Sokół and Trojnar-Spelina Journal of Inequalities and Applications 2013, 2013:383

Theorem 1.1. [8] If $0 < b \le 1$ , then <span id="page-1-4"></span> $$\mathcal{G}_b \subset \mathcal{S}^* \left( \frac{2}{1 + \sqrt{1 + 8b}} \right).$$ The result is sharp for all b. In [9] the authors obtained the following.

Theorem 1.2. [9] Iff belongs to the class $\mathcal{G}_{b(\beta)}$ with <span id="page-1-0"></span> $$b(\beta) = \frac{\beta}{\sqrt{(1-\beta)^{1-\beta}(1+\beta)^{1+\beta}}},$$ <span id="page-1-3"></span>then $f \in \mathcal{SS}^*(\beta)$ . In this work, we consider the analogous problem for the classes $\mathcal{G}_b$ and $\mathcal{SS}^(\alpha, \beta)$ . Namely, given $\alpha$ , $\beta$ , we look for possible great b such that $\mathcal{G}_b \subset \mathcal{SS}^(\alpha, \beta)$ . To obtain the main theorem, we need the following version of the well-known Jack's lemma.

Theorem 1.3. Let p be analytic in $\mathbb{U}$ with p(0) = 1 and $p(z) \neq 0$ . If there exist two points $z_1 \in \mathbb{U}$ and $z_2 \in \mathbb{U}$ such that $|z_1| = |z_2| = r$ and for $z \in \mathbb{U}_r = \{z : |z| < r\}$ <span id="page-1-1"></span> $$-\frac{\pi\beta}{2} = \arg p(z_1) < \arg p(z) < \arg p(z_2) = \frac{\pi\alpha}{2},$$ (1.2) with some $0 < \alpha \le 2$ , $0 < \beta \le 2$ , then we have <span id="page-1-2"></span> $$\frac{z_1 p'(z_1)}{p(z_1)} = -i \frac{\alpha + \beta}{2} m_1 \tag{1.3}$$ and $$\frac{z_2 p'(z_2)}{p(z_2)} = i \frac{\alpha + \beta}{2} m_2, \tag{1.4}$$ where $$m_1 \ge \frac{1-t}{1+t}, \qquad m_2 \ge \frac{1+t}{1-t},$$ and where <span id="page-2-1"></span><span id="page-2-0"></span> $$t = \tan\frac{\pi}{4} \left( \frac{\alpha - \beta}{\alpha + \beta} \right). \tag{1.5}$$ Proof The assumption (1.2) says that the domain $p(\mathbb{U}_r)$ lies in a sector between two rays $\arg\{w\} = -\pi\beta/2$ and $\arg\{w\} = \pi\alpha/2$ , and it contacts with the rays at $p(z_1)$ and at $p(z_2)$ . The idea of this proof is that we transform this sector into the unit disc, and then we will use Jack's lemma. We restrict our considerations to proving (1.3), the proof of (1.4) runs analogously as that of (1.3). The function $$q(z) = \exp\left\{-i\frac{\pi(\alpha - \beta)}{2(\alpha + \beta)}\right\} \left\{p(z)\right\}^{\frac{2}{\alpha + \beta}} \quad (z \in \mathbb{U}_r)$$ (1.6) maps $\mathbb{U}_r$ onto the set $q(\mathbb{U}_r)$ on the right half-plane $\mathfrak{Re}\{\omega\} > 0$ . The boundary $\partial q(\mathbb{U}_r)$ is tangent to the imaginary axis at $q(z_1)$ and at $q(z_2)$ because $\partial p(\mathbb{U}_r)$ is tangent to the sector $-\pi\beta/2 < \arg w < \pi\alpha/2$ at $p(z_1)$ and at $p(z_2)$ . Moreover, $q(z_1)$ lies on the negative imaginary axis, while $q(z_2)$ lies on the positive imaginary axis. Denote $q(z_1) = -ix_1$ , $x_1 > 0$ . The function $$\phi(z) = \frac{q(z) - 1}{q(z) + 1} \quad (z \in \mathbb{U}_r)$$ maps the disc $\mathbb{U}_r$ onto the domain $\phi(\mathbb{U}_r)$ , contained in the unit disc $\mathbb{U}$ . Since $$\phi(z_1) = \frac{q(z_1) - 1}{q(z_1) + 1} = \frac{-ix_1 - 1}{-ix_1 + 1} = \frac{x_1^2 - 1}{x_2^2 + 1} - \frac{2x_1i}{x_2^2 + 1}$$ then $\mathfrak{Im}\{\phi(z_1)\}\$ < 0, because $x_1 > 0$ . Moreover, <span id="page-2-2"></span> $$\left|\phi(z_1)\right| = \left(\frac{x_1^2 - 1}{x_1^2 + 1}\right)^2 + \frac{4x_1^2}{(x_1^2 + 1)^2} = 1,$$ hence $\phi(z_1) = e^{i\gamma}$ with some $\gamma \in (\pi, 2\pi)$ such that $$\sin \gamma = \frac{-2x_1}{1+x_1^2}, \quad x_1 > 0. \tag{1.7}$$ Notice that $$\phi(0) = -i \tan \frac{\pi}{4} \left( \frac{\alpha - \beta}{\alpha + \beta} \right) = -it, \tag{1.8}$$ with t given by (1.5), $t \in (-1,1)$ . The following fractional transformation obtained from $\phi(z)$ $$F(z) = \frac{\phi(z) + it}{1 + it\phi(z)} \quad (z \in \mathbb{U}_r)$$ maps the disc $\mathbb{U}_r$ onto a domain contained in the unit disc $\mathbb{U}$ and tangent to the unit circle at the points $F(z_1)$ and at $F(z_2)$ . Since F(0)=0 and |F(z)| attains its maximum at the point $z_1$ , then by Jack's lemma, there exists $k \ge 1$ such that <span id="page-3-0"></span> $$\frac{z_1 F'(z_1)}{F(z_1)} = k$$ or, equivalently, <span id="page-3-1"></span> $$\frac{z_1\phi'(z_1)(1-|it|^2)}{(1+\overline{it}\phi(z_1))(\phi(z_1)+it)} = k. \tag{1.9}$$ Taking logarithmic derivative in (1.6), we find that $$\frac{zp'(z)}{p(z)} = \frac{\alpha + \beta}{2} \frac{zq'(z)}{q(z)}.$$ (1.10) Taking logarithmic derivative in <span id="page-3-2"></span> $$q(z) = \frac{1 + \phi(z)}{1 - \phi(z)},$$ <span id="page-3-3"></span>we obtain $$\frac{zq'(z)}{q(z)} = \frac{2z\phi'(z)}{1 - \phi^2(z)}. (1.11)$$ Using together (1.9), (1.10) and (1.11), we get $$\frac{z_1 p'(z_1)}{p(z_1)} = k(\alpha + \beta) \frac{(1 + it\phi(z_1))(\phi(z_1) + it)}{(1 - itt^2)(1 - \phi^2(z_1))}$$ $$= k(\alpha + \beta) \frac{(1 - ite^{i\gamma})(e^{i\gamma} + it)}{(1 - t^2)(1 - e^{2i\gamma})}$$ $$= k(\alpha + \beta) \frac{(1 + t^2)e^{i\gamma} - it(e^{2i\gamma} - 1)}{(1 - t^2)(1 - e^{2i\gamma})}$$ $$= k(\alpha + \beta) i \frac{1 + 2t\sin\gamma + t^2}{2(1 - t^2)\sin\gamma}$$ $$= -i \frac{\alpha + \beta}{2} \frac{(1 + t^2)(-1/\sin\gamma) - 2t}{1 - t^2} k.$$ (1.12) Since $(-1/\sin \gamma) > 1$ for $\gamma \in (\pi, 2\pi)$ , and since $k \ge 1$ , then $$\frac{z_1p'(z_1)}{p(z_1)}=-i\frac{\alpha+\beta}{2}m_1,$$ where $$m_1 = \frac{(1+t^2)(-1/\sin\gamma) - 2t}{1-t^2}k$$ $$\geq \frac{1+t^2 - 2t}{1-t^2}$$ $$= \frac{1-t}{1+t}.$$ Analogously, we may find that $$\frac{z_2p'(z_2)}{p(z_2)}=i\frac{\alpha+\beta}{2}m_2,$$ where $$m_2 \ge \frac{1+t}{1-t}.$$ If we denote a = it, where by (1.5) $t \in (-1, 1)$ , then $$\min\left\{\frac{1-t}{1+t}, \frac{1+t}{1-t}\right\} = \frac{1-|a|}{1+|a|}.$$ Therefore, under the assumptions of Theorem 1.3, there exists $$m \ge \frac{1-|a|}{1+|a|}, \quad |a| = \tan \frac{\pi}{4} \left( \frac{\alpha-\beta}{\alpha+\beta} \right),$$ such that $$\frac{z_1p'(z_1)}{p(z_1)} = -i\frac{\alpha+\beta}{2}m$$ and $$\frac{z_2p'(z_2)}{p(z_2)}=i\frac{\alpha+\beta}{2}m.$$ <span id="page-4-1"></span>The above result is a corollary of Theorem 1.3 but it was given earlier in [5], [9] without a proof. For a proof the authors of [5] refereed to the paper [10], but it probably has not been published yet.

Theorem 2.1. Assume that $0 < \alpha \le 1$ , $0 < \beta \le 1$ . If $f \in \mathcal{G}_{b(\alpha,\beta)}$ with <span id="page-4-0"></span> $$b(\alpha,\beta) = \min \left\{ \frac{\delta(\widetilde{x}_1^{1-\delta} - 2\widetilde{x}_1^{-\delta}\sin\theta + \widetilde{x}_1^{-1-\delta})}{2\cos\theta}, \frac{\delta(\widetilde{x}_2^{1-\delta} + 2\widetilde{x}_2^{-\delta}\sin\theta + \widetilde{x}_2^{-1-\delta})}{2\cos\theta} \right\},$$ where $$\delta = \frac{\alpha + \beta}{2}, \qquad \theta = \frac{\pi}{2} \left( \frac{\alpha - \beta}{\alpha + \beta} \right),$$ $$\widetilde{x}_1 = \frac{\sqrt{1 - \delta^2 \cos^2 \theta} - \delta \sin \theta}{1 - \delta}, \qquad \widetilde{x}_2 = \frac{\sqrt{1 - \delta^2 \cos^2 \theta} + \delta \sin \theta}{1 - \delta},$$ (2.1) then $f \in \mathcal{SS}^*(\alpha, \beta)$ . Proof Assume that $f \in \mathcal{G}_{b(\alpha,\beta)}$ . Let us define the function p(z) = zf'(z)/f(z). Then we have <span id="page-5-0"></span> $$\frac{1 + \frac{zf''(z)}{f'(z)}}{\frac{zf'(z)}{f(z)}} - 1 = \frac{zp'(z)}{p^2(z)}.$$ If $f \notin \mathcal{SS}^*(\alpha, \beta)$ , then $f(\mathbb{U})$ is not contained in the sector $-\pi\beta/2 < \arg w < \pi\alpha/2$ , hence, there exists a point $z_1 \in \mathbb{U}$ such that $f(|z| < |z_1|)$ is contained in this sector, while $f(z_1)$ lies on the ray $\arg w = -\pi\beta/2$ or on the ray $\arg w = \pi\alpha/2$ . To fix the next considerations, suppose that $\arg p(z_1) = -\pi\beta/2$ . We shall apply the considerations from the proof of Theorem 1.3. Using (1.12) with $\sin \gamma$ given in (1.7) we obtain <span id="page-5-1"></span> $$\left| \frac{z_1 p'(z_1)}{p(z_1)} \right| = \left| -i \frac{\alpha + \beta}{2} \frac{(1+t^2)^{\frac{1+x_1^2}{2}} - 2t}{1-t^2} k \right|,\tag{2.2}$$ <span id="page-5-2"></span>where $k \ge 1$ and where $$q(z_1) = -ix_1 = \exp\left\{-i\frac{\pi(\alpha - \beta)}{2(\alpha + \beta)}\right\} \left\{p(z_1)\right\}^{2/(\alpha + \beta)}, \quad x_1 > 0.$$ (2.3) Applying (2.2) together with (2.3), we get $$\left| \frac{z_1 p'(z_1)}{p^2(z_1)} \right| = \left| -i \frac{\alpha + \beta}{2} \cdot \frac{(1 + t^2) \frac{1 + x_1^2}{2x_1} - 2t}{1 - t^2} \cdot k \cdot \left( -ix_1 \exp\left(i \frac{\pi}{2} \cdot \frac{\alpha - \beta}{\alpha + \beta}\right)\right)^{-\frac{\alpha + \beta}{2}} \right| = \left| \frac{\alpha + \beta}{2} \frac{(1 + t^2) \frac{1 + x_1^2}{2x_1} - 2t}{1 - t^2} k x_1^{\frac{\alpha + \beta}{2}} \right| = \frac{\delta(1 + t^2)}{2(1 - t^2)} \left( x_1^{1 - \delta} - \frac{4t x_1^{-\delta}}{1 + t^2} + x_1^{-1 - \delta} \right) k,$$ (2.4) where $$\delta = \frac{\alpha + \beta}{2} \in (0,1].$$ To estimate (2.4), let us consider the function $$g_1(x) = x^{1-\delta} - \frac{4t}{1+t^2}x^{-\delta} + x^{-1-\delta}, \quad x > 0.$$ Then we have $$g_1'(x) = x^{-2-\delta} \left( (1-\delta)x^2 + \frac{4t\delta}{1+t^2}x - (1+\delta) \right), \quad x > 0,$$ and $$\{g_1'(x) = 0, x > 0\}$$ $\Leftrightarrow$ $x = \widetilde{x}_1 = \frac{\sqrt{4t^2\delta^2 + (1 - \delta^2)(1 + t^2)^2} - 2t\delta}{(1 - \delta)(1 + t^2)}.$ Hence $g_1(x)$ takes its minimum at $\widetilde{x}_1$ , and so, (2.4) attains its minimum at $\widetilde{x}_1$ too. Since $t = \tan(\theta/2)$ , then after some standard calculations, we get $$\widetilde{x}_1 = \frac{\sqrt{4t^2\delta^2 + (1 - \delta^2)(1 + t^2)^2} - 2t\delta}{(1 - \delta)(1 + t^2)} = \frac{\sqrt{1 - \delta^2\cos^2\theta} - \delta\sin\theta}{1 - \delta},$$ the same as in (2.1). Therefore, $$\left|\frac{z_1 p'(z_1)}{p^2(z_1)}\right| \geq \frac{\delta(1+t^2)}{2(1-t^2)} \bigg(\widetilde{x_1}^{1-\delta} - \frac{4t\widetilde{x_1}^{-\delta}}{1+t^2} + \widetilde{x_1}^{-1-\delta}\bigg).$$ Applying again $t = \tan(\theta/2)$ , we obtain <span id="page-6-0"></span> $$\left|\frac{z_1p'(z_1)}{p^2(z_1)}\right| \geq \frac{\delta(1+t^2)}{2(1-t^2)} \left(\widetilde{x_1}^{1-\delta} - \frac{4t\widetilde{x_1}^{-\delta}}{1+t^2} + \widetilde{x_1}^{-1-\delta}\right) = \frac{\delta(\widetilde{x_1}^{1-\delta} - 2\widetilde{x_1}^{-\delta}\sin\theta + \widetilde{x_1}^{-1-\delta})}{2\cos\theta} \geq b(\alpha,\beta).$$ This contradicts the assumption that $f \in \mathcal{G}_{b(\alpha,\beta)}$ . If $\arg p(z_2) = \pi \alpha/2$ similar argument also leads to the contradiction. Namely, assume that $f(|z| < |z_2|)$ is contained in the sector $-\pi \beta/2 < \arg w < \pi \alpha/2$ , while $f(z_2)$ lies on the ray $\arg w = \pi \alpha/2$ . Applying the previous considerations, we obtain <span id="page-6-1"></span> $$\left| \frac{z_2 p'(z_2)}{p(z_2)} \right| = \left| -i \frac{\alpha + \beta}{2} \frac{(1 + t^2) \frac{1 + x_2^2}{-2x_2} - 2t}{1 - t^2} k \right|,\tag{2.5}$$ <span id="page-6-2"></span>where $k \ge 1$ and where $$q(z_2) = ix_2 = \exp\left\{-i\frac{\pi(\alpha - \beta)}{2(\alpha + \beta)}\right\} \left\{p(z_2)\right\}^{2/(\alpha + \beta)}, \quad x_2 > 0.$$ (2.6) Applying (2.5) and (2.6), we get $$\left| \frac{z_2 p'(z_2)}{p^2(z_2)} \right| = \left| i \frac{\alpha + \beta}{2} \cdot \frac{(1 + t^2) \frac{1 + x_2^2}{2x_2} + 2t}{1 - t^2} \cdot k \cdot \left( i x_2 \exp\left(i \frac{\pi}{2} \cdot \frac{\alpha - \beta}{\alpha + \beta}\right) \right)^{-\frac{\alpha + \beta}{2}} \right| = \left| \frac{\alpha + \beta}{2} \frac{(1 + t^2) \frac{1 + x_2^2}{2x_2} + 2t}{1 - t^2} k x_2^{\frac{\alpha + \beta}{2}} \right| = \frac{\delta(1 + t^2)}{2(1 - t^2)} \left( x_2^{1 - \delta} + \frac{4t x_2^{-\delta}}{1 + t^2} + x_2^{-1 - \delta} \right) k,$$ (2.7) where $$\delta = \frac{\alpha + \beta}{2} \in (0,1].$$ To estimate (2.7), let us consider the function $$g_2(x) = x^{1-\delta} + \frac{4t}{1+t^2}x^{-\delta} + x^{-1-\delta}, \quad x > 0.$$ Then we have $$g_2'(x) = x^{-2-\delta} \left( (1-\delta)x^2 - \frac{4t\delta}{1+t^2}x - (1+\delta) \right), \quad x > 0,$$ and $$\left\{ g_2'(x) = 0, x > 0 \right\} \quad \Leftrightarrow \quad x = \widetilde{x}_2 = \frac{\sqrt{4t^2\delta^2 + (1 - \delta^2)(1 + t^2)^2} + 2t\delta}{(1 - \delta)(1 + t^2)}.$$ Hence, $g_2(x)$ takes its minimum at $\widetilde{x}_2$ , given in (2.1), and so, (2.4) attains its minimum at $\widetilde{x}_2$ too. Because $t = \tan(\theta/2)$ , we obtain $$\widetilde{x}_2 = \frac{\sqrt{4t^2\delta^2 + (1-\delta^2)(1+t^2)^2} + 2t\delta}{(1-\delta)(1+t^2)} = \frac{\sqrt{1-\delta^2\cos^2\theta} + \delta\sin\theta}{1-\delta}.$$ Therefore, $$\left|\frac{z_2p'(z_2)}{p^2(z_2)}\right| \geq \frac{\delta(1+t^2)}{2(1-t^2)}\left(\widetilde{x}_2^{1-\delta} + \frac{4t\widetilde{x}_2^{-\delta}}{1+t^2} + \widetilde{x}_2^{-1-\delta}\right) = \frac{\delta(\widetilde{x}_2^{1-\delta} + 2\widetilde{x}_2^{-\delta}\sin\theta + \widetilde{x}_2^{-1-\delta})}{2\cos\theta} \geq b(\alpha,\beta).$$ This contradicts the assumption that $f \in \mathcal{G}_{b(\alpha,\beta)}$ . If $\alpha = \beta$ in the theorem above, then we get the following corollary.

Corollary 2.2. Assume that $0 < \alpha < 1$ . If $f \in \mathcal{G}_{b(\alpha)}$ with $$b(\alpha) = \frac{\alpha}{2} \left\{ \left( \frac{1+\alpha}{1-\alpha} \right)^{\frac{-\alpha-1}{2}} + \left( \frac{1+\alpha}{1-\alpha} \right)^{\frac{-\alpha+1}{2}} \right\} = \frac{\alpha}{\sqrt{(1-\alpha)^{1-\alpha}(1+\alpha)^{1+\alpha}}},$$ then $f \in \mathcal{SS}^*(\alpha)$ . This is the result from Theorem 1.2. Putting $\alpha = 1/2$ , $\beta = 1/2$ in Theorem 2.1, we obtain $$\delta = 1/2$$ , $\theta = 0$ , $\widetilde{x}_1 = \widetilde{x}_2 = \sqrt{3}$ and $$b(1/2, 1/2) = \min\left\{\frac{\sqrt[4]{3}}{3}, \frac{\sqrt[4]{3}}{3}\right\} = \frac{\sqrt[4]{3}}{3}.$$ Therefore, we may write the following corollary. Corollary 2.3 If $$\left| \frac{1 + \frac{zf''(z)}{f'(z)}}{\frac{zf'(z)}{f(z)}} - 1 \right| < \frac{\sqrt[4]{3}}{3}, \quad z \in \mathbb{U}, \tag{2.8}$$ then f is strongly starlike of order 1/2. Putting $\alpha = 3/4$ , $\beta = 1/4$ in Theorem 2.1, we obtain $$\delta = \frac{1}{2}, \qquad \theta = \frac{\pi}{4}, \qquad \widetilde{x}_1 = \frac{\sqrt{2}(\sqrt{7} - 1)}{2}, \qquad \widetilde{x}_2 = \frac{3\sqrt{2}(\sqrt{7} + 1)}{2}$$ and $$b(\alpha, \beta) = \min \left\{ \frac{\sqrt[4]{2}(\sqrt{7} - 2)}{3\sqrt{\sqrt{7} + 1}}, \frac{\sqrt[4]{2}(\sqrt{7} + 5)}{6\sqrt{\sqrt{7} + 1}} \right\} = \frac{\sqrt[4]{2}(\sqrt{7} - 2)}{3\sqrt{\sqrt{7} + 1}}.$$ Therefore, we may write the following corollary.

Corollary 2.4. If $f \in \mathcal{G}_{b(\alpha,\beta)}$ with $$b(\alpha, \beta) = \frac{\sqrt[4]{2}(\sqrt{7} - 2)}{3\sqrt{\sqrt{7} + 1}} \approx 0.134,$$ then $f \in SS^*(3/4, 1/4)$ . For some related sufficient conditions for starlikeness of order $\alpha$ , we refer to the recent papers [11] and [12].

Lemma 3.1. [14], [15, p.28] Let $\Omega$ be a set in the complex plane $\mathbb{C}$ . Assume that $\psi : \mathbb{C}^2 \times \mathbb{U} \to \mathbb{C}$ satisfies $$\psi(q(\zeta), m\zeta q'(\zeta); z) \notin \Omega, \tag{3.3}$$ <span id="page-8-0"></span>when $m \ge 1$ , $z \in \mathbb{U}$ and $\zeta \in \partial \mathbb{U} \setminus \{\zeta \in \partial \mathbb{U} : \lim_{z \to \zeta} q(z) = \infty\}$ . If p, q are analytic in $\mathbb{U}$ and $$p(0) = q(0) \quad and \quad \psi\left(p(z), mzp'(z); z\right) \in \Omega,\tag{3.4}$$ then $p \prec q$ .

Theorem 3.2. Assume that $-1 \le B < A \le 1$ and that $b(1 + |A|)^2 \le |A - B|$ . If $f \in \mathcal{G}_b$ , then $f \in \mathcal{S}^*(A, B)$ . Proof Note that $$f \in \mathcal{G}_b \Leftrightarrow \left| \frac{zp'(z)}{p^2(z)} \right| < b \quad \Leftrightarrow \quad \frac{zp'(z)}{p^2(z)} < bz,$$ where p(z) = zf(z)/f(z). If $b(1 + |A|)^2 \le |A - B|$ , then $$|1 + \zeta A| \le \sqrt{\frac{|A - B|}{h}}$$ , for all $|\zeta| = 1$ . Hence, $$\left| \frac{A - B}{(1 + A\zeta)^2} \right| \ge b$$ , for all $|\zeta| = 1$ , $\zeta \ne \frac{-1}{A}$ , and so, $$\left| \frac{m\zeta (A - B)}{(1 + A\zeta)^2} \right| \ge b, \quad \text{for all } |\zeta| = 1, \zeta \ne \frac{-1}{A} \text{ and for all } m \ge 1.$$ Therefore, $$\left|\frac{m\zeta\left(\frac{1+A\zeta}{1+B\zeta}\right)'}{\frac{\left(\frac{1+A\zeta}{1+B\zeta}\right)^2}{1+B\zeta}}\right| \ge b, \quad \text{for all } |\zeta| = 1, \zeta \ne \frac{-1}{B} \text{ and for all } m \ge 1$$ or, equivalently, $$\left| \frac{m\zeta q'(\zeta)}{q'(\zeta)} \right| \ge b$$ , for all $|\zeta| = 1, \zeta \ne \frac{-1}{B}$ and for all $m \ge 1$ . Hence, $$p(z) = \frac{zf'(z)}{f(z)} \prec q(z) = \frac{1 + Az}{1 + Bz}$$ or, equivalently, $f \in \mathcal{S}^*(A, B)$ . The function $$q(z) = \frac{1+z/2}{1-z/2}, \quad z \in \mathbb{U}$$ maps the unit disc onto the disc D(C, R) with the center C = 5/3 and the radius R = 4/3. Hence, putting A = 1/2, B = -1/2, b = 4/9 in Theorem 3.2, we obtain the following corollary.

Corollary 3.3. If $f \in \mathcal{G}_{4/9}$ , then $$\left|\frac{zf'(z)}{f(z)} - \frac{5}{3}\right| < \frac{4}{3}.$$