Radius of Starlikeness for Bloch Functions

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

$$R^2 + 3\sqrt{3}R - 3 = 0.$$

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

$$\frac{zf'(z)}{f(z)} = \frac{1}{3}$$
$$= \phi_c(-1).$$

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

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  $2(\sqrt{2} - 1) < c < 2$ ,

$$r_c = \begin{cases} c - 2(\sqrt{2} - 1) & \text{if } 2(\sqrt{2} - 1) < c \le \sqrt{2} \\ 2 - c & \text{if } \sqrt{2} \le c < 2 \end{cases}$$
 (2.10)

then  $\{w : |w - c| < r_c\} \subseteq \psi(\mathbb{D})$