Starlikeness of certain analytic functions

Theorem 1.1. The following radius results hold for the class $\Pi_1$ : - (1) The $S^(\alpha)$ radius is $R_{S^(\alpha)} = (1-\alpha)/(3+\sqrt{8+\alpha^2}), \quad 0 \le \alpha < 1.$ - (2) The $S_L$ radius is $R_{S_L} = (\sqrt{2} 1)(\sqrt{10} 3) \approx 0.067217$ . - (3) The $S_P$ radius is $R_{S_P} = (6 \sqrt{33})/3 \approx 0.0851$ . - (4) The $S_e$ radius is $R_{S_e} = (e-1)/(3e + \sqrt{8e^2 + 1}) \approx 0.1080$ . - (5) The $S_c$ radius is $R_{S_c} = (9 \sqrt{73})/4 \approx 0.1140$ . (6) The $S_{\sin}$ radius is $R_{S_{\sin}} = \sin(1)/(\sqrt{9 + \sin^2(1) + 2\sin(1)} + 3) \approx 0.1320$ . - (7) The $S_{\mathbb{Q}}$ radius is $R_{S_{\mathbb{Q}}} = 3/\sqrt{2} \sqrt{1/2(11 2\sqrt{2})} \approx 0.09999$ . - (8) The $S_R$ radius $R_{S_R} = (3 2\sqrt{5 2\sqrt{2}})/(2\sqrt{2} 1) \approx 0.0289$ . The functions $f_2, f_3 : \mathbb{D} \to \mathbb{C}$ defined by $$f_2(z) = \frac{z}{1-z}$$ and $f_3(z) = \frac{z(1+z)^2}{(1-z)^3}$ , (1.2) satisfy the conditions $|f_i(z)/g_i(z)-1|<1$ and $Re((1-z)^2g_i(z))>0$ for i=2,3 with $g_2, g_3: \mathbb{D} \to \mathbb{C}$ defined by $$g_2(z) = \frac{z}{1-z^2}$$ and $g_3(z) = \frac{z(1+z)}{(1-z)^3}$ , (1.3) and hence $f_2, f_3 \in \Pi_2$ . This proves that the class $\Pi_2$ is non-empty. The Taylor series $f_3(z) = z + 5z^2 + 13z^3 + 25z^4 + \cdots$ shows that it is not univalent. It is an extremal function for the radius problems we consider. The derivative of $f_3$ is given by <span id="page-2-2"></span> $$f_3'(z) = \frac{(1+5z)(1+z)}{(1-z)^4}.$$ Since $f_3'(-1/5) = 0$ and, by Theorem 1.2 (1), the radius of starlikeness of the class $\Pi_1$ is 1/5, it follows that the radius of univalence of this class is also 1/5. The other radius results for class $\Pi_2$ are given in the following theorem.

Theorem 1.2. The following radius results hold for the class $\Pi_2$ : - (1) The $S^(\alpha)$ radius is $R_{S^(\alpha)} = 2(1-\alpha)/(5+\sqrt{4\alpha^2-4\alpha+25}), \quad 0 \le \alpha < 1.$ (2) The $S_L$ radius is at least $R_{S_L} = (\sqrt{4\sqrt{2}+25}-5)/(2(\sqrt{2}+2)) \approx 0.0786.$ - (3) The $S_p$ radius is $R_{S_p} = 5 2\sqrt{6} \approx 0.1010$ . - (4) The $S_e$ radius is $S_e^ = (2(e-1))/(5e + \sqrt{25e^2 4e + 4}) \approx 0.1276$ . (5) The $S_c$ radius is $R_{S_c} = (15 \sqrt{217})/2 \approx 0.1345$ . (6) The $S_{sin}$ radius is at least $S_{sin}^ = (\sqrt{25 + 4(3 + \sin(1))\sin(1)} 5)/(2(3 + \cos(1))) \approx 0.1500$ $\sin(1)) \approx 0.1508.$ - (7) The $S_{\mathbb{C}}$ radius is $R_{S_{\mathbb{C}}} = (5 \sqrt{41 12\sqrt{2}})/(2(\sqrt{2} 1)) \approx 0.1183$ . - (8) The $S_R$ radius is $R_{S_R} = (5 \sqrt{81 40\sqrt{2}})/(4(\sqrt{2} 1)) \approx 0.0345$ . It is worth to point out that $R_{\mathcal{S}_{P}^{}} = R_{\mathcal{S}^{}(1/2)}$ and $R_{\mathcal{S}_{e}^{}} = R_{\mathcal{S}^{}(1/e)}$ in both theorems.

Lemma 2.1. [1, Lemma 2.2, p.4] For $0 < \alpha < \sqrt{2}$ , let $\mathbf{r}_a$ be given by $$r_a = \begin{cases} (\sqrt{1-a^2} - (1-a^2))^{\frac{1}{2}}, & 0 < a \le 2\sqrt{2}/3\\ \sqrt{2} - a, & 2\sqrt{2}/3 \le a < \sqrt{2}. \end{cases}$$ Then $\{\omega : |\omega - a| < r_a\} \subseteq \{\omega : |\omega^2 - 1| < 1\}.$

Lemma 2.2. [17, Lemma 1, p. 321] For $a > \frac{1}{2}$ , let $r_a$ be given by $$r_a = \begin{cases} a - \frac{1}{2}, & \frac{1}{2} < a \le \frac{3}{2} \\ \sqrt{2a - 2}, & a \ge \frac{3}{2} \end{cases}$$ Then $\{w : |w - a| < r_a\} \subseteq \{w : \text{Re } w > |w - 1|\}.$ <span id="page-3-3"></span>Lemma 2.3. [13, Lemma 2.2, p.368] For $e^{-1} < a < e$ , let $\boldsymbol{r}_a$ be given by $$r_a = \begin{cases} a - e^{-1}, & e^{-1} < a \le \frac{e + e^{-1}}{2} \\ e - a, & \frac{e + e^{-1}}{2} \le a \le e. \end{cases}$$ Then $\{w : |w - a| < r_a\} \subseteq \{w : |\log w| < 1\} = \Omega_e$ .

Lemma 2.4. [18, Lemma 2.2, p. 926] For $\frac{1}{3} < a < 3$ , let $\mathbf{r}_a$ be given by $$\mathbf{r}_a = \begin{cases} a - \frac{1}{3}, & \frac{1}{3} < a \le \frac{5}{3} \\ 3 - a, & \frac{5}{3} \le a \le 3. \end{cases}$$ Then $\{w : |w - a| < r_a\} \subseteq \Omega_c$ , where $\Omega_c$ is the region bonded by the cardioid given $\{x + iy : (9x^2 + 9y^2 - 18x + 5)^2 - 16(9x^2 + 9y^2 - 6x + 1) = 0\}$ .

Lemma 2.5. [2, Lemma 3.3, p.7] For $1 - \sin 1 < a < 1 + \sin 1$ , let $r_a = \sin 1 - |a - 1|$ . Then $\{w : |\omega - a| < r_a\} \subseteq \Omega_s$ ; $\Omega_s$ is the image of the unit disk $\mathbb D$ under $1 + \sin z$ .

Lemma 2.6. [4, Lemma 2.1, p. 3]. For $\sqrt{2} - 1 < a < \sqrt{2} + 1$ , let $\mathbf{r}_a = 1 - |\sqrt{2} - a|$ . Then $$\{w: |w-a| < r_a\} \subseteq \{w: |w^2-1| < 2|w|\}.$$

Lemma 2.7. [8, Lemma 2.2, p. 202] For $2(\sqrt{2}-1) < a < 2$ , let $r_a$ be given by $$r_a = \begin{cases} a - 2(\sqrt{2} - 1), & 2(\sqrt{2} - 1) < a \le \sqrt{2} \\ 2 - a, & \sqrt{2} \le a \le 2. \end{cases}$$ Then $\{w: |w-a| < r_a\}$ , where $\Omega_r$ is the image of the disk $\mathbb{D}$ under the function $1 + (zk + z^2)/(k^2 - kz)$ , $k = \sqrt{2} + 1$ . Proof of Theorem 1.1. Let the function $f \in \Pi_1$ . Then there is a function $g : \mathbb{D} \to \mathbb{C}$ satisfying <span id="page-3-0"></span> $$\operatorname{Re}\left(\frac{f(z)}{g(z)}\right) > 0 \quad \text{and} \quad \operatorname{Re}\left(\frac{(1-z)^2 g(z)}{z}\right) \quad \forall z \in \mathbb{D}.$$ (2.1) Define functions $p_1, p_2 : \mathbb{D} \to \mathbb{C}$ as the following. $$p_1(z) = \frac{(1-z)^2 g(z)}{z}$$ and $p_2(z) = \frac{f(z)}{g(z)}$ . (2.2) By using (2.1) and (2.2), we have $p_1, p_2 \in \mathcal{P}$ , and $f(z) = zp_1(z)p_2(z)/(1-z)^2$ . Then it follows that $$\frac{zf'(z)}{f(z)} = \frac{zp'_1(z)}{p_2(z)} + \frac{zp'_2(z)}{p_2(z)} + \frac{1+z}{1-z}.$$ (2.3) The bilinear transformation (1+z)/(1-z) maps the disk $|z| \le r$ onto the disk <span id="page-4-3"></span><span id="page-4-2"></span><span id="page-4-1"></span><span id="page-4-0"></span> $$\left| \frac{1+z}{1-z} - \frac{1-r^2}{1+r^2} \right| \le \frac{2r}{1-r^2}.$$ (2.4) For $p \in \mathcal{P}(\alpha)$ , we have $$\left| \frac{zp'(z)}{p(z)} \right| \le \frac{2(1-\alpha)r}{(1-r)(1+(1-2\alpha)r)}, \quad |z| \le r.$$ (2.5) By using (2.3), (2.4) and (2.5), function f maps disk |z| < r onto disk <span id="page-4-4"></span> $$\left| \frac{zf'(z)}{f(z)} - \frac{1+r^2}{1-r^2} \right| \le \frac{6r}{1-r^2}.$$ (2.6) From (2.6), it follows that <span id="page-4-5"></span> $$\operatorname{Re} \frac{zf'(z)}{f(z)} \ge \frac{1 - 6r + r^2}{1 - r^2} \ge 0,$$ (2.7) for all $0 \le r \le 3 - 2\sqrt{2}$ . Therefore, the function $f \in \Pi_1$ is starlike in $|z| \le 3 - 2\sqrt{2} \approx 0.171573$ . Hence, all our radii found here must be less than $3 - 2\sqrt{2}$ . (1) The number $\rho = R_{\mathcal{S}}(\alpha)$ , is the smallest positive root of the equation $(1+\alpha)r^2 - 6r + 1 - \alpha = 0$ in [0,1]. For $0 < r \le R_{\mathcal{S}^(\alpha)}$ , from (2.7), it follows that Re $$\frac{zf'(z)}{f(z)} \ge \frac{1 - 6r + r^2}{1 - r^2} \ge \frac{1 - 6\rho + \rho^2}{1 - \rho^2} = \alpha$$ . This shows that the radius of starlikeness of order $\alpha$ is at least $R_{\mathcal{S}^*}(\alpha)$ . To show that it is sharp, consider the function $f_0 \in \Pi_1$ given in (1.1). For this function $f_0$ , we have $$\frac{zf_0'(z)}{f_0(z)} = \frac{1+6z+z^2}{1-z^2}.$$ At $z = -\rho$ , we have <span id="page-4-6"></span>Re $$\frac{zf_0'(z)}{f_0(z)} = \frac{1 - 6\rho + \rho^2}{1 - \rho^2} = \alpha$$ , proving the sharpness of the radius. (2) We can give a proof using Lemma 2.1 but we give a different proof here. The number $\rho := R_{\mathcal{S}_L}$ is the smallest positive root of the equation $(1 + \sqrt{2})r^2 + 6r + 1 - \sqrt{2} = 0$ in interval (0, 1), and, from (2.6), it is clear that, for $0 \le r \le \rho$ , $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \left| \frac{zf'(z)}{f(z)} - \frac{1+r^2}{1-r^2} \right| + \frac{2r^2}{1-r^2} \le \frac{6r + 2r^2}{1-r^2} \le \frac{6\rho + \rho^2}{1-\rho^2} = \sqrt{2} - 1 \quad (2.8)$$ and $$\left| \frac{zf'(z)}{f(z)} + 1 \right| \le 2 + \left| \frac{zf'(z)}{f(z)} - 1 \right| \le \sqrt{2} + 1.$$ (2.9) Thus, from (2.8) and (2.9), it follows that, for $0 \le r \le \rho$ , $$\left| \left( \frac{zf'(z)}{f(z)} \right)^2 - 1 \right| = \left| \frac{zf'(z)}{f(z)} - 1 \right| \left| \frac{zf'(z)}{f(z)} + 1 \right| \le (\sqrt{2} + 1)(\sqrt{2} - 1) = 1.$$ For the function $f_0 \in \Pi_1$ given in (1.1), we have, at $z = \rho$ , <span id="page-5-0"></span> $$\frac{zf_0'(z)}{f_0(z)} = 1 + \frac{6\rho + 2\rho^2}{1 - \rho^2} = \sqrt{2}$$ and so, at $z = \rho$ , $$\left| \left( \frac{z f_0'(z)}{f_0(z)} \right)^2 - 1 \right| = 1.$$ This proves the sharpness. (3) For $\rho := R_{S_P} = (6 - \sqrt{33})/3$ , we have $$\frac{1}{2} < 1 \le a = \frac{1+r^2}{1-r^2} \le \frac{1+\rho^2}{1-\rho^2} = \frac{3\sqrt{33}-1}{16} \approx 1.0146 < 3/2.$$ Also, for $\rho = R_{\mathcal{S}_P}$ , we have $$\frac{6\rho}{(1-\rho^2)} \le \frac{1+\rho^2}{1-\rho^2} - \frac{1}{2}$$ and the disk in (2.6) for $r = \rho$ becomes $$\left| \frac{zf'(z)}{f(z)} - a \right| = \left| \frac{zf'(z)}{f(z)} - \frac{1+\rho^2}{1-\rho^2} \right| \le \frac{1+\rho^2}{1-\rho^2} - \frac{1}{2} = a - \frac{1}{2}.$$ By Lemma 2.2, it follows that the disk in (2.6) lies inside region $\Omega_{PAR}$ . This proves that the radius of parabolic starlikeness is at least $R_{S_P}$ . The radius is sharp for the function $f_0 \in \Pi_1$ . At the point $z = -\rho = -R_{\mathcal{S}_P}$ , we have $$\operatorname{Re}\left(\frac{zf_0'(z)}{f_0(z)}\right) = \frac{1 - 6\rho + \rho^2}{1 - \rho^2} = \frac{1}{2} = \frac{6\rho - 2\rho^2}{1 - \rho^2} = \left|\frac{zf_0'(z)}{f_0(z)} - 1\right|.$$ (4) For $e^{-1} < a \le \frac{e + e^{-1}}{2}$ , Lemma 2.3 gives $$\{w \in \mathbb{C} : |w - a| < a - e^{-1}\} \subseteq \{w \in \mathbb{C} : |\log w| < 1\} =: \Omega_e,$$ (2.10) For $\rho = R_{\mathcal{S}_e^*}$ , we have $$e^{-1} < a := \frac{1 + \rho^2}{1 - \rho^2} = \frac{1 + 9e^2}{e(1 + 3\sqrt{1 + 8e^2})} \approx 1.0236 \le \frac{e + e^{-1}}{2} \approx 1.5430$$ and, $\rho$ being smallest positive root of the equation $(1+e)r^2 - 6er + e - 1 = 0$ , <span id="page-5-1"></span> $$\frac{6\rho}{1-\rho^2} \le \frac{1+\rho^2}{1-\rho^2} - \frac{1}{e} = a - e^{-1}.$$ Consequently, the disk in (2.6) for $r = \rho$ becomes $$\left| \frac{zf'(z)}{f(z)} - a \right| = \left| \frac{zf'(z)}{f(z)} - \frac{1+\rho^2}{1-\rho^2} \right| \le \frac{1+\rho^2}{1-\rho^2} - \frac{1}{e} = a - e^{-1}.$$ By (2.10) the above disk is inside $\Omega_e$ proving that the $\mathcal{S}_e$ radius for the class $\Pi_1$ is at least $R_{\mathcal{S}_{e}^{}}$ . The result is sharp for the function $f_{0}$ given in (1.1). Indeed, at $z = -\rho$ where $\rho = R_{\mathcal{S}_{e}^{*}}$ , we have $$\left| \log \left( \frac{z f_0'(z)}{f_0(z)} \right) \right| = \left| \log \left( \frac{1 - 6\rho + \rho^2}{1 - \rho^2} \right) \right| = 1.$$ (5) For $\frac{1}{3} < a \le \frac{5}{3}$ , by an application of Lemma 2.4, it follows that $$\left\{ w \in \mathbb{C} : |w - a| < a - \frac{1}{3} \right\} \subseteq \Omega_c, \tag{2.11}$$ where $\Omega_c$ is the domain bounded by the cardioid $\{x+iy: (9x^2+9y^2-18x+5)^2\}$ $-16(9x^2 + 9y^2 - 6x + 1) = 0$ . For $\rho = R_{\mathcal{S}_c^*}$ , we have $$\frac{1}{3} < a := \frac{1+\rho^2}{1-\rho^2} = \frac{3\sqrt{73}-1}{24} \approx 1.0263 \le \frac{5}{3}$$ and, $\rho$ being the samllest positive root of the equation $2r^2 - 9r + 1 = 0$ , $$\frac{6\rho}{1-\rho^2} = \frac{1+\rho^2}{1-\rho^2} - \frac{1}{3}.$$ Therefore, the disk in (2.6) becomes $$\left| \frac{zf'(z)}{f(z)} - a \right| = \left| \frac{zf'(z)}{f(z)} - \frac{1 + \rho^2}{1 - \rho^2} \right| \le \frac{1 + \rho^2}{1 - \rho^2} - \frac{1}{3} = a - \frac{1}{3}$$ and this disk is inside $\Omega_c$ . This shows that $\mathcal{S}_c$ radius is at least $R_{\mathcal{S}_c}$ . For the function $f_0$ given in (1.1), at $z = \rho = R_{\mathcal{S}_c^*}$ , we have $$\frac{zf_0'(z)}{f_0(z)} = \frac{1 - 6\rho + \rho^2}{1 - \rho^2} = \frac{1}{3} = \varphi_c(-1) \in \partial \varphi_c(\mathbb{D})$$ where $\varphi_c(z) = 1 + 4z/3 + 2z^2/3$ . (6) For $\rho = R_{\mathcal{S}_{\sin}^*}$ , and $a := (1 + r^2)/(1 - r^2)$ , we have $$|a-1| = \frac{2\rho^2}{1-\rho^2} \approx 0.13199 < \sin 1 \approx 0.8414.$$ and $$\frac{6\rho}{1-\rho^2} \le \sin 1 - \frac{2\rho^2}{1-\rho^2}.$$ The disk in (2.6) for $r = \rho$ becomes $$\left| \frac{zf'(z)}{f(z)} - a \right| = \left| \frac{zf'(z)}{f(z)} - \frac{1 + \rho^2}{1 - \rho^2} \right| \le \sin 1 - \frac{2\rho^2}{1 - \rho^2} = \sin 1 - |1 - a|.$$ Lemma 2.5 shows that the disk in (2.6) is inside $\Omega_s$ where $\Omega_s =: \varphi_s(\mathbb{D})$ is the image of the unit disk $\mathbb{D}$ under the mapping $\varphi_s(z) = 1 + \sin z$ . This proves that the $\mathcal{S}_{\sin}$ radius is at least $R_{\mathcal{S}_{\sin}}$ . For the function $f_0$ given in (1.1), with $\rho = R_{\mathcal{S}_{\sin}^*}$ , we have $$\left(\frac{zf'(z)}{f(z)}\right) = \frac{1 + 6\rho + \rho^2}{1 - \rho^2} = 1 + \sin 1 \in \varphi_s(1) \in \partial \varphi_s(\mathbb{D}).$$ (7) For $\rho = R_{\mathcal{S}_{\mathcal{O}}^*}$ , we have $$a := \frac{1+\rho^2}{1-\rho^2} \approx 1.0202 \in (\sqrt{2}-1, \sqrt{2}+1)$$ and $$\frac{1 - 6\rho + \rho^2}{1 - \rho^2} = \sqrt{2} - 1.$$ The disk in (2.6) becomes $$\left| \frac{zf'(z)}{f(z)} - a \right| \le 1 - |\sqrt{2} - a|$$ and by Lemma 2.6 it lies inside $\{w: |w^2-1| < 2|w|\}$ . This shows that $\mathcal{S}^_{\mathbb{Q}}$ radius is at least $R_{\mathcal{S}^_{\mathbb{Q}}}$ . The sharpness follows as the function $f_0$ defined in (1.1) satisfies, at $z = \rho = R_{\mathcal{S}^*_{\mathbb{Q}}}$ , $$\left| \left( \frac{z f_0'(z)}{f_0(z)} \right)^2 - 1 \right| = \left| \left( \frac{1 - 6\rho + \rho^2}{1 - \rho^2} \right)^2 - 1 \right| = 2(\sqrt{2} - 1)$$ $$= 2 \frac{1 - 6\rho + \rho^2}{1 - \rho^2} = 2 \left| \frac{z f_0'(z)}{f_0(z)} \right|.$$ (8) For $\rho = R_{\mathcal{S}_{R}^{*}}$ , we have $$2(\sqrt{2}-1) < a := \frac{1+\rho^2}{1-\rho^2} \approx 1.00167 \le \sqrt{2} < 2,$$ and $$\frac{1 - 6\rho + \rho^2}{1 - \rho^2} = 2 - 2\sqrt{2}.$$ The disk (2.6) becomes $$\left| \frac{zf'(z)}{f(z)} - a \right| < a - 2(\sqrt{2} - 1).$$ (2.12) By Lemma 2.7, this disk lies inside the domain $\Omega_r$ . This proves that $\mathcal{S}_R$ radius is at least $R_{\mathcal{S}_R}$ . To prove sharpness, consider the function $f_0 \in \Pi_1$ given in (1.1). At $z = -\rho = -R_{\mathcal{S}_R^*}$ , we have $$\frac{zf'(z)}{f(z)} = \frac{1 - 6\rho + \rho^2}{1 - \rho^2} = 2(\sqrt{2} - 1) = \varphi_r(-1) \in \partial \varphi_r(\mathbb{D})$$ where $$\varphi_r(z) = 1 + (kz + z^2)/(k^2 - kz), k = \sqrt{2} + 1.$$ Proof of Theorem 1.2. Since |w-1| < 1 is equivalent to Re(1/w) > 1/2, the condition |f(z)/g(z) - 1| < 1 is the same as the condition Re(g(z)/f(z)) > 1/2. Let the function $f \in \Pi_2$ . Let $g : \mathbb{D} \to \mathbb{C}$ be chosen such that $$\operatorname{Re}\left(\frac{g(z)}{f(z)}\right) > \frac{1}{2} \quad \text{and} \quad \operatorname{Re}\left(\frac{(1-z)^2}{z}g(z)\right).$$ (2.13) Define $p_1, p_2 : \mathbb{D} \to \mathbb{C}$ as <span id="page-8-1"></span><span id="page-8-0"></span> $$p_1(z) = \frac{(1-z)^2}{z}g(z), \quad p_2(z) = \frac{g(z)}{f(z)}.$$ (2.14) From (2.13) and (2.14) it follows that $p_1 \in \mathcal{P}$ , and $p_2 \in \mathcal{P}(1/2)$ , and $f(z) = z/(1-z)^2 p_1(z)/p_2(z)$ . A calculation shows that $$\frac{zf'(z)}{f(z)} = \frac{zp'_1(z)}{p_1(z)} - \frac{zp'_2(z)}{p_2(z)} + \frac{1+z}{1-z}.$$ (2.15) The bilinear transformation $\omega = (1+z)/(1-z)$ maps the disk $|z| \leq r$ onto disk <span id="page-8-4"></span><span id="page-8-3"></span><span id="page-8-2"></span> $$\left| \frac{1+z}{1-z} - \frac{1+r^2}{1-r^2} \right| \le \frac{2r}{1-r^2}.$$ (2.16) Recall that for $p \in \mathcal{P}(\alpha)$ , we have $$\left| \frac{zp'(z)}{p(z)} \right| \le \frac{2(1-\alpha)r}{(1-r)(1+(1-2\alpha)r)}, \quad |z| \le r.$$ (2.17) Using (2.16) and (2.17) in (2.15), we get <span id="page-8-5"></span> $$\left| \frac{zf'(z)}{f(z)} - \frac{1+r^2}{1-r^2} \right| \le \frac{5r+r^2}{1-r^2}.$$ (2.18) From (2.18), it follows that $$\operatorname{Re}\left(\frac{zf'(z)}{f(z)}\right) \ge \frac{1-5r}{1-r^2} \ge 0 \tag{2.19}$$ for $0 \le r \le 1/5$ . For the function $f_3$ given in (1.3), we have <span id="page-8-6"></span> $$\frac{zf_3'(z)}{f_3(z)} = \frac{1+5z}{1-z^2} = 0$$ for z = -1/5. Thus, the radius of starlikeness of the class $\Pi_2$ is 1/5. All radius values to be computed here will be less than 1/5. (1) The number $\rho := R_{\mathcal{S}}(\alpha)$ is the smallest positive root of the equation $\alpha r^2 - 5r + 1 - \alpha = 0$ . For $0 < r \le R_{\mathcal{S}}(\alpha)$ , from (2.19), we have $$\operatorname{Re}\left(\frac{zf'(z)}{f(z)}\right) \ge \frac{1-5r}{1-r^2} \ge \frac{1-5\rho}{1-\rho^2} = \alpha.$$ For the function $f_2 \in \Pi_2$ given in (1.3), we have, at $z = -\rho = -R_{\mathcal{S}^*}(\alpha)$ , $$\frac{zf_3'(z)}{f_3(z)} = \frac{1 - 5\rho}{1 - \rho^2} = \alpha.$$ This proves that the radius of starlikeness of order $\alpha$ is $R_{\mathcal{S}^*}(\alpha)$ . (2) From (2.18), it follows that $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \left| \frac{zf'(z)}{f(z)} - \frac{1+r^2}{1-r^2} \right| + \frac{2r^2}{1-r^2} \le \frac{5r+3r^2}{1-r^2}. \tag{2.20}$$ The number $\rho = R_{\mathcal{S}_L}$ , is the positive root of the equation $5r + 3r^2 - (1 - r^2)(\sqrt{2} - 1) = 0$ . For $0 < r \le \rho = R_{\mathcal{S}_I}$ , we have $$\frac{5r+3r^2}{1-r^2} \le \frac{5\rho+3\rho^2}{1-\rho^2} = \sqrt{2}-1. \tag{2.21}$$ Therefore, by (2.20), (2.21), and for $0 < r \le \rho = R_{\mathcal{S}_L^*}$ , it follows that <span id="page-9-1"></span><span id="page-9-0"></span> $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \sqrt{2} - 1,$$ (2.22) and $$\left| \frac{zf'(z)}{f(z)} + 1 \right| \le \sqrt{2} + 1.$$ (2.23) The last two inequalities immediately yields $$\left| \left( \frac{zf'(z)}{f(z)} \right)^2 - 1 \right| \le \left| \frac{zf'(z)}{f(z)} + 1 \right| \left| \frac{zf'(z)}{f(z)} - 1 \right| \le (\sqrt{2} + 1)(\sqrt{2} - 1) = 1.$$ This proves that $\mathcal{S}_L$ is at least $R_{\mathcal{S}_L}$ . (3) For $0 \le r \le \rho := R_{\mathcal{S}_P} = 5 - 2\sqrt{6}$ , we have for $$\frac{1}{2} < 1 \le a = \frac{1 + \rho^2}{1 - \rho^2} = \frac{5\sqrt{6}}{12} < 3/2$$ and, $\rho$ being the smallest positive root of the equation $r^2 - 10r + 1 = 0$ , $$\frac{5\rho + \rho^2}{(1 - \rho^2)} \le \frac{1 + \rho^2}{1 - \rho^2} - \frac{1}{2}.$$ The disk in (2.18) becomes $$\left| \frac{zf'(z)}{f(z)} - \frac{1+\rho^2}{1-\rho^2} \right| \le \frac{1+\rho^2}{1-\rho^2} - \frac{1}{2}.$$ By Lemma 2.2, the disk in (2.18) is inside the region $\Omega_{PAR}$ . Thus, the radius of parabolic starlikeness of the class $\Pi_2$ is at least $R_{\mathcal{S}_p}$ . For the function $f_3$ given in (1.3) at $z = -\rho$ where $\rho = R_{\mathcal{S}_p}$ , we have $$\operatorname{Re}\left(\frac{zf_3'(z)}{f_3(z)}\right) = \frac{1 - 5\rho}{1 - \rho^2} = \frac{5\rho - \rho^2}{1 - \rho^2} = \left|\frac{zf_3'(z)}{f_3(z)} - 1\right|.$$ (4) For $\rho = R_{\mathcal{S}_e^*}$ , we have $1/e < a := (1 + \rho^2)/(1 - \rho^2) \approx 1.0331 \le (e + e^{-1})/2$ and $$\frac{5\rho + \rho^2}{1 - \rho^2} = \frac{1 + \rho^2}{1 - \rho^2} - \frac{1}{e}.$$ The disk in (2.18) becomes $$\left| \frac{zf'(z)}{f(z)} - \frac{1+\rho^2}{1-\rho^2} \right| \le \frac{1+\rho^2}{1-\rho^2} - \frac{1}{e}.$$ By Lemma 2.3, this disk is inside the region $\Omega_e$ , proving that $\mathcal{S}_e$ radius is at least $R_{\mathcal{S}_e}$ . The result is sharp for the function $f_3$ given in (1.3). For this function, we have, at $z = -\rho$ where $\rho = R_{\mathcal{S}_e^*}$ , $$\left| \log \left( \frac{z f_3'(z)}{f_3(z)} \right) \right| = \left| \log \left( \frac{1 - 5\rho}{1 - \rho^2} \right) \right| = \left| \log(e^{-1}) \right| = 1.$$ (5) For $\rho = R_{\mathcal{S}_c^*}$ , we have $1/3 < a := (1+\rho^2)/(1-\rho^2) = \frac{1}{72}(1+5\sqrt{217}) \approx 1.03686 \le 5/2$ and, $\rho$ being the smallest positive root of $r^2 - 15r + 2 = 0$ , $$\frac{5\rho + \rho^2}{1 - \rho^2} = \frac{1 + \rho^2}{1 - \rho^2} - \frac{1}{3}.$$ The disk in (2.18) becomes $$\left| \frac{zf'(z)}{f(z)} - \frac{1+\rho^2}{1-\rho^2} \right| \le \frac{1+\rho^2}{1-\rho^2} - \frac{1}{3}.$$ By Lemma 2.3, this disk is inside the region $\Omega_c$ , proving that $\mathcal{S}_c$ radius is at least $R_{\mathcal{S}_c}$ . The radius is sharp for the function $f_3$ given in (1.3). At $z = -\rho$ where $\rho = R_{\mathcal{S}_c^*}$ , we have $$\frac{zf_3'(z)}{f_3(z)} = \frac{1-5\rho}{1-\rho^2} = \frac{1}{3} = \varphi_c(-1) \in \partial \varphi_c(\mathbb{D})$$ where $\varphi_c(z) = 1 + 4z/3 + 2z^2/3$ . (6) For $\rho = R_{\mathcal{S}_{\sin}^*}$ , and $a := (1 + \rho^2)/(1 - \rho^2)$ , we have $$|a-1| = \frac{2\rho^2}{1-\rho^2} \approx 0.0465396 < \sin 1 \approx 0.8414.$$ and $$\frac{5\rho + \rho^2}{1 - \rho^2} \le \sin 1 - \frac{2\rho^2}{1 - \rho^2}.$$ The disk in (2.6) for $r = \rho$ becomes $$\left| \frac{zf'(z)}{f(z)} - a \right| = \left| \frac{zf'(z)}{f(z)} - \frac{1 + \rho^2}{1 - \rho^2} \right| \le \sin 1 - \frac{2\rho^2}{1 - \rho^2} = \sin 1 - |1 - a|.$$ Lemma 2.5 shows that the disk in (2.18) is inside $\Omega_s$ where $\Omega_s =: \varphi_s(\mathbb{D})$ is the image of the unit disk $\mathbb{D}$ under the mapping $\varphi_s(z) = 1 + \sin z$ . This proves that the $\mathcal{S}_{\sin}$ radius is at least $R_{\mathcal{S}_{\sin}}$ . (7) For $\rho = R_{\mathcal{S}_{\mathcal{O}}^*}$ , we have $$a := \frac{1+\rho^2}{1-\rho^2} \approx 1.02839 \in (\sqrt{2}-1, \sqrt{2}+1)$$ and $$\frac{5\rho + \rho^2}{1 - \rho^2} = \frac{1 + \rho^2}{1 - \rho^2} + 1 - \sqrt{2}.$$ The disk in (2.18) becomes $$\left| \frac{zf'(z)}{f(z)} - a \right| \le 1 - |\sqrt{2} - a|$$ and by Lemma 2.6 it lies inside $\{w: |w^2-1|<2|w|\}$ . This shows that $\mathcal{S}^_{\mathbb{Q}}$ radius is at least $R_{\mathcal{S}^_{\mathbb{Q}}}$ . The sharpness follows as the function $f_3$ defined in (1.1) satisfies, at $z=\rho=R_{\mathcal{S}^*_{\mathbb{Q}}}$ , $$\left| \left( \frac{zf_3'(z)}{f_3(z)} \right)^2 - 1 \right| = \left| \left( \frac{1 - 5\rho}{1 - \rho^2} \right)^2 - 1 \right| = 2(\sqrt{2} - 1)$$ $$= 2\frac{1 - 5\rho}{1 - \rho^2} = 2 \left| \frac{zf_3'(z)}{f_3(z)} \right|.$$ (8) For $\rho = R_{\mathcal{S}_{\mathcal{B}}^*}$ , we have $$2(\sqrt{2}-1) < a := \frac{1+\rho^2}{1-\rho^2} \approx 1.00238 \le \sqrt{2} < 2,$$ and $$\frac{5\rho + \rho^2}{1 - \rho^2} = \frac{1 + \rho^2}{1 - \rho^2} - 2(\sqrt{2} - 1).$$ The disk (2.18) becomes $$\left| \frac{zf'(z)}{f(z)} - a \right| < a - 2(\sqrt{2} - 1).$$ By Lemma 2.7, this disk lies inside the domain $\Omega_r$ . This proves that $\mathcal{S}_R$ radius is at least $R_{\mathcal{S}_R}$ . To prove sharpness, consider the function $f_3 \in \Pi_1$ given in (1.1). At $z = -\rho = -R_{\mathcal{S}_{\mathcal{P}}^*}$ , we have $$\frac{zf_3'(z)}{f_3(z)} = \frac{1 - 5\rho}{1 - \rho^2} = 2(\sqrt{2} - 1) = \varphi_r(-1) \in \partial \varphi_r(\mathbb{D})$$ where $\varphi_r(z) = 1 + (kz + z^2)/(k^2 - kz), k = \sqrt{2} + 1.$ We have only obtained lower bounds for the $\mathcal{S}_L$ and $\mathcal{S}_{sin}$ radii for the class $\Pi_2$ and we believe the bounds are sharp but unable to prove it.