Radius of Starlikeness for Bloch Functions

Theorem 2.1. The $S_e^*$ -radius of the class $\mathcal{B}$ of Bloch functions is $$\mathcal{R}_{S_e^*}(\mathcal{B}) = \frac{1}{4}\sqrt{3}\left(3 - 3e + \sqrt{1 - 10e + 9e^2}\right) \approx 0.517387.$$ The obtained radius is sharp. PROOF. For functions $f \in \mathcal{B}$ , Bonk [1] proved the following inequality <span id="page-1-0"></span> $$\left| \frac{zf'(z)}{f(z)} - \frac{\sqrt{3}}{\sqrt{3} - r} \right| \le \frac{\sqrt{3}r}{(\sqrt{3} - r)(\sqrt{3} - 2r)}, \quad |z| = r < \frac{1}{\sqrt{3}}.$$ (2.1) The function $$h(r) := \frac{\sqrt{3}}{\sqrt{3} - r} - \frac{\sqrt{3}r}{(\sqrt{3} - r)(\sqrt{3} - 2r)} = \frac{3 - 3\sqrt{3}r}{(\sqrt{3} - r)(\sqrt{3} - 2r)}$$ is a decreasing function of r for $0 \le r < 1/\sqrt{3} = \mathcal{R}_{S}(\mathcal{B})$ . The number $R = \mathcal{R}_{S_e}(\mathcal{B}) < 1/\sqrt{3} = \mathcal{R}_{S^*}(\mathcal{B})$ is the smallest positive root of the polynomial <span id="page-1-3"></span> $$2R^{2} + 3\sqrt{3}(e-1)R + 3(1-e) = 0$$ (2.2) or h(R) = 1/e. Therefore, for $0 \le r < R$ , it follows that 1/e = h(R) < h(r) and hence <span id="page-1-2"></span><span id="page-1-1"></span> $$\frac{\sqrt{3}r}{(\sqrt{3}-r)(\sqrt{3}-2r)} < \frac{\sqrt{3}}{\sqrt{3}-r} - \frac{1}{e}.$$ (2.3) Thus (2.1) and (2.3) give $$\left| \frac{zf'(z)}{f(z)} - \frac{\sqrt{3}}{\sqrt{3} - r} \right| < \frac{\sqrt{3}}{\sqrt{3} - r} - \frac{1}{e}, \quad |z| = r < R. \tag{2.4}$$ The function $C(r) = \sqrt{3}/(\sqrt{3} - r)$ is an increasing function of r, so for $r \in [0, R)$ , it follows that $C(r) \in [1, C(R)) \subseteq [1, C(0.6)) \approx [1, 1.53001) \subseteq (0.367879, 1.54308) \approx (1/e, (e + e^{-1})/2)$ . By [2, Lemma 2.2], for 1/e < c < e, we have $\{w : |w - c| < r_c\} \subseteq \{w : |\log(w)| < 1\}$ when $r_c$ is given by $$r_c = \begin{cases} c - e^{-1} & \text{if } e^{-1} < c \le \frac{e + e^{-1}}{2}, \\ e - c & \text{if } \frac{e + e^{-1}}{2} \le c < e. \end{cases}$$ (2.5) By (2.4), we see that w = zf'(z)/f(z), |z| < R, satisfies $|w - c| < c - e^{-1}$ and hence it follows that $|\log(w)| < 1$ . This shows that $S_e^*$ -radius of the class $\mathcal{B}$ is at least R. We now show that R is the exact $S_e^*$ -radius of the class $\mathcal{B}$ . The function $f_0: \mathbb{D} \to \mathbb{C}$ defined by $$f(z) = \frac{\sqrt{3}}{4} \left\{ 1 - 3 \left( \frac{z - \sqrt{1/3}}{1 - \sqrt{1/3}z} \right)^2 \right\} = \frac{3z(3 - 2\sqrt{3}z)}{(3 - \sqrt{3}z)^2}$$ is an example of function in the class $\mathcal{B}$ and it serves as an extremal function for the various problems. For this function, we have $$\frac{zf'(z)}{f(z)} = \frac{3\sqrt{3} - 9z}{2\sqrt{3}z^2 - 9z + 3\sqrt{3}}$$ Using the equation (2.2), we get $2\sqrt{3}R^2 - 9R + 3\sqrt{3} = e(3\sqrt{3} - 9R)$ , thus, for z = R $$\left|\log\left(\frac{zf'(z)}{f(z)}\right)\right| = \left|\log\left(\frac{3\sqrt{3} - 9z}{2\sqrt{3}z^2 - 9z + 3\sqrt{3}}\right)\right| = \left|\log\left(\frac{1}{e}\right)\right| = 1.$$ This proves that R is the exact $S_e^*$ -radius of the class $\mathcal{B}$ . Sharma et al. studied the class $S_c^ = S^(\phi_c) = S^*(1 + (4/3)z + (2/3)z^2)$ and gave [3, Lemma 2.5] For 1/3 < c < 3, $$r_c = \begin{cases} \frac{3c-1}{3} & \text{if } \frac{1}{3} < c \le \frac{5}{3} \\ 3-c & \text{if } \frac{5}{3} \le c < 3 \end{cases}$$ (2.6) then $\{w : |w-c| < r_c\} \subseteq \Omega_c$ . Here $\Omega_c$ is the region bounded by the cadioid $\{x + \iota y : (9x^2 + 9y^2 - 18x + 5)^2 - 16(9x^2 + 9y^2 - 6x + 1) = 0\}$ .

Theorem 2.2. The $S_c$ -radius $\mathcal{R}_{S_c} \approx 0.524423$ . This radius is sharp. PROOF. $R = \mathcal{R}_{S_c^*}$ is the smallest positive root of the equation <span id="page-2-1"></span><span id="page-2-0"></span> $$R^2 + 3\sqrt{3}R - 3 = 0.$$ The function $$h(r) := \frac{\sqrt{3}}{\sqrt{3} - r} - \frac{\sqrt{3}r}{(\sqrt{3} - r)(\sqrt{3} - 2r)} = \frac{3 - 3\sqrt{3}r}{(\sqrt{3} - r)(\sqrt{3} - 2r)}$$ is a decreasing function of r for $0 \le r < 1/\sqrt{3} = \mathcal{R}_{S}$ . [1, Corollary, P.455] Note that the class $\mathcal{S}_c$ is a subclass of the parabolic starlike class $S$ , Also since, $R = \mathcal{R}_{S_c}$ is the smallest positive root of the equation h(r) = 1/3. For $0 \le r < R$ , we have $$\frac{\sqrt{3}r}{(\sqrt{3}-r)(\sqrt{3}-2r)} < \frac{\sqrt{3}}{\sqrt{3}-r} - \frac{1}{3}$$ (2.7) Thus (2.1) and (2.7) give $$\left| \frac{zf'(z)}{f(z)} - \frac{1}{1 - ar} \right| < \frac{1}{1 - ar} - \frac{1}{3}; \ |z| \le r, \ a = \frac{1}{\sqrt{3}}.$$ The center C(r) of (2.1) is an increasing function of r, so for $r \in [0, R)$ , $C(r) \in [1, C(R)) \subseteq [1, c(0.6)) \approx [1, 1.53001) \subseteq (1/3, 5/3)$ . Now, by (2.6) we get that the disc $\{w : |w - c| < 0.6\}$ $$c-1/3\} \subset \Omega_c$$ . For proving sharpness, consider the function $$f(z) = \frac{\sqrt{3}}{4} \left\{ 1 - 3 \left( \frac{z - \sqrt{1/3}}{1 - \sqrt{1/3}z} \right)^2 \right\}$$ for this function, $\frac{zf'(z)}{f(z)} = \frac{3\sqrt{3} - 9z}{2\sqrt{3}z^2 - 9z + 3\sqrt{3}}$ , and using the equation for R, we get $2\sqrt{3}r^2 - 9r + 3\sqrt{3} = 3(3\sqrt{3} - 9r)$ , thus for z = R <span id="page-3-1"></span> $$\frac{zf'(z)}{f(z)} = \frac{1}{3}$$ $$= \phi_c(-1).$$ The class $\mathcal{S}^_{\mathbb{Q}} = \mathcal{S}^(z + \sqrt{1+z^2})$ was introduced in 2015 by Rain and Sokól [4] in 2015 and proved that $f \in \mathcal{S}^*_{\mathbb{Q}} \iff zf'(z)/f(z)$ lies in the lune region $\{w : |w^2 - 1| < 2|w|\}$ . Gandhi and Ravichandran [5, Lemma 2.1] proved that for $\sqrt{2} - 1 < c \le \sqrt{2} + 1$ , $$\{w: |w-c| < 1 - |\sqrt{2} - c|\} \subseteq \{w: |w^2 - 1| < 2|w|\}$$ (2.8)

Theorem 2.3. The $\mathcal{S}_{\mathcal{A}}$ radius, $\mathcal{R}_{\mathcal{S}_{\mathcal{A}}} \approx 0.507306$ . The radius is sharp. PROOF. For $R = \mathcal{R}_{\mathcal{S}_{\mathbb{Q}}^*}$ , the center of (2.1), $C(R) = \sqrt{2}$ ; since C(r) is an increasing function of r, thus for $0 \le r < R$ , $1 \le C(r) < \sqrt{2}$ , or <span id="page-3-0"></span>for $$0 \le r < R$$ , $\sqrt{2} - C(r) \ge 0$ . So, $R = \mathcal{R}_{\mathcal{S}_{\mathcal{I}}^*}$ is the smallest positive root of the equation $$(2 - 2\sqrt{2})R^2 + \sqrt{3}(3\sqrt{2} - 6)R + 3(2 - \sqrt{2}) = 0.$$ The function $$h(r) := \frac{\sqrt{3}}{\sqrt{3} - r} - \frac{\sqrt{3}r}{(\sqrt{3} - r)(\sqrt{3} - 2r)} = \frac{3 - 3\sqrt{3}r}{(\sqrt{3} - r)(\sqrt{3} - 2r)}$$ is a decreasing funciton of r for $0 \le r < 1/\sqrt{3} = \mathcal{R}_{S}$ . [1, Corollary, P.455] Note that the class $\mathcal{S}^_{\mathbb{Q}}$ is a subclass of the parabolic starlike class $S$ . Also since, $R = \mathcal{R}_{\mathcal{S}^_{\mathbb{Q}}}$ is the smallest positive root of the equation $h(r) = \sqrt{2} - 1$ . For $0 \le r < R$ , we have $$\frac{\sqrt{3}r}{(\sqrt{3}-r)(\sqrt{3}-2r)} < 1 - \sqrt{2} + \frac{\sqrt{3}}{\sqrt{3}-r} = 1 - \left|\sqrt{2} - \frac{\sqrt{3}}{\sqrt{3}-r}\right|. \tag{2.9}$$ Thus (2.1) and (2.9) give $$\left| \frac{zf'(z)}{f(z)} - \frac{1}{1 - ar} \right| < 1 - \left| \sqrt{2} - \frac{1}{1 - ar} \right|; \ |z| \le r, \ a = \frac{1}{\sqrt{3}}.$$ The center C(r) of (2.1) is an increasing function of r, so for $r \in [0, R)$ , $C(r) \in [1, C(R)) \subseteq [1, c(0.6)) \approx [1, 1.53001) \subseteq (\sqrt{2} - 1, \sqrt{2} + 1)$ . Now, by (2.8) we get that the R is the required radius. Consider the $$f(z) = \frac{\sqrt{3}}{4} \left\{ 1 - 3 \left( \frac{z - \sqrt{1/3}}{1 - \sqrt{1/3}z} \right)^2 \right\}$$ for this function, $\frac{zf'(z)}{f(z)} = \frac{3\sqrt{3} - 9z}{2\sqrt{3}z^2 - 9z + 3\sqrt{3}}$ and we can easily see that for $z = \frac{1}{2}[2\sqrt{3} - \sqrt{6}],$ $$\left| \left( \frac{zf'(z)}{f(z)} \right)^2 - 1 \right| = 2 \left( \frac{zf'(z)}{f(z)} \right) = 2(\sqrt{2} - 1).$$ Thus, the result is sharp. The next class that we consider is the class of starlike functions associated with a rational function. Kumar and Ravichandran [6] introduced the class of starlike functions associated with the rational function $\psi(z) = 1 + ((z^2 + kz)/(k^2 - kz))$ where $k = \sqrt{2} + 1$ , denoted by $\mathcal{S}_R^ = \mathcal{S}^(\psi(z))$ They proved[6, Lemma 2.2] that for $2(\sqrt{2} - 1) < c < 2$ , <span id="page-4-1"></span> $$r_c = \begin{cases} c - 2(\sqrt{2} - 1) & \text{if } 2(\sqrt{2} - 1) < c \le \sqrt{2} \\ 2 - c & \text{if } \sqrt{2} \le c < 2 \end{cases}$$ (2.10) then $\{w : |w - c| < r_c\} \subseteq \psi(\mathbb{D})$

Theorem 2.4. The $S_R$ radius is the smallest positive root of the polynomial $4(1 - \sqrt{2})r^2 + 3\sqrt{3}(2\sqrt{2} - 3)r + 3(3 - 2\sqrt{2})$ that is $\mathcal{R}_{S_R} \approx 0.349865$ . The result is sharp. PROOF. $R = \mathcal{S}_R^*$ is the smallest positive root of the equation $$4(1 - \sqrt{2})R^2 + 3\sqrt{3}(2\sqrt{2} - 3)R + 3(3 - 2\sqrt{2}) = 0.$$ The function $$h(r) := \frac{\sqrt{3}}{\sqrt{3} - r} - \frac{\sqrt{3}r}{(\sqrt{3} - r)(\sqrt{3} - 2r)} = \frac{3 - 3\sqrt{3}r}{(\sqrt{3} - r)(\sqrt{3} - 2r)}$$ is a decreasing function of r for $0 \le r < 1/\sqrt{3} = \mathcal{R}_{S}$ . [1, Corollary, P.455] Note that the class $\mathcal{S}_R$ is a subclass of the parabolic starlike class $S$ . Also since, $R = \mathcal{S}_R$ is the smallest positive root of the equation $h(r) = 2(\sqrt{2} - 1)$ . For $0 \le r < R$ , we have <span id="page-4-0"></span> $$\frac{\sqrt{3}r}{(\sqrt{3}-r)(\sqrt{3}-2r)} < \frac{\sqrt{3}}{\sqrt{3}-r} - 2(\sqrt{2}-1) \tag{2.11}$$ Thus (2.1) and (2.11) give $$\left| \frac{zf'(z)}{f(z)} - \frac{1}{1 - ar} \right| < \frac{1}{1 - ar} - 2(\sqrt{2} - 1); \ |z| \le r, \ a = \frac{1}{\sqrt{3}}.$$ The center C(r) of (2.1) is an increasing function of r, so for $r \in [0, R)$ , $C(r) \in [1, C(R)) \subseteq [1, c(0.4)) \approx [1, 1.30029) \subseteq (2(\sqrt{2} - 1), \sqrt{2})$ . Now, by (2.10) we get that the disc $\{w : |w - c| < x - 2(\sqrt{2} - 1)\} \subseteq \psi(\mathbb{D})$ . To show that the result is sharp, consider the function $$f(z) = \frac{\sqrt{3}}{4} \left\{ 1 - 3 \left( \frac{z - \sqrt{1/3}}{1 - \sqrt{1/3}z} \right)^2 \right\}$$ for this function, $\frac{zf'(z)}{f(z)} = \frac{3\sqrt{3} - 9z}{2\sqrt{3}z^2 - 9z + 3\sqrt{3}}$ , and using the equation for R we get $3\sqrt{3} - 9r = (2\sqrt{2} - 2)(2\sqrt{3}r^2 - 9r + 3\sqrt{3})$ thus for z = R $$\frac{zf'(z)}{f(z)} = 2\sqrt{2} - 2$$ $$= \psi(-1).$$

Theorem 2.5. For the class $\mathcal{B}$ the following results hold: - (1) The Lemniscate starlike radius, $R_{S_r^*} = \frac{2\sqrt{3}-\sqrt{6}}{4} \approx 0.253653$ . - (2) The starlike radius associated with the sine function, $R_{S_{sin}^*} = \frac{\sqrt{3}\sin 1}{2+2\sin 1} \approx 0.395735$ . - (3) The nephroid radius, $R_{\mathcal{S}_{Ne}^*} = \frac{\sqrt{3}}{5} \approx 0.34641$ . - (4) The sigmoid radius, $R_{\mathcal{S}_{SG}^*} = \frac{\sqrt{3(e-1)}}{4e} \approx 0.273716$ .