Radius of Starlikeness of Certain Analytic Functions

Theorem 2.1. For the class $\mathcal{F}_1$ , the following results hold: (i) The $$S^*(\alpha)$$ radius $R_{S^*(\alpha)} = 2(1-\alpha)/\left(5+\sqrt{25-(4\alpha(1-\alpha))}\right)$ , $0 \le \alpha < 1$ . (ii) The $$S_L^*$$ radius $R_{S_L^*} = (2\sqrt{2} - 2)/(5 + \sqrt{33 - 4\sqrt{2}}) \approx 0.0809$ . (iii) The $$S_p$$ radius $R_{S_p} = 5 - 2\sqrt{6} \approx 0.1010$ . (iv) The $$S_e^*$$ radius $R_{S_e^*} = (2e - 2)/\left(5e + \sqrt{25e^2 + 4(1 - e)}\right) \approx 0.1276$ . (v) The $S_c$ radius $R_{S_c} = (15 - \sqrt{217})/2 \approx 0.1345$ . (vi) The $$S_{sin}^*$$ radius $R_{S_{sin}^*} = 2\sin 1/\left(5 + \sqrt{25 + 4\sin 1(1 + \sin 1)}\right) \approx 0.1589$ . (vii) The $$S_{\mathbb{C}}^*$$ radius $R_{S_{\mathbb{C}}^*} = (4 - 2\sqrt{2})/(5 + \sqrt{41 - 12\sqrt{2}}) \approx 0.1183$ . (viii) The $$S_R^*$$ radius $R_{S_R^*} = (6 - 4\sqrt{2}) / (5 + \sqrt{81 - 40\sqrt{2}}) \approx 0.0342$ . (ix) The $\mathcal{S}_{RL}$ radius $R_{\mathcal{S}_{RL}}$ is the root ( $\approx 0.0566$ ) in [0,1] of the equation $$25r^{2} + (r^{2} - 1)^{2} + (r^{2} - 1)\sqrt{(r^{2} + \sqrt{2})(2 - \sqrt{2} - r^{2})}$$ $$-(1 + \sqrt{2}(r^{2} - 1))^{2} = 0.$$ (x) The $S_{\gamma}$ radius of strong starlikeness $R_{S_{\gamma}} \ge \sin(\pi \gamma/2)/5$ , $0 < \gamma \le 1$ . PROOF. Let the function $f \in \mathcal{F}_1$ . Let the function $g : \mathbb{D} \longrightarrow \mathbb{C}$ be chosen such that <span id="page-3-0"></span> $$\operatorname{Re} \frac{f(z)}{g(z)} > 0 \quad \text{and} \quad \operatorname{Re} \left(\frac{1+z}{z}g(z)\right) > 0 \quad (z \in \mathbb{D}).$$ (2.2) Define the functions $p_1, p_2 : \mathbb{D} \longrightarrow \mathbb{C}$ by <span id="page-3-1"></span> $$p_1(z) = \frac{1+z}{z}g(z)$$ and $p_2(z) = \frac{f(z)}{g(z)}$ . (2.3) By (2.2) and (2.3), we have $p_1, p_2 \in \mathcal{P}$ and $f(z) = zp_1(z)p_2(z)/(1+z)$ . Then, a calculation involving logarithmic derivative of the function f shows that <span id="page-3-4"></span> $$\frac{zf'(z)}{f(z)} = \frac{zp_1'(z)}{p_1(z)} + \frac{zp_2'(z)}{p_2(z)} + \frac{1}{1+z}.$$ (2.4) The bilinear transformation 1/(1+z) maps the disk $|z| \leq r$ onto the disk <span id="page-3-2"></span> $$\left| \frac{1}{1+z} - \frac{1}{1-r^2} \right| \le \frac{r}{1-r^2}. \tag{2.5}$$ For $p \in \mathcal{P}(\alpha) := \{ p \in \mathcal{P} | \operatorname{Re} p > \alpha \}$ , by [41, Lemma 2], we have <span id="page-3-3"></span> $$\left| \frac{zp'(z)}{p(z)} \right| \le \frac{2(1-\alpha)r}{(1-r)(1+(1-2\alpha)r)} \quad (|z| \le r). \tag{2.6}$$ Using (2.5) and (2.6), it follows from (2.4) that the function f maps the disk $|z| \leq r$ onto the disk <span id="page-3-5"></span> $$\left| \frac{zf'(z)}{f(z)} - \frac{1}{1 - r^2} \right| \le \frac{5r}{1 - r^2}.$$ (2.7) The classes we discuss here are all subclasses of starlike functions. These classes are described by the quantity zf'(z)/f(z) lying in some region in the right half plane. The radius problems are solved by finding r such that the disk in (2.7) is contained in the corresponding regions. By (2.7), we have <span id="page-3-6"></span>Re $$\frac{zf'(z)}{f(z)} \ge \frac{1-5r}{1-r^2} \ge 0$$ , $(r \le 1/5)$ , (2.8) then the function $f \in \mathcal{F}_1$ is starlike in $|z| \leq 1/5$ . Hence all the radii that we estimate will be less than 1/5. Note that, for $0 < r \leq 1/5$ , the centre of disk in (2.7) lies in the interval $[1, 25/24] \approx [1, 1.0416]$ . (i) The number $r = R_{\mathcal{S}^(\alpha)}$ is the root of $\alpha r^2 - 5r + 1 - \alpha = 0$ in [0, 1] and hence, for $0 < r \le R_{\mathcal{S}^(\alpha)}$ , it follows from (2.8) that Re $$\frac{zf'(z)}{f(z)} \ge \frac{1-5r}{1-r^2} \ge \alpha$$ . For the functions $f_1 \in \mathcal{F}_1$ given by (2.1) we have $$\frac{zf_1'(z)}{f_1(z)} = \frac{1 - 5z}{1 - z^2} = \frac{1 - 5r}{1 - r^2} = \alpha, \quad (z = r = R_{\mathcal{S}^*(\alpha)})$$ and this shows that the radius is sharp. (ii) It follows from (2.7) that <span id="page-4-0"></span> $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \left| \frac{zf'(z)}{f(z)} - \frac{1}{1 - r^2} \right| + \frac{r^2}{1 - r^2} \le \frac{5r + r^2}{1 - r^2}.$$ (2.9) The number $r = R_{\mathcal{S}_L}$ is the root in [0,1] of $(5r+r^2) = (\sqrt{2}-1)(1-r^2)$ and for $0 < r \le R_{\mathcal{S}_L}$ , we have <span id="page-4-1"></span> $$\frac{5r+r^2}{1-r^2} \le \sqrt{2}-1. \tag{2.10}$$ Therefore, by (2.9) and (2.10), for $0 < r \le R_{\mathcal{S}_{t}^{*}}$ , we have <span id="page-4-2"></span> $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{5r + r^2}{1 - r^2} \le \sqrt{2} - 1. \tag{2.11}$$ For $0 < r \le R_{\mathcal{S}_{L}^{*}}$ , using triangle inequality and (2.11), we have <span id="page-4-3"></span> $$\left| \frac{zf'(z)}{f(z)} + 1 \right| \le 2 + \left| \frac{zf'(z)}{f(z)} - 1 \right| \le \sqrt{2} + 1$$ (2.12) and hence by (2.11) and (2.12), $$\left| \left( \frac{zf'(z)}{f(z)} \right)^2 - 1 \right| \le \left| \frac{zf'(z)}{f(z)} + 1 \right| \left| \frac{zf'(z)}{f(z)} - 1 \right| \le (\sqrt{2} + 1)(\sqrt{2} - 1) = 1.$$ The number $\rho = R_{\mathcal{S}_L^*}$ satisfies $(1+5\rho)/(1-\rho^2) = \sqrt{2}$ . Using this, we see that the function $f_1$ defined in (2.1) satisfies $$\left| \left( \frac{z f_1'(z)}{f_1(z)} \right)^2 - 1 \right| = \left| \left( \frac{1 - 5z}{1 - z^2} \right)^2 - 1 \right| = \left| \left( \frac{1 + 5\rho}{1 - \rho^2} \right)^2 - 1 \right| = 1, \quad (z := -\rho = -R_{\mathcal{S}_L^*}).$$ This shows that the radius is sharp (See Figure 1.(a)). (iii) Let $\Omega_{PAR} = \{w = u + i v : v^2 < 2u - 1\} = \{w : \text{Re } w > |w - 1|\}$ . Note that $\Omega_{PAR}$ is the interior of a parabola in the right half plane which is symmetric about real axis and has vertex at (1/2, 0). By Lemma [43, pp.321], for 1/2 < a < 3/2, we have <span id="page-4-4"></span> $$\{w \in \mathbb{C} : |w - a| < a - 1/2\} \subseteq \Omega_{PAR}. \tag{2.13}$$ If $0 < r \le R_{S_p}$ , then $a = 1/(1 - r^2) \le 3/2$ and $$\frac{5r}{1-r^2} \le \frac{1}{1-r^2} - \frac{1}{2}.$$ <span id="page-5-0"></span> FIGURE 1. Sharpness of starlike functions associated with lemniscate and parabolic starlike functions. Thus, by (2.13), we see that the disk in (2.7) lies inside the parabolic region $\Omega_{PAR}$ . Sharpness follows for the function $f_1$ defined in (2.1) (See Figure 1.(b)). At $z := \rho = R_{\mathcal{S}_p}$ , we have Re $$\frac{zf'(z)}{f(z)} = \frac{1-5\rho}{1-\rho^2} = \frac{5\rho-\rho^2}{1-\rho^2} = \left|\frac{\rho^2-5\rho}{1-\rho^2}\right| = \left|\frac{zf'(z)}{f(z)}-1\right|.$$ (iv) For $e^{-1} \le a \le (e + e^{-1})/2$ , by [34, Lemma 2.2], we have <span id="page-5-1"></span> $$\{w \in \mathbb{C} : |w - a| < a - e^{-1}\} \subseteq \{w \in \mathbb{C} : |\log w| < 1\} =: \Omega_e.$$ (2.14) For $0 < r \le R_{\mathcal{S}_e^*}$ , we have $1/e \le a = 1/(1-r^2) \le (e+e^{-1})/2$ and $$\frac{5r}{1-r^2} \le \frac{1}{1-r^2} - \frac{1}{e}.$$ By (2.14), the disk in (2.7) lies inside $\Omega_e$ for $0 < r \le R_{\mathcal{S}_e}$ proving that the $\mathcal{S}_e$ radius for the class $\mathcal{F}_1$ is $R_{\mathcal{S}_e}$ . The sharpness follows for the function $f_1$ defined in (2.1) (See Figure 2.(a). Indeed at $z := \rho = R_{\mathcal{S}_e}$ , we have $$\left| \log \frac{z f_1'(z)}{f_1(z)} \right| = \left| \log \frac{1 - 5\rho}{1 - \rho^2} \right| = 1.$$ (v) For $1/3 < a \le 5/3$ , by [44, Lemma 2.5], we have <span id="page-5-2"></span> $$\{w \in \mathbb{C} : |w - a| < (3a - 1)/3\} \subseteq \Omega_c \tag{2.15}$$ where $\Omega_c$ is the region bounded by the cardioid $\{x+iy: (9x^2+9y^2-18x+5)^2-16(9x^2+9y^2-6x+1)=0\}$ . If $0 < r \le R_{\mathcal{S}_c^*}$ , then $$\frac{5r}{1-r^2} \le \frac{1}{1-r^2} - \frac{1}{3}.$$ By (2.15), we see that the disk in (2.7) lies inside $\Omega_c$ , if $0 < r \le R_{\mathcal{S}_c}$ . The result is sharp for the function $f_1$ defined in (2.1). At $z := \rho = R_{\mathcal{S}_c}$ , $$\left| \frac{zf'(z)}{f(z)} \right| = \left| \frac{1 - 5\rho}{1 - \rho^2} \right| = \frac{1}{3} = \Omega_c(-1) \in \partial_c(\mathbb{D}).$$ (vi) For $|a-1| \leq \sin 1$ , by [7, Lemma 3.3], we have <span id="page-5-3"></span> $$\{w \in \mathbb{C} : |w - a| < \sin 1 - |a - 1|\} \subseteq \Omega_s \tag{2.16}$$ <span id="page-6-0"></span>   (B) Sharpness of class $S_{sin}^*$ FIGURE 2. Sharpness for starlike functions associated with exponential and sine functions. where $\Omega_s := q_0(\mathbb{D})$ is the image of the unit disk $\mathbb{D}$ under the mappings $q_0(z) = 1 + \sin z$ . It is evident from (2.7) and (2.16) that $$\frac{5r}{1-r^2} \le \sin 1 - \frac{r^2}{1-r^2}.$$ Hence the disk in (2.7) lies inside $\Omega_s$ provided $0 < r \le R_{\mathcal{S}_{sin}}$ . For the function $f_1$ defined in (2.1) (See Figure 2.(b)), at $z := -\rho = -R_{\mathcal{S}_{sin}}$ , $$\left|\frac{zf'(z)}{f(z)}\right| = \left|\frac{1+5\rho}{1-\rho^2}\right| = 1 + \sin 1 = q_0(1) \in \partial\Omega_s(\mathbb{D}).$$ (vii) Let $\mathcal{S}^_{\mathbb{Q}} = \{f \in \mathcal{S}^ : |(zf'(z)/f(z))^2 - 1| < 2|zf'(z)/f(z)|\}$ . In 2015, Sokół [36] proved that if $f \in \mathcal{S}^_{\mathbb{Q}}$ then the function f is a starlike function and $|zf'(z)/f(z) - 1| < \sqrt{2}$ and $|zf'(z)/f(z) + 1| > \sqrt{2}$ . Interpreting these conditions geometrically, for $f \in \mathcal{S}^_{\mathbb{Q}}$ we see that w = zf'(z)/f(z) lies in the right half plane, inside the intersection of disks $\{w : |w - 1| < \sqrt{2}\}$ and $\{w : |w + 1| < \sqrt{2}\}$ . According to [9, Lemma 2.1], we have <span id="page-6-1"></span> $$\{w \in \mathbb{C} : |w - a| < 1 - |\sqrt{2} - a|\} \subseteq \{w \in \mathbb{C} : |w^2 - 1| < 2|w|\}$$ (2.17) or, $(1 - \sqrt{2})r^2 + 5r + \sqrt{2} - 2 \le 0$ . If $0 < r \le R_{\mathcal{S}^*_{(\!(})}$ , then $$\frac{5r-1}{1-r^2} \le 1 - \sqrt{2}.$$ Thus, by (2.17), the disk in (2.7) lies inside $\{w \in \mathbb{C} : |w^2 - 1| < 2 |w|\}$ and hence $f \in \mathcal{S}^_{\mathbb{C}}$ . The sharpness follows from the functions defined in (2.1). At $z := \rho = R_{\mathcal{S}^_{\mathbb{C}}}$ , we have $$\left| \left( \frac{zf_1'(z)}{f_1(z)} \right)^2 - 1 \right| = \left| \left( \frac{1 - 5\rho}{1 - \rho^2} \right)^2 - 1 \right| = 2 \left| \frac{1 - 5\rho}{1 - \rho^2} \right| = 2 \left| \frac{zf_1'(z)}{f_1(z)} \right|.$$ (viii) For $2(\sqrt{2}-1) < a \le \sqrt{2}$ , by [26, Lemma 2.2], <span id="page-6-2"></span> $$\{w \in \mathbb{C} : |w - a| < a - 2(\sqrt{2} - 1)\} \subseteq \psi(\mathbb{D}) \tag{2.18}$$ where $\psi$ is given by $\psi(z) = 1 + (z^2k + z^2/(k^2 - kz))$ , $k = \sqrt{2} + 1$ . If $0 < r \le R_{\mathcal{S}_R^*}$ , $2(\sqrt{2} - 1) < a = 1/(1 - r^2) \le \sqrt{2}$ and $$\frac{5r - 1}{1 - r^2} \le 2 - 2\sqrt{2}, \quad (0 < r \le R_{\mathcal{S}_R^*}).$$ Then, by (2.18), the disk in (2.7) lies inside $\psi(\mathbb{D})$ . The result is sharp for the function defined in (2.1). At $z := \rho = R_{\mathcal{S}_p^*}$ , $$\left| \frac{zf'(z)}{f(z)} \right| = \left| \frac{1 - 5\rho}{1 - \rho^2} \right| = 2(\sqrt{2} - 1) = \psi(1) \in \partial \psi(\mathbb{D}).$$ (ix) For $\sqrt{2}/3 \le a < \sqrt{2}$ , by [33, Lemma 3.2], we have <span id="page-7-0"></span> $$\{w \in \mathbb{C} : |w - a| < r_{RL}\} \subseteq \{w \in \mathbb{C} : |(w - \sqrt{2})^2 - 1| < 1\},$$ (2.19) provided $r_{RL} = \left( \left( 1 - \left( \sqrt{2} - a \right)^2 \right)^{1/2} - \left( 1 - \left( \sqrt{2} - a \right)^2 \right) \right)^{1/2}$ . If $0 < r \le R_{\mathcal{S}_{RL}}^*$ , then it follows that $\sqrt{2}/3 \le a = 1/(1 - r^2) < \sqrt{2}$ , and $$25r^{2} - (1 - r^{2})\sqrt{(1 - r^{2})^{2} - ((\sqrt{2} - \sqrt{2}r^{2}) - 1)^{2}} + (1 - r^{2})^{2} - ((\sqrt{2} - \sqrt{2}r^{2}) - 1)^{2} \le 0.$$ Then, by (2.19), the disk in (2.7) lies inside the region $\{w : |(w - \sqrt{2})^2 - 1| < 1\}$ . The result is sharp for the function defined in (2.1). (x) If a function f(z) is strongly starlike of order $\gamma, 0 < \gamma \le 1$ , then $|\arg\{zf'(z)/f(z)\}| \le \pi\gamma/2$ . In other words the values of |zf'(z)/f(z)| are in the sector $|y| \le \tan(\pi\gamma/2)x$ , $x \ge 0$ . In 1997, Gangadharan et al. [10, Lemma 3.1] proved that <span id="page-7-1"></span> $$\{w \in \mathbb{C} : |w - a| < a\sin(\pi\gamma/2)\} \subseteq \{w : |\arg w| \le (\pi\gamma)/2\}, \quad 0 < \gamma \le 1.$$ (2.20) If $0 < r \le R_{\mathcal{S}_{\gamma}}$ , $0 < \gamma \le 1$ , then $r \le (\sin(\pi\gamma)/2)/5$ . It is evident from (2.20) that the disk in (2.7) is contained in the sector $|\arg w| \le (\pi\gamma)/2$ , $0 < \gamma \le 1$ if $0 < r \le R_{\mathcal{S}_{\gamma}}$ . Remark 2.2. (i) The class $$\mathcal{S}^*(c,d) = \left\{ f \in \mathcal{A} : \left| \frac{zf'(z)}{f(z)} - c \right| < d \right\}$$ is very closely related to the class of Janowski starlike functions. For the class $\mathcal{F}_1$ , the $\mathcal{S}_p$ radius, $\mathcal{S}^(1/2)$ radius and $\mathcal{S}^(1,1/2)$ radius are all equal. It is also clear that $\mathcal{S}^(1,\sqrt{2}-1)\subseteq \mathcal{S}_L$ . The $\mathcal{S}_L$ radius and $\mathcal{S}^(1,\sqrt{2}-1)$ radius are also equal. (ii) The radius of strong starlikeness is clearly sharp for $\gamma = 0$ and for other cases, we have given a lower bound only.

Theorem 2.4. For the class $\mathcal{F}_2$ the following results hold: (i) The $$S^*(\alpha)$$ radius $R_{S^*(\alpha)} = (1-\alpha)/\left(2+\sqrt{4+(\alpha-1)^2}\right)$ , $0 \le \alpha < 1$ . - (ii) The $S_L$ radius $R_{S_L} \ge (\sqrt{5} 2)/(1 + \sqrt{2}) \approx 0.0977$ . - (iii) The $S_p$ radius $R_{S_p} = \sqrt{17} 4 \approx 0.1231$ . - (iv) The $S_e$ radius $R_{S_e} = (2e 2)/(4e + \sqrt{20e^2 8e + 4}) \approx 0.1543$ . - (v) The $S_c$ radius $R_{S_c} = \sqrt{10} 3 \approx 0.1623$ . - (vi) The $S_{sin}$ radius $R_{S_{sin}} \ge \sin 1/\left(2 + \sqrt{4 + \sin 1(2 + \sin 1)}\right) \approx 0.1858$ . - (vii) The $S_{\mathbb{C}}$ radius $R_{S_{\mathbb{C}}} = (2 \sqrt{2}) / (2 + \sqrt{10 4\sqrt{2}}) \approx 0.1434$ . - (viii) The $\mathcal{S}_R$ radius $R_{\mathcal{S}_R} = (3 2\sqrt{2})/(2 + \sqrt{21 12\sqrt{2}}) \approx 0.0428$ . (ix) The $\mathcal{S}_{RL}$ radius $R_{\mathcal{S}_{RL}}$ is at least the smallest root ( $\approx 0.0692$ ) in [0, 1] of the equation $$r^{2}(r+4)^{2} + (r^{2}-1)^{2} + (r^{2}-1)\sqrt{(r^{2}+\sqrt{2})(2-\sqrt{2}-r^{2})} - (1+\sqrt{2}(r^{2}-1))^{2} = 0.$$ (x) The $$S_{\gamma}^*$$ radius $R_{S_{\gamma}^*} \ge \sin(\pi \gamma/2)/(2 + \sqrt{4 + \sin(\pi \gamma/2)}, \quad 0 < \gamma \le 1.$ PROOF. Let the function $f \in \mathcal{F}_2$ . Then |f(z)/g(z)-1|<1 if and only if $\operatorname{Re}(g(z)/f(z))>$ 1/2. Let the function $g: \mathbb{D} \longrightarrow \mathbb{C}$ be chosen such that <span id="page-8-0"></span> $$\operatorname{Re} \frac{g(z)}{f(z)} > 1/2 \quad \text{and} \quad \operatorname{Re} \left(\frac{1+z}{z}g(z)\right) > 0 \quad (z \in \mathbb{D}).$$ (2.22) Define the functions $p_1, p_2 : \mathbb{D} \longrightarrow \mathbb{C}$ by <span id="page-8-1"></span> $$p_1(z) = \frac{1+z}{z}g(z)$$ and $p_2(z) = \frac{g(z)}{f(z)}$ . (2.23) By (2.22) and (2.23), we have $p_1 \in \mathcal{P}$ , $p_2 \in \mathcal{P}(1/2)$ and $f(z) = zp_1(z)/((1+z)p_2(z))$ . It can be shown by calculation that <span id="page-8-2"></span> $$\frac{zf'(z)}{f(z)} = \frac{zp'_1(z)}{p_1(z)} - \frac{zp'_2(z)}{p_2(z)} + \frac{1}{1+z}.$$ (2.24) Using (2.5) and (2.6), it follows from (2.24) that the function f maps the disk $|z| \leq r$ onto the disk <span id="page-9-0"></span> $$\left| \frac{zf'(z)}{f(z)} - \frac{1}{1 - r^2} \right| \le \frac{4r + r^2}{1 - r^2}.$$ (2.25) The classes we discuss here are all subclasses of starlike functions. By (2.25), we have <span id="page-9-1"></span>Re $$\frac{zf'(z)}{f(z)} \ge \frac{1 - 4r - r^2}{1 - r^2} \ge 0$$ , $(r \le \sqrt{5} - 2)$ . (2.26) Hence all the radii that we estimate will be less than $\sqrt{5} - 2 \approx 0.2361$ . Also, for $0 < r \le \sqrt{5} - 2$ , the centre of disk in (2.25) lies in the interval $[1, 1/4(\sqrt{5} - 2)] \approx [1, 1.05902]$ . (i) The number $r = R_{\mathcal{S}^(\alpha)}$ is the root of $(\alpha - 1)r^2 - 4r + 1 - \alpha = 0$ in [0, 1] and hence, for $0 < r \le R_{\mathcal{S}^(\alpha)}$ , it follows by (2.26) that Re $$\frac{zf'(z)}{f(z)} \ge \frac{1 - 4r - r^2}{1 - r^2} \ge \alpha$$ . For the functions $f_2 \in \mathcal{F}_2$ given by (2.21), we have $$\frac{zf_2'(z)}{f_2(z)} = \frac{1 - 4z - z^2}{1 - z^2} = \frac{1 - 4r - r^2}{1 - r^2} = \alpha, \quad (z = r = R_{\mathcal{S}^*(\alpha)})$$ and this shows that the radius is sharp. (ii) It follows from (2.25) that <span id="page-9-2"></span> $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \left| \frac{zf'(z)}{f(z)} - \frac{1}{1 - r^2} \right| + \frac{r^2}{1 - r^2} \le \frac{4r + 2r^2}{1 - r^2}. \tag{2.27}$$ The number $r = R_{\mathcal{S}_L}$ is the root in [0,1] of $(4r+2r^2) = \sqrt{2} - 1(1-r^2)$ and for $0 < r \le R_{\mathcal{S}_I}$ , we have <span id="page-9-3"></span> $$\frac{4r + 2r^2}{1 - r^2} \le \sqrt{2} - 1. \tag{2.28}$$ Therefore, by (2.27) and (2.28), for $0 < r \le R_{\mathcal{S}_L^*}$ , we have <span id="page-9-4"></span> $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{4r + 2r^2}{1 - r^2} \le \sqrt{2} - 1. \tag{2.29}$$ For $0 < r \le R_{\mathcal{S}_L^*}$ , using triangle inequality and (2.29), we have <span id="page-9-5"></span> $$\left| \frac{zf'(z)}{f(z)} + 1 \right| \le \sqrt{2} + 1$$ (2.30) and hence by (2.29) and (2.30) $$\left| \left( \frac{zf'(z)}{f(z)} \right)^2 - 1 \right| \le \left| \frac{zf'(z)}{f(z)} + 1 \right| \left| \frac{zf'(z)}{f(z)} - 1 \right| \le (\sqrt{2} + 1)(\sqrt{2} - 1) = 1.$$ The result obtained is not sharp. (iii) If $0 < r \le R_{S_p}$ , then $a = 1/(1 - r^2) \le 3/2$ and $$\frac{4r+r^2}{1-r^2} \le \frac{1}{1-r^2} - \frac{1}{2}.$$ Thus, by (2.13), we see that the disk in (2.25) lies inside the parabolic region $\Omega_{PAR}$ . Sharpness follows for the function $f_2$ defined in (2.21). At $z := \rho = R_{\mathcal{S}_p}$ , we have $$\operatorname{Re} \frac{zf'(z)}{f(z)} = \frac{1 - 4\rho - \rho^2}{1 - \rho^2} = \frac{4\rho}{1 - \rho^2} = \left| \frac{-4\rho}{1 - \rho^2} \right| = \left| \frac{zf'(z)}{f(z)} - 1 \right|.$$ (iv) For $0 < r \le R_{\mathcal{S}_e^*}$ , we have $1/e \le a = 1/(1-r^2) \le (e+e^{-1})/2$ and $$\frac{4r+r^2}{1-r^2} \le \frac{1}{1-r^2} - \frac{1}{e}.$$ By (2.14), the disk in (2.25) lies inside $\Omega_e$ for $0 < r \le R_{\mathcal{S}_e}$ proving that the $\mathcal{S}_e$ radius for the class $\mathcal{F}_2$ is $R_{\mathcal{S}_e}$ . The sharpness follows for the function $f_2$ defined in (2.21). Indeed at $z := \rho = R_{\mathcal{S}_e}$ , we have $$\left| \log \frac{z f_2'(z)}{f_2(z)} \right| = \left| \log \frac{1 - 4\rho - \rho^2}{1 - \rho^2} \right| = 1.$$ (v) If $0 < r \le R_{\mathcal{S}_c^*}$ , then $$\frac{4r+r^2}{1-r^2} \le \frac{1}{1-r^2} - \frac{1}{3}.$$ By (2.15), we see that the disk in (2.25) lies inside $\Omega_c$ , if $0 < r \le R_{\mathcal{S}_c}$ . The result is sharp for the function $f_2$ defined in (2.21) (See Figure 3.(a)). At $z := \rho = R_{\mathcal{S}_c}$ , $$\left| \frac{zf'(z)}{f(z)} \right| = \left| \frac{1 - 4\rho - \rho^2}{1 - \rho^2} \right| = \frac{1}{3} = \Omega_c(-1) \in \partial\Omega_c(\mathbb{D}).$$ (vi) For $0 < r \le R_{\mathcal{S}^*_{sin}}$ , $|a - 1| \le \sin 1$ and $$\frac{4r+r^2}{1-r^2} \le \sin 1 - \frac{r^2}{1-r^2},$$ then it is evident from (2.16) that the disk in (2.25) lies inside $\Omega_s$ . The radius is not sharp. (vii) If $0 < r \le R_{\mathcal{S}_{\mathcal{I}}^*}$ , then $$\frac{r^2 + 4r - 1}{1 - r^2} \le 1 - \sqrt{2}.$$ Thus by (2.17), the disk in (2.25) lies inside $\{w \in \mathbb{C} : |w^2 - 1| < 2 |w|\}$ and hence $f \in \mathcal{S}^_{\mathbb{C}}$ . The sharpness follows for the functions defined in (2.21) (See Figure 3.(b)). At $z := \rho = R_{\mathcal{S}^_{\mathcal{D}}}$ , we have $$\left| \left( \frac{z f_2'(z)}{f_2(z)} \right)^2 - 1 \right| = \left| \left( \frac{\rho^2 + 4\rho - 1}{1 - \rho^2} \right)^2 - 1 \right| = 2 \left| \frac{\rho^2 + 4\rho - 1}{1 - \rho^2} \right| = 2 \left| \frac{z f_1'(z)}{f_1(z)} \right|.$$ (viii) If $0 < r \le R_{\mathcal{S}_R^*}$ , $2(\sqrt{2}-1) < a = 1/(1-r^2) \le \sqrt{2}$ and $$\frac{r^2 + 4r - 1}{1 - r^2} \le 2 - 2\sqrt{2}, \quad 0 < r \le R_{\mathcal{S}_R^*}.$$ <span id="page-11-0"></span>  (B) Sharpness of class $\mathcal{S}_{\mathcal{A}}^*$ FIGURE 3. Sharpness of starlike functions associated with cardioid and lune. Then, by (2.18), the disk in (2.25) lies inside $\psi(\mathbb{D})$ . For the function defined in (2.21), at $z := \rho = R_{\mathcal{S}_R^*}$ , $$\left| \frac{zf'(z)}{f(z)} \right| = \left| \frac{1 - 4\rho - \rho^2}{1 - \rho^2} \right| = 2(\sqrt{2} - 1) = \psi(1) \in \partial \psi(\mathbb{D}).$$ (ix) If $$0 < r \le R_{\mathcal{S}_{RL}^*}$$ , $\sqrt{2}/3 \le a = 1/(1-r^2) < \sqrt{2}$ , and $$(4r+r^2)^2 - (1-r^2)\sqrt{(1-r^2)^2 - ((\sqrt{2}-\sqrt{2}r^2)-1)^2} + (1-r^2)^2 - ((\sqrt{2}-\sqrt{2}r^2)-1)^2 \le 0.$$ Then by (2.19) the disk in (2.25) lies inside the region $\{w: |(w-\sqrt{2})^2-1|<1\}$ . The result obtained is not sharp. (x) If $0 < r \le R_{\mathcal{S}^_{\gamma}}$ and $0 < \gamma \le 1$ , then $r^2 + 4r \le (\sin(\pi\gamma)/2)$ . It is evident from (2.20) that the disk in (2.25) is contained in the sector $|\arg w| \le (\pi\gamma)/2, 0 < \gamma \le 1$ if $0 < r \le R_{\mathcal{S}^_{\gamma}}$ . The problem of determination of the exact radii of starlikeness associated with lemniscate of Bernoulli, sine function, reverse lemniscate and strongly starlike function is open.

Theorem 2.6. For the class $\mathcal{F}_3$ , the following results hold: (i) The $$S^*(\alpha)$$ radius $R_{S^*(\alpha)} = 2(1-\alpha)/\left(3+\sqrt{9-4\alpha(1-\alpha)}\right)$ , $0 \le \alpha < 1$ . (ii) The $$S_L^*$$ radius $R_{S_L^*} = (2\sqrt{2} - 2) / \left(3 + \sqrt{9 - 4\sqrt{2}(1 - \sqrt{2})}\right) \approx 0.1301$ . (iii) The $$S_p$$ radius $R_{S_p} = 3 - 2\sqrt{2} \approx 0.1716$ . (iv) The $$S_e^*$$ radius $R_{S_e^*} = (2e - 2)/(3e + \sqrt{9e^2 + 4(1 - e)}) \approx 0.2165$ . (v) The $$S_c^*$$ radius $R_{S_c^*} = (9 - \sqrt{73})/2 \approx 0.2279$ . (vi) The $$S_{sin}^*$$ radius $R_{S_{sin}^*} = 2\sin 1/\left(3 + \sqrt{9 + 4\sin 1(1 + \sin 1)}\right) \approx 0.2439$ . (vii) The $$S_{\mathbb{C}}^*$$ radius $R_{S_{\mathbb{C}}^*} = (4 - 2\sqrt{2})/(3 + \sqrt{25 - 12\sqrt{2}}) \approx 0.2008$ . (viii) The $$S_R^*$$ radius $R_{S_R^*} = (6 - 4\sqrt{2})/(3 + \sqrt{65 - 40\sqrt{2}}) \approx 0.0581$ . (ix) The $\mathcal{S}_{RL}$ radius $R_{\mathcal{S}_{RL}}$ is the root ( $\approx 0.0926$ ) in [0,1] of the equation $$9r^{2} + (r^{2} - 1)^{2} + (r^{2} - 1)\sqrt{(r^{2} + \sqrt{2})(2 - \sqrt{2} - r^{2})} - (1 + \sqrt{2}(r^{2} - 1))^{2} = 0.$$ (x) The $S_{\gamma}$ radius $R_{S_{\gamma}} \ge \sin(\pi \gamma/2)/3$ , $0 < \gamma \le 1$ . PROOF. Let the function $f \in \mathcal{F}_3$ . Then <span id="page-12-0"></span> $$\operatorname{Re}\left(\frac{1+z}{z}f(z)\right) > 0 \quad (z \in \mathbb{D}).$$ (2.32) Define the function $h: \mathbb{D} \longrightarrow \mathbb{C}$ by <span id="page-12-1"></span> $$h(z) = \frac{1+z}{z}f(z).$$ (2.33) By (2.32) and (2.33) we have $h \in \mathcal{P}$ and f(z) = zh(z)/(1+z). Therefore, by calculation it can be shown that <span id="page-12-2"></span> $$\frac{zf'(z)}{f(z)} = \frac{zh'(z)}{h(z)} + \frac{1}{1+z}. (2.34)$$ Using (2.5) and (2.6), it follows from (2.34) that f maps the disk $|z| \le r$ onto the disk <span id="page-12-3"></span> $$\left| \frac{zf'(z)}{f(z)} - \frac{1}{1 - r^2} \right| \le \frac{3r}{1 - r^2}.$$ (2.35) As the classes we discuss here are all subclasses of starlike functions. By (2.35), we have <span id="page-12-4"></span>Re $$\frac{zf'(z)}{f(z)} \ge \frac{1-3r}{1-r^2} \ge 0$$ , $(r \le 1/3)$ . (2.36) Hence all the radii that we estimate will be less than 1/3. Note that, for $0 < r \le 1/3$ , the centre of disk in (2.35) lies in the interval $[1, 9/8] \approx [1, 1.125]$ . (i) The number $r = R_{\mathcal{S}^(\alpha)}$ is the root of $\alpha r^2 - 3r + 1 - \alpha = 0$ in [0, 1] and hence, for $0 < r \le R_{\mathcal{S}^(\alpha)}$ , it follows by (2.36) that Re $$\frac{zf'(z)}{f(z)} \ge \frac{1-3r}{1-r^2} \ge \alpha$$ . For the function $f_3 \in \mathcal{F}_3$ given by (2.31), we have $$\frac{zf_3'(z)}{f_3(z)} = \frac{1-3z}{1-z^2} = \frac{1-3r}{1-r^2} = \alpha, \quad (z = r = R_{\mathcal{S}^*(\alpha)})$$ and this shows that the radius is sharp (See Figure 4.(a)). (ii) It follows from (2.35) that <span id="page-13-0"></span> $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \left| \frac{zf'(z)}{f(z)} - \frac{1}{1 - r^2} \right| + \frac{r^2}{1 - r^2} \le \frac{3r + r^2}{1 - r^2}.$$ (2.37) The number $r = R_{\mathcal{S}_L}$ is the root in [0,1] of $(3r+r^2) = (\sqrt{2}-1)(1-r^2)$ and for $0 < r \le R_{\mathcal{S}_L}$ , we have <span id="page-13-1"></span> $$\frac{3r+r^2}{1-r^2} \le \sqrt{2} - 1. \tag{2.38}$$ Therefore, by (2.37) and (2.38), for $0 < r \le R_{\mathcal{S}_L^*}$ , we have <span id="page-13-2"></span> $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{3r + r^2}{1 - r^2} \le \sqrt{2} - 1. \tag{2.39}$$ For $0 < r \le R_{\mathcal{S}_L^*}$ , using triangle inequality and (2.39), we have <span id="page-13-3"></span> $$\left| \frac{zf'(z)}{f(z)} + 1 \right| \le \sqrt{2} + 1$$ (2.40) and hence by (2.39) and (2.40) $$\left| \left( \frac{zf'(z)}{f(z)} \right)^2 - 1 \right| \le \left| \frac{zf'(z)}{f(z)} + 1 \right| \left| \frac{zf'(z)}{f(z)} - 1 \right| \le (\sqrt{2} + 1)(\sqrt{2} - 1) = 1.$$ The number $\rho = R_{\mathcal{S}_L^*}$ satisfies $(1+3\rho)/(1-\rho^2) = \sqrt{2}$ . Using this, we see that the function $f_3$ defined in (2.31) satisfies $$\left| \left( \frac{z f_3'(z)}{f_3(z)} \right)^2 - 1 \right| = \left| \left( \frac{1 - 3z}{1 - z^2} \right)^2 - 1 \right| = \left| \left( \frac{1 + 3\rho}{1 - \rho^2} \right)^2 - 1 \right| = 1, \quad (z := -\rho = -R_{\mathcal{S}_L^*}).$$ This shows that the radius is sharp. (iii) If $0 < r \le R_{S_p}$ , then $a = 1/(1 - r^2) \le 3/2$ and $$\frac{3r}{1-r^2} \le \frac{1}{1-r^2} - \frac{1}{2}.$$ Thus, by (2.13), we see that the disk in (2.35) lies inside the parabolic region $\Omega_{PAR}$ . Sharpness follows for the function $f_3$ defined in (2.31). At $z := \rho = R_{\mathcal{S}_p}$ , we have Re $$\frac{zf'(z)}{f(z)} = \frac{1-3\rho}{1-\rho^2} = \frac{3\rho-\rho^2}{1-\rho^2} = \left|\frac{\rho^2-3\rho}{1-\rho^2}\right| = \left|\frac{zf'(z)}{f(z)}-1\right|.$$ <span id="page-14-0"></span>   (B) Sharpness of class $\mathcal{S}_c^*$ FIGURE 4. Sharpness of starlike functions associated with lemniscate and cardioid. (iv) For $0 < r \le R_{S_e^*}$ , we have $1/e \le a = 1/(1-r^2) \le (e+e^{-1})/2$ and $$\frac{3r}{1-r^2} \le \frac{1}{1-r^2} - \frac{1}{e}.$$ By (2.14), the disk in (2.35) lies inside $\Omega_e$ for $0 < r \le R_{\mathcal{S}_e}$ and it proves that the $\mathcal{S}_e$ radius for the class $\mathcal{F}_3$ is $R_{\mathcal{S}_e}$ . The sharpness follows for the function $f_3$ defined in (2.31). Indeed at $z := \rho = R_{\mathcal{S}_e}$ , we have $$\left| \log \frac{z f_1'(z)}{f_1(z)} \right| = \left| \log \frac{1 - 3\rho}{1 - \rho^2} \right| = 1.$$ (v) If $0 < r \le R_{\mathcal{S}_c^*}$ , then $$\frac{3r}{1-r^2} \leq \frac{1}{1-r^2} - \frac{1}{3}.$$ By (2.15), we see that the disk in (2.35) lies inside $\Omega_c$ , if $0 < r \le R_{\mathcal{S}_c}$ . The result is sharp for the function $f_3$ defined in (2.31) (See Figure 4.(b)). At $z := \rho = R_{\mathcal{S}_c}$ , $$\left| \frac{zf'(z)}{f(z)} \right| = \left| \frac{1 - 3\rho}{1 - \rho^2} \right| = \frac{1}{3} = \Omega_c(-1) \in \partial \Omega_c(\mathbb{D}).$$ (vi) For $0 < r \le R_{\mathcal{S}_{sin}^*}$ $$\frac{3r}{1-r^2} \le \sin 1 - \frac{r^2}{1-r^2}.$$ It is evident from (2.16) that the disk in (2.35) lies inside $\Omega_s$ . For the function $f_3$ defined in (2.31), at $z := -\rho = -R_{\mathcal{S}_{sin}^*}$ , $$\left| \frac{zf'(z)}{f(z)} \right| = \left| \frac{1+3\rho}{1-\rho^2} \right| = 1 + \sin 1 = q_0(1) \in \partial\Omega_s(\mathbb{D}).$$ (vii) If $0 < r \le R_{\mathcal{S}_{0}^{*}}$ , then $$\frac{3r-1}{1-r^2} \le 1 - \sqrt{2}.$$ Thus, by (2.17), the disk in (2.35) lies inside $\{w \in \mathbb{C} : |w^2 - 1| < 2|w|\}$ and hence $f \in \mathcal{S}^*_{\mathbb{C}}$ . The sharpness follows for the function defined in (2.31) (See Figure 5.(a)). <span id="page-15-0"></span>   (B) Sharpness of class $\mathcal{S}_{RL}^*$ FIGURE 5. Sharpness of starlike functions associated with lune and reverse lemniscate. At $z := \rho = R_{\mathcal{S}_{\mathcal{O}}^*}$ , we have $$\left| \left( \frac{zf_3'(z)}{f_3(z)} \right)^2 - 1 \right| = \left| \left( \frac{1 - 3\rho}{1 - \rho^2} \right)^2 - 1 \right| = 2 \left| \frac{1 - 3\rho}{1 - \rho^2} \right| = 2 \left| \frac{zf_3'(z)}{f_3(z)} \right|.$$ (viii) If $0 < r \le R_{\mathcal{S}_p^*}$ , $2(\sqrt{2} - 1) < a = 1/(1 - r^2) \le \sqrt{2}$ and $$\frac{3r-1}{1-r^2} \le 2 - 2\sqrt{2}, \quad 0 < r \le R_{\mathcal{S}_R^*}.$$ Then, by (2.18), the disk in (2.35) lies inside $\psi(\mathbb{D})$ . The result is sharp for the function defined in (2.31). At $z := \rho = R_{\mathcal{S}_{\mathcal{R}}^*}$ , $$\left| \frac{zf'(z)}{f(z)} \right| = \left| \frac{1 - 3\rho}{1 - \rho^2} \right| = 2(\sqrt{2} - 1) = \psi(1) \in \partial \psi(\mathbb{D}).$$ (ix) If $0 < r \le R_{\mathcal{S}_{RL}^*}$ , $\sqrt{2}/3 \le a = 1/(1-r^2) < \sqrt{2}$ , and $$9r^{2} - (1 - r^{2})\sqrt{(1 - r^{2})^{2} - ((\sqrt{2} - \sqrt{2}r^{2}) - 1)^{2}} + (1 - r^{2})^{2} - ((\sqrt{2} - \sqrt{2}r^{2}) - 1)^{2} \le 0.$$ Then, by (2.19), the disk in (2.35) lies inside the region $\{w : |(w - \sqrt{2})^2 - 1| < 1\}$ . The result is sharp for the function defined in (2.31) (See Figure 5.(b)). (x) If $0 < r \le R_{\mathcal{S}_{\gamma}}$ , $0 < \gamma \le 1$ , then $r \le (\sin(\pi\gamma)/2)/3$ . It is evident from (2.20) that the disk in (2.35) is contained in the sector $|\arg w| \le (\pi\gamma)/2$ , $0 < \gamma \le 1$ if $0 < r \le R_{\mathcal{S}_{\gamma}}$ .

Theorem 2.8. For the class $\mathcal{F}_4$ , the following results hold: - (i) The $S^(\alpha)$ radius $R_{S^(\alpha)} = (1 \alpha)/(2 + \sqrt{3 + \alpha^2}), \quad 0 \le \alpha < 1$ . - (ii) The $S_L$ radius $R_{S_L} = (\sqrt{5} 2)/(1 + \sqrt{2}) \approx 0.9778$ . - (iii) The $S_p$ radius $R_{S_p} = (4 \sqrt{13})/3 \approx 0.1315$ . - (iv) The $S_e$ radius $R_{S_e} = (e-1)/(2e+\sqrt{3e^2+1}) \approx 0.1676$ . - (v) The $S_c$ radius $R_{S_c} = (3 \sqrt{7})/2 \approx 0.1771$ . - (vi) The $S_{sin}$ radius $R_{S_{sin}} = \sin 1/\left(2 + \sqrt{4 + \sin 1(2 + \sin 1)}\right) \approx 0.1858$ . - (vii) The $\mathcal{S}_{\mathbb{Q}}$ radius $R_{\mathcal{S}_{\mathbb{Q}}} = \sqrt{2} \sqrt{(3 \sqrt{2})} \approx 0.1549$ . - (viii) The $S_R$ radius $R_{S_R} = (3 2\sqrt{2}) / (2 + \sqrt{15 8\sqrt{2}}) \approx 0.0438$ . - (ix) The $\mathcal{S}_{RL}$ radius $R_{\mathcal{S}_{RL}}$ is the root ( $\approx 0.0694$ ) in [0,1] of the equation $$16r^{2} + (r^{2} - 1)^{2} - (1 - \sqrt{2} + (1 + \sqrt{2})r^{2})^{2} + (r^{2} - 1)\sqrt{2\sqrt{2} - 2 - 2(1 + \sqrt{2})r^{4}} = 0.$$ (x) The $$S_{\gamma}^*$$ radius $R_{S_{\gamma}^*} \ge (\sin(\pi\gamma/2))/\left(2 + \sqrt{4 - \sin^2(\pi\gamma/2)}\right)$ , $0 < \gamma \le 1$ . PROOF. Let the function $f \in \mathcal{F}_4$ . Then <span id="page-16-0"></span> $$\operatorname{Re}\left(\frac{(1+z)^2}{z}f(z)\right) > 0 \quad (z \in \mathbb{D}).$$ (2.42) Define the function $h: \mathbb{D} \longrightarrow \mathbb{C}$ by <span id="page-16-1"></span> $$h(z) = \frac{(1+z)^2}{z} f(z). \tag{2.43}$$ By (2.42) and (2.43) we have $h \in \mathcal{P}$ and $f(z) = zh(z)/(1+z)^2$ . Therefore, by calculation it can be shown that <span id="page-16-3"></span> $$\frac{zf'(z)}{f(z)} = \frac{zh'(z)}{h(z)} + \frac{1-z}{1+z}. (2.44)$$ The bilinear transformation (1-z)/(1+z) maps the disk $|z| \le r$ onto the disk <span id="page-16-2"></span> $$\left| \frac{1-z}{1+z} - \frac{1+r^2}{1-r^2} \right| \le \frac{2r}{1-r^2}.$$ (2.45) Using (2.45) and (2.6), it follows from (2.44) that f maps the disk $|z| \le r$ onto the disk <span id="page-17-0"></span> $$\left| \frac{zf'(z)}{f(z)} - \frac{1+r^2}{1-r^2} \right| \le \frac{4r}{1-r^2}.$$ (2.46) The classes we discuss here are all subclasses of starlike functions. By (2.46), we have <span id="page-17-1"></span>Re $$\frac{zf'(z)}{f(z)} \ge \frac{1 - 4r + r^2}{1 - r^2} \ge 0$$ , $(r \le 2 - \sqrt{3})$ . (2.47) Hence all the radii that we estimate will be less than $2 - \sqrt{3} \approx 0.26794$ . Note that for $0 < r \le (2 - \sqrt{3})$ , the centre of disk in (2.46) lies in the interval $[1, (4 - 2\sqrt{3})/(2\sqrt{3} - 3)] \approx [1, 1.1547]$ . (i) The number $r = R_{S^(\alpha)}$ is the root of $(1+\alpha)r^2 - 4r + 1 - \alpha = 0$ in [0,1] and hence, for $0 < r \le R_{S^(\alpha)}$ , it follows by (2.47) that Re $$\frac{zf'(z)}{f(z)} \ge \frac{1 - 4r + r^2}{1 - r^2} \ge \alpha$$ . For the function $f_4 \in \mathcal{F}_4$ given by (2.41) we have $$\frac{zf_4'(z)}{f_4(z)} = \frac{1 - 4z + z^2}{1 - z^2} = \frac{1 - 4r + r^2}{1 - r^2} = \alpha, \quad (z = r = R_{\mathcal{S}^*(\alpha)})$$ and this shows that the radius is sharp. (ii) It follows from (2.46) that <span id="page-17-2"></span> $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \left| \frac{zf'(z)}{f(z)} - \frac{1+r^2}{1-r^2} \right| + \frac{2r^2}{1-r^2} \le \frac{4r+2r^2}{1-r^2}. \tag{2.48}$$ The number $r = R_{\mathcal{S}_L}$ is the root in [0,1] of $(4r+2r^2) \leq (\sqrt{2}-1)(1-r^2)$ and for $0 < r \leq R_{\mathcal{S}_L}$ , we have <span id="page-17-3"></span> $$\frac{4r + 2r^2}{1 - r^2} \le \sqrt{2} - 1. \tag{2.49}$$ Therefore, by (2.48) and (2.49), for $0 < r \le R_{\mathcal{S}_r^*}$ , we have <span id="page-17-4"></span> $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{4r + 2r^2}{1 - r^2} \le \sqrt{2} - 1. \tag{2.50}$$ For $0 < r \le R_{\mathcal{S}_L^*}$ , using triangle inequality and (2.50), we have <span id="page-17-5"></span> $$\left| \frac{zf'(z)}{f(z)} + 1 \right| \le \sqrt{2} + 1$$ (2.51) and hence by (2.50) and (2.51) $$\left| \left( \frac{zf'(z)}{f(z)} \right)^2 - 1 \right| \le \left| \frac{zf'(z)}{f(z)} + 1 \right| \left| \frac{zf'(z)}{f(z)} - 1 \right| \le (\sqrt{2} + 1)(\sqrt{2} - 1) = 1.$$ The number $\rho = R_{\mathcal{S}_L}$ satisfies $(1 + 4\rho - \rho^2)/(1 - \rho^2) = \sqrt{2}$ . Using this, we see that at $z := -\rho = -R_{\mathcal{S}_L}$ , the function $f_4$ defined in (2.41) satisfies $$\left| \left( \frac{z f_4'(z)}{f_4(z)} \right)^2 - 1 \right| = \left| \left( \frac{1 - 4z + z^2}{1 - z^2} \right)^2 - 1 \right| = \left| \left( \frac{1 + 4\rho + \rho^2}{1 - \rho^2} \right)^2 - 1 \right| = 1.$$ <span id="page-18-0"></span>   (B) Sharpness of class $S_{sin}^*$ FIGURE 6. Sharpness of parabolic starlike functions and starlike functions associated with sine function This shows that the radius is sharp. (iii) If $0 < r \le R_{S_p}$ , then $a = 1/(1 - r^2) \le 3/2$ and $$\frac{4r}{1-r^2} \le \frac{1+r^2}{1-r^2} - \frac{1}{2}.$$ Thus, by (2.13), we see that the disk in (2.46) lies inside the parabolic region $\Omega_{PAR}$ . Sharpness follows for the function $f_4$ defined in (2.41) (See Figure 6.(a)). At $z := \rho = R_{S_p}$ , we have $$\operatorname{Re} \frac{zf'(z)}{f(z)} = \frac{1 - 4\rho + \rho^2}{1 - \rho^2} = \frac{4\rho - 2\rho^2}{1 - \rho^2} = \left| \frac{2\rho^2 - 4\rho}{1 - \rho^2} \right| = \left| \frac{zf'(z)}{f(z)} - 1 \right|.$$ (iv) For $0 < r \le R_{\mathcal{S}_e^*}$ , we have $1/e \le a = 1/(1-r^2) \le (e+e^{-1})/2$ and $$\frac{4r}{1-r^2} \le \frac{1+r^2}{1-r^2} - \frac{1}{e}.$$ By (2.14), the disk in (2.46) lies inside $\Omega_e$ for $0 < r \le R_{\mathcal{S}_e}$ proving that the $\mathcal{S}_e$ radius for the class $\mathcal{F}_4$ is $R_{\mathcal{S}_e}$ . The sharpness follows for the function $f_4$ defined in (2.41). Indeed at $z := \rho = R_{\mathcal{S}_e}$ , we have $$\left| \log \frac{z f_4'(z)}{f_4(z)} \right| = \left| \log \frac{1 - 4\rho + \rho^2}{1 - \rho^2} \right| = 1.$$ (v) If $0 < r \le R_{\mathcal{S}_c^*}$ , then $$\frac{4r}{1-r^2} \le \frac{1+r^2}{1-r^2} - \frac{1}{3}.$$ By (2.15), we see that the disk in (2.46) lies inside $\Omega_c$ , if $0 < r \le R_{\mathcal{S}_c}$ . The result is sharp for the function $f_4$ defined in (2.41). At $z := \rho = R_{\mathcal{S}_c}$ , $$\left| \frac{zf'(z)}{f(z)} \right| = \left| \frac{1 - 4\rho + \rho^2}{1 - \rho^2} \right| = \frac{1}{3} = \Omega_c(-1) \in \partial\Omega_c(\mathbb{D}).$$ (vi) For $0 < r \le R_{\mathcal{S}_{sin}^*}$ $$\frac{4r}{1-r^2} \le \sin 1 - \frac{2r^2}{1-r^2}.$$ <span id="page-19-0"></span> FIGURE 7. Sharpness of starlike functions associated with a rational function and reverse lemniscate. It is evident from (2.16) that the disk in (2.46) lies inside $\Omega_s$ provided $0 < r \le R_{\mathcal{S}_{sin}}$ . For the function $f_4$ defined in (2.41) (See Figure 6.(b)), at $z := -\rho = -R_{\mathcal{S}_{sin}}$ , $$\left|\frac{zf'(z)}{f(z)}\right| = \left|\frac{1+4\rho+\rho^2}{1-\rho^2}\right| = 1 + \sin 1 = q_0(1) \in \partial\Omega_s(\mathbb{D}).$$ (vii) If $0 < r \le R_{\mathcal{S}^*_{\mathcal{O}}}$ , then $$\frac{4r - 1 - r^2}{1 - r^2} \le 1 - \sqrt{2}.$$ Thus, by (2.17), the disk in (2.46) lies inside $\{w \in \mathbb{C} : |w^2 - 1| < 2|w|\}$ and hence $f \in \mathcal{S}^_{\mathbb{C}}$ . The sharpness follows from the function defined in (2.41). At $z := \rho = R_{\mathcal{S}^_{\mathbb{C}}}$ , we have $$\left| \left( \frac{z f_4'(z)}{f_4(z)} \right)^2 - 1 \right| = \left| \left( \frac{\rho^2 - 4\rho + 1}{1 - \rho^2} \right)^2 - 1 \right| = 2 \left| \left( \frac{\rho^2 - 4\rho + 1}{1 - \rho^2} \right) \right| = 2 \left| \left( \frac{z f_4'(z)}{f_4(z)} \right) \right|.$$ (viii) If $$0 < r \le R_{\mathcal{S}_R^*}, \, 2(\sqrt{2}-1) < a = 1/(1-r^2) \le \sqrt{2}$$ and $$\frac{4r - 1 - r^2}{1 - r^2} \le 2 - 2\sqrt{2}, \quad 0 < r \le R_{\mathcal{S}_R^*}.$$ Then, by (2.18), the disk in (2.46) lies inside $\psi(\mathbb{D})$ . The result is sharp for the function defined in (2.41) (See Figure 7.(a)). At $z := \rho = R_{\mathcal{S}_R^*}$ , $$\left| \frac{zf'(z)}{f(z)} \right| = \left| \frac{1 - 4\rho + \rho^2}{1 - \rho^2} \right| = 2(\sqrt{2} - 1) = \psi(1) \in \partial \psi(\mathbb{D}).$$ (ix) If $0 < r \le R_{\mathcal{S}_{RL}^*}$ , $\sqrt{2}/3 \le a = 1/(1 - r^2) < \sqrt{2}$ , and $$16r^{2} - (1 - r^{2})\sqrt{(1 - r^{2})^{2} - ((\sqrt{2} - \sqrt{2}r^{2}) - (1 + r^{2})^{2}} + (1 - r^{2})^{2} - ((\sqrt{2} - \sqrt{2}r^{2}) - (1 + r^{2})^{2} \le 0.$$ Then, by (2.19), the disk in (2.46) lies inside the region $\{w : |(w - \sqrt{2})^2 - 1| < 1\}$ . The result is sharp for the function defined in (2.41) (See Figure 7.(b)). (x) If $0 < r \le R_{\mathcal{S}^_{\gamma}}$ , $0 < \gamma \le 1$ , then $r^2 \sin(\pi \gamma/2) - 4r + \sin(\pi \gamma)/2 \le 0$ . It is evident from (2.20) that the disk (2.46) is contained in the sector $|\arg w| \le (\pi \gamma)/2$ , if $0 < r \le R_{\mathcal{S}^_{\gamma}}$ .

Definition 2.3. Let $\mathcal{F}_2$ be the class of functions $f \in \mathcal{A}$ satisfying the inequality $$\left| \frac{f(z)}{g(z)} - 1 \right| < 1 \quad (z \in \mathbb{D})$$ for some $g \in \mathcal{A}$ with $$\operatorname{Re}\left(\frac{1+z}{z}g(z)\right) > 0 \quad (z \in \mathbb{D}).$$ The functions $f_2, g_2 : \mathbb{D} \longrightarrow \mathbb{C}$ defined by <span id="page-7-2"></span> $$f_2(z) = \frac{z(1-z)^2}{(1+z)^2}$$ and $g_2(z) = \frac{z(1-z)}{(1+z)^2}$ (2.21) satisfy $$\left| \frac{f_2(z)}{g_2(z)} - 1 \right| = |z| < 1, \quad \operatorname{Re} \frac{(1+z)}{z} g_2(z) = \operatorname{Re} \frac{1-z}{1+z} > 0$$ and hence the function $f_2 \in \mathcal{F}_2$ . This proves that the class $\mathcal{F}_2$ is non-empty and the function $f_2$ is extremal function for the radius problems we consider. As $$f_2(z) = z - 4z^2 + 8z^3 - 12z^4 + \dots,$$ the functions in $\mathcal{F}_2$ are not necessarily univalent. Since $$f_2'(z) = \frac{1 - 5z + 3z^2 + z^3}{(1+z)^3},$$ we have $f_2'(\sqrt{5}-2)=0$ and it follows by the first part of the following theorem, that the radius of univalence of the functions in class $\mathcal{F}_2$ is $\sqrt{5} - 2 \approx 0.2361$ .

Definition 2.5. Let $\mathcal{F}_3$ be the class of functions $f \in \mathcal{A}$ satisfying the inequality $$\operatorname{Re}\left(\frac{1+z}{z}f(z)\right) > 0 \quad (z \in \mathbb{D}).$$ The function $f_3: \mathbb{D} \longrightarrow \mathbb{C}$ defined by <span id="page-11-1"></span> $$f_3(z) = \frac{z(1-z)}{(1+z)^2} \tag{2.31}$$ satisfy Re $$\frac{(1+z)f_3(z)}{z}$$ = Re $\frac{1-z}{1+z} > 0$ and hence the function $f_3 \in \mathcal{F}_3$ . This proves that the class $\mathcal{F}_3$ is non-empty. This function $f_3$ is extremal function for the radii problem we consider. As $$f_3(z) = z - 3z^2 + 5z^3 - 7z^4 + \dots,$$ the functions in $\mathcal{F}_3$ are not necessarily univalent. Since $$f_3'(z) = \frac{1 - 3z}{(1+z)^3},$$ we have $f_3'(1/3) = 0$ and it follows, by the first part of the following theorem, that the radius of univalence of the functions in class $\mathcal{F}_3$ is $1/3 \approx 0.3333$ .

Definition 2.7. Let $\mathcal{F}_4$ be the class of functions $f \in \mathcal{A}$ satisfying the inequality $$\operatorname{Re}\left(\frac{(1+z)^2}{z}f(z)\right) > 0 \quad (z \in \mathbb{D}).$$ The functions $f_4: \mathbb{D} \longrightarrow \mathbb{C}$ defined by <span id="page-15-1"></span> $$f_4(z) = \frac{z(1-z)}{(1+z)^3}. (2.41)$$ satisfy $$\operatorname{Re} \frac{(1+z)}{z} f_4(z) = \operatorname{Re} \frac{1-z}{1+z} > 0$$ and hence the function $f_4 \in \mathcal{F}_4$ . This proves that the class $\mathcal{F}_4$ is non-empty. Also, the function $f_4$ is extremal function for the radii problem we consider. Since $$f_4(z) = z - 4z^2 + 9z^3 - 16z^4 + \dots,$$ the functions in $\mathcal{F}_4$ are not necessarily univalent. Since $$f_4'(z) = \frac{1 - 4z + z^2}{(1+z)^4},$$ we have $f_4'(2-\sqrt{3})=0$ and it follows by the first part of the following theorem, that the radius of univalence of the functions in class $\mathcal{F}_4$ is $2-\sqrt{3}\approx 0.276949$ .