_{1}} - p\left(\frac{B_{2}}{B_{1}} - 1\right)^{2} + \frac{B_{3}}{B_{1}} \right] - 4pc_{2}^{2} - 4c_{1}^{2}c_{2} \left[ 1 - \frac{B_{2}}{B_{1}} + p\left(\frac{B_{2}}{B_{1}} - 1\right) \right] \right|,$$ (3.3) <span id="page-9-0"></span>where <span id="page-9-1"></span> $$T = \frac{|\tau|^2 B_1^2}{128(1+\gamma)(1+3\gamma)} \quad \text{and} \quad p = \frac{8}{9} \frac{(1+\gamma)(1+3\gamma)}{(1+2\gamma)^2}.$$ It can be easily verified that $p \in \left[\frac{64}{81}, \frac{8}{9}\right]$ for $0 \le \gamma \le 1$ . Let (3.4) $$d_1 = 4, \quad d_2 = -4 \left[ 1 - \frac{B_2}{B_1} + p \left( \frac{B_2}{B_1} - 1 \right) \right],$$ $$d_3 = -4p, \quad d_4 = 1 - 2\frac{B_2}{B_1} - p \left( \frac{B_2}{B_1} - 1 \right)^2 + \frac{B_3}{B_1}.$$ Then (3.3) becomes $$|a_2a_4 - a_3^2| = T|d_1c_1c_3 + d_2c_1^2c_2 + d_3c_2^2 + d_4c_1^4|.$$ It follows that $$|a_2a_4 - a_3^2| = \frac{T}{4} |c^4(d_1 + 2d_2 + d_3 + 4d_4) + 2xc^2(4 - c^2)(d_1 + d_2 + d_3)$$ $$+ (4 - c^2)x^2(-d_1c^2 + d_3(4 - c^2)) + 2d_1c(4 - c^2)(1 - |x|^2)z|.$$ An application of triangle inequality, replacement of |x| by $\mu$ and substituting the values of $d_1$ , $d_2$ , $d_3$ and $d_4$ from (3.4) yield $$|a_{2}a_{4} - a_{3}^{2}| \leq \frac{T}{4} \left[ 4c^{4} \left| \frac{B_{3}}{B_{1}} - p \frac{B_{2}^{2}}{B_{1}^{2}} \right| + 8 \left| \frac{B_{2}}{B_{1}} \right| \mu c^{2} (4 - c^{2}) (1 - p) \right.$$ $$\left. + (4 - c^{2}) \mu^{2} (4c^{2} + 4p(4 - c^{2})) + 8c(4 - c^{2}) (1 - \mu^{2}) \right]$$ $$= T \left[ c^{4} \left| \frac{B_{3}}{B_{1}} - p \frac{B_{2}^{2}}{B_{1}^{2}} \right| + 2c(4 - c^{2}) + 2\mu \left| \frac{B_{2}}{B_{1}} \right| c^{2} (4 - c^{2}) (1 - p) + \mu^{2} (4 - c^{2}) (1 - p) (c - \alpha) (c - \beta) \right]$$ $$\equiv F(c, \mu)$$ where $\alpha = 2$ , $\beta = 2p/(1-p) > 2$ . Similarly as in the previous proofs, it can be shown that $F(c, \mu)$ is an increasing function of $\mu$ for $0 < \mu < 1$ . So for fixed $c \in [0, 2]$ , let $$\max F(c, \mu) = F(c, 1) \equiv G(c)$$ , which is $$G(c) = T \left\{ c^4 \left[ \left| \frac{B_3}{B_1} - p \frac{B_2^2}{B_1^2} \right| - (1 - p) \left( 2 \left| \frac{B_2}{B_1} \right| + 1 \right) \right] + 4c^2 \left[ 2 \left| \frac{B_2}{B_1} \right| (1 - p) + 1 - 2p \right] + 16p \right\}.$$ Let <span id="page-10-0"></span>(3.7) $$P = \left| \frac{B_3}{B_1} - p \frac{B_2^2}{B_1^2} \right| - (1 - p) \left( 2 \left| \frac{B_2}{B_1} \right| + 1 \right),$$ $$Q = 4 \left[ 2 \left| \frac{B_2}{B_1} \right| (1 - p) + 1 - 2p \right],$$ $$R = 16p.$$ Using (2.11), we have $$|a_2a_4 - a_3^2| \le T \begin{cases} R, & Q \le 0, P \le -\frac{Q}{4}; \\ 16P + 4Q + R, & Q \ge 0, P \ge -\frac{Q}{8} \text{ or } Q \le 0, P \ge -\frac{Q}{4}; \\ \frac{4PR - Q^2}{4P}, & Q > 0, P \le -\frac{Q}{8} \end{cases}$$ where P, Q, R are given in (3.7). REMARK 3.1. For the choice $\varphi(z) := (1+Az)/(1+Bz)$ with $-1 \le B < A \le 1$ , Theorem 3.1 reduces to [6, Theorem 2.1].

Definition 3.2. Let $\varphi : \mathbb{D} \to \mathbb{C}$ be analytic and $\varphi(z)$ as given in (2.1). For fixed real number $\alpha$ , function $f \in \mathscr{A}$ is in the class $\mathscr{G}_{\alpha}(\varphi)$ if it satisfies the following subordination: $$(1-\alpha)f'(z) + \alpha\left(1 + \frac{zf''(z)}{f'(z)}\right) \prec \varphi(z).$$ Al-Amiri and Reade [1] introduced the class $\mathscr{G}_{\alpha} := \mathscr{G}_{\alpha}((1+z)/(1-z))$ and they have shown that $\mathscr{G}_{\alpha} \subset \mathscr{S}$ for $\alpha < 0$ . Univalence of the functions in the class $\mathscr{G}_{\alpha}$ was also investigated in [35,36]. Singh et al. also obtained the bound for the second Hankel determinant of functions in $\mathscr{G}_{\alpha}$ . The following theorem provides a bound for the second Hankel determinant of the functions in the class $\mathscr{G}_{\alpha}(\varphi)$ . <span id="page-11-0"></span>THEOREM 3.2. Let the function f given by (1.1) be in the class $\mathscr{G}_{\alpha}(\varphi)$ , $0 \le \alpha \le 1$ . Also, let $$p = \frac{8}{9} \frac{(1+2\alpha)}{(1+\alpha)}.$$ (1) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$B_1^2\alpha(3-2p)+2|B_2|(1+\alpha-p)+B_1(1+\alpha-2p)\leq 0,$$ $$B_1^4 \alpha (2\alpha - 1 - p\alpha) + \alpha B_1^2 |B_2| (3 - 2p) + (\alpha + 1)B_1 |B_3| - p(B_1^2 + B_2^2) \le 0,$$ then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \le \frac{B_1^2}{9(1+\alpha)^2}.$$ (2) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$\begin{split} B_1^2\alpha(3-2p) + 2|B_2|(1+\alpha-p) + B_1(1+\alpha-2p) &\geq 0, \\ 2B_1^4\alpha(2\alpha-1-p\alpha) + 2\alpha B_1^2|B_2|(3-2p) - B_1^3\alpha(3-2p) \\ + 2(\alpha+1)B_1|B_3| - 2(1+\alpha-p)B_1|B_2| - (1+\alpha)B_1^2 - 2pB_2^2 &\geq 0, \end{split}$$ or $$B_1^2\alpha(3-2p)+2|B_2|(1+\alpha-p)+B_1(1+\alpha-2p)\leq 0,$$ $$B_1^4 \alpha (2\alpha - 1 - p\alpha) + \alpha B_1^2 |B_2| (3 - 2p) + (\alpha + 1)B_1 |B_3| - p(B_1^2 + B_2^2) \ge 0,$$ then the second Hankel determinant satisfies $$|a_2a_4 - a_3^2| \le \frac{B_1^4\alpha(2\alpha - 1 - p\alpha) + \alpha B_1^2|B_2|(3 - 2p) + (\alpha + 1)B_1|B_3| + p(B_2^2 - B_1^2)}{8(1 + \alpha)(1 + 2\alpha)}.$$ (3) If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$\begin{split} B_1^2\alpha(3-2p) + 2|B_2|(1+\alpha-p) + B_1(1+\alpha-2p) &> 0, \\ 2B_1^4\alpha(2\alpha-1-p\alpha) + 2\alpha B_1^2|B_2|(3-2p) - B_1^3\alpha(3-2p) \\ + 2(\alpha+1)B_1|B_3| - 2(1+\alpha-p)B_1|B_2| - (1+\alpha)B_1^2 - 2pB_2^2 &\leq 0, \end{split}$$ then the second Hankel determinant satisfies $$|a_2a_4-a_3^2| \leq \frac{B_1^2}{32(1+\alpha)(1+2\alpha)} \left[ 4p - \frac{[B_1^2\alpha(3-2p)+2|B_2|(1+\alpha-p)+B_1(1+\alpha-2p)]^2}{B_1^4\alpha(2\alpha-1-p\alpha)+\alpha B_1^2|B_2|(3-2p)-B_1^3\alpha(3-2p)} + (\alpha+1)B_1|B_3| - (1+\alpha-p)B_1(2|B_2|+1)-pB_2^2 \right].$$ PROOF. For $f \in \mathscr{G}_{\alpha}(\varphi)$ , a calculation shows that $$|a_{2}a_{4} - a_{3}^{2}| = T \left| 4(1+\alpha)B_{1}c_{1}c_{3} + c_{1}^{4} \left[ -3\alpha B_{1}^{2} + \alpha(2\alpha - 1)B_{1}^{3} + B_{1}(1+\alpha) + 3\alpha B_{1}B_{2} \right] + (1+\alpha)(B_{3} - 2B_{2}) - p \frac{(\alpha B_{1}^{2} - B_{1} + B_{2})^{2}}{B_{1}} \right] - 4pB_{1}c_{2}^{2}$$ $$(3.8) + 2c_{1}^{2}c_{2} \left[ -2(1+\alpha)B_{1} + 3\alpha B_{1}^{2} + 2(1+\alpha)B_{2} - 2p(\alpha B_{1}^{2} - B_{1} + B_{2}) \right]$$ where $$T = \frac{B_1}{128(1+\alpha)(1+2\alpha)}$$ and $p = \frac{8(1+2\alpha)}{9(1+\alpha)}$ . It can be easily verified that for $0 \le \alpha \le 1$ , $p \in \left[\frac{8}{9}, \frac{4}{3}\right]$ . Let (3.9) $d_1 = 4(1+\alpha)B_1,$ $$d_1 = 4(1+\alpha)B_1$$ $$d_2 = 2 \left[ -2(1+\alpha)B_1 + 3\alpha B_1^2 + 2(1+\alpha)B_2 - 2p(\alpha B_1^2 - B_1 + B_2) \right],$$ $$d_3 = -4pB_1$$ $$d_4 = -3\alpha B_1^2 + \alpha(2\alpha - 1)B_1^3 + B_1(1 + \alpha) + 3\alpha B_1B_2 + (1 + \alpha)(B_3 - 2B_2) - p\frac{(\alpha B_1^2 - B_1 + B_2)^2}{B_1},$$ Then $$(3.10) |a_2a_4 - a_3^2| = T|d_1c_1c_3 + d_2c_1^2c_2 + d_3c_2^2 + d_4c_1^4|.$$ Similar as in earlier theorems, it follows that $$(3.11) \quad |a_{2}a_{4} - a_{3}^{2}| = \frac{T}{4} \left| c^{4} (d_{1} + 2d_{2} + d_{3} + 4d_{4}) + 2xc^{2} (4 - c^{2}) (d_{1} + d_{2} + d_{3}) \right. \\ \left. + (4 - c^{2})x^{2} (-d_{1}c^{2} + d_{3}(4 - c^{2})) + 2d_{1}c(4 - c^{2})(1 - |x|^{2})z \right| \\ \leq T \left[ c^{4} \left[ B_{1}^{3}\alpha(2\alpha - 1 - p\alpha) + \alpha B_{1}|B_{2}|(3 - 2p) + (\alpha + 1)|B_{3}| - p\frac{B_{2}^{2}}{B_{1}} \right] \right. \\ \left. + \mu c^{2} (4 - c^{2})[B_{1}^{2}\alpha(3 - 2p) + 2|B_{2}|(1 + \alpha - p)] + 2c(4 - c^{2})B_{1}(1 + \alpha) \right. \\ \left. + \mu^{2} (4 - c^{2})B_{1}(1 + \alpha - p)(c - 2) \left( c - \frac{2p}{1 + \alpha - p} \right) \right] \\ \equiv F(c, \mu),$$ and for fixed $c \in [0,2]$ , $\max F(c,\mu) = F(c,1) \equiv G(c)$ with $$\begin{split} G(c) &= T \left[ c^4 \left[ B_1^3 \alpha (2\alpha - 1 - p\alpha) + \alpha B_1 | B_2 | (3 - 2p) - B_1^2 \alpha (3 - 2p) + (\alpha + 1) | B_3 | \right. \\ &- (1 + \alpha - p) (2|B_2| + B_1) - p \frac{B_2^2}{B_1} \right] + 4c^2 [B_1^2 \alpha (3 - 2p) + 2|B_2| (1 + \alpha - p) \\ &+ B_1 (1 + \alpha - 2p)] + 16p B_1 \right]. \end{split}$$ Let $$P = B_1^3 \alpha (2\alpha - 1 - p\alpha) + \alpha B_1 |B_2| (3 - 2p) - B_1^2 \alpha (3 - 2p) + (\alpha + 1) |B_3|$$ $$- (1 + \alpha - p)(2|B_2| + B_1) - p \frac{B_2^2}{B_1}$$ $$Q = 4 \left[ B_1^2 \alpha (3 - 2p) + 2|B_2| (1 + \alpha - p) + B_1 (1 + \alpha - 2p) \right],$$ $$R = 16pB_1,$$ <span id="page-13-8"></span>By using (2.11), we have $$|a_2a_4 - a_3^2| \le T \begin{cases} R, & Q \le 0, P \le -\frac{Q}{4}; \\ 16P + 4Q + R, & Q \ge 0, P \ge -\frac{Q}{8} \text{ or } Q \le 0, P \ge -\frac{Q}{4}; \\ \frac{4PR - Q^2}{4P}, & Q > 0, P \le -\frac{Q}{8} \end{cases}$$ where P, O, R are given in (3.13) REMARK 3.2. For $\alpha=0$ , Theorem 3.2 reduces to Theorem 2.2. For $0\leq\alpha<1$ , let $\varphi(z):=(1+(1-2\alpha)z)/(1-z)$ . For this function $\varphi$ , $B_1=B_2=B_3=2(1-\alpha)$ . In this case, Theorem 3.2 reduces to [39, Theorem 3.1]. П