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

$$\operatorname{Re} \frac{f(z)}{g(z)} > 0 \quad \text{and} \quad \operatorname{Re} \left(\frac{1+z}{z}g(z)\right) > 0 \quad (z \in \mathbb{D}).$$
 (2.2)

Define the functions  $p_1, p_2 : \mathbb{D} \longrightarrow \mathbb{C}$  by

$$p_1(z) = \frac{1+z}{z}g(z)$$
 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

$$\frac{zf'(z)}{f(z)} = \frac{zp_1'(z)}{p_1(z)} + \frac{zp_2'(z)}{p_2(z)} + \frac{1}{1+z}.$$
 (2.4)

The bilinear transformation 1/(1+z) maps the disk  $|z| \leq r$  onto the disk

$$\left| \frac{1}{1+z} - \frac{1}{1-r^2} \right| \le \frac{r}{1-r^2}. \tag{2.5}$$

For  $p \in \mathcal{P}(\alpha) := \{ p \in \mathcal{P} | \operatorname{Re} p > \alpha \}$ , by [41, Lemma 2], we have

$$\left| \frac{zp'(z)}{p(z)} \right| \le \frac{2(1-\alpha)r}{(1-r)(1+(1-2\alpha)r)} \quad (|z| \le r). \tag{2.6}$$

Using (2.5) and (2.6), it follows from (2.4) that the function f maps the disk  $|z| \leq r$  onto the disk

$$\left| \frac{zf'(z)}{f(z)} - \frac{1}{1 - r^2} \right| \le \frac{5r}{1 - r^2}.$$
 (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

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

$$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \left| \frac{zf'(z)}{f(z)} - \frac{1}{1 - r^2} \right| + \frac{r^2}{1 - r^2} \le \frac{5r + r^2}{1 - r^2}.$$
 (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

$$\frac{5r+r^2}{1-r^2} \le \sqrt{2}-1. \tag{2.10}$$

Therefore, by (2.9) and (2.10), for  $0 < r \le R_{\mathcal{S}_{t}^{*}}$ , we have

$$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{5r + r^2}{1 - r^2} \le \sqrt{2} - 1. \tag{2.11}$$

For  $0 < r \le R_{\mathcal{S}_{L}^{*}}$ , using triangle inequality and (2.11), we have

$$\left| \frac{zf'(z)}{f(z)} + 1 \right| \le 2 + \left| \frac{zf'(z)}{f(z)} - 1 \right| \le \sqrt{2} + 1$$
 (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

$$\{w \in \mathbb{C} : |w - a| < a - 1/2\} \subseteq \Omega_{PAR}. \tag{2.13}$$

If  $0 < r \le R_{S_p}$ , then  $a = 1/(1 - r^2) \le 3/2$  and

$$\frac{5r}{1-r^2} \le \frac{1}{1-r^2} - \frac{1}{2}.$$

![](_page_5_Figure_2.jpeg)

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

$$\{w \in \mathbb{C} : |w - a| < a - e^{-1}\} \subseteq \{w \in \mathbb{C} : |\log w| < 1\} =: \Omega_e.$$
 (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

$$\{w \in \mathbb{C} : |w - a| < (3a - 1)/3\} \subseteq \Omega_c \tag{2.15}$$

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

$$\{w \in \mathbb{C} : |w - a| < \sin 1 - |a - 1|\} \subseteq \Omega_s \tag{2.16}$$

![](_page_6_Figure_2.jpeg)

![](_page_6_Figure_3.jpeg)

![](_page_6_Figure_4.jpeg)

(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

$$\{w \in \mathbb{C} : |w - a| < 1 - |\sqrt{2} - a|\} \subseteq \{w \in \mathbb{C} : |w^2 - 1| < 2|w|\}$$
 (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],

$$\{w \in \mathbb{C} : |w - a| < a - 2(\sqrt{2} - 1)\} \subseteq \psi(\mathbb{D}) \tag{2.18}$$

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

$$\{w \in \mathbb{C} : |w - a| < r_{RL}\} \subseteq \{w \in \mathbb{C} : |(w - \sqrt{2})^2 - 1| < 1\},$$
 (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

$$\{w \in \mathbb{C} : |w - a| < a\sin(\pi\gamma/2)\} \subseteq \{w : |\arg w| \le (\pi\gamma)/2\}, \quad 0 < \gamma \le 1.$$
 (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

$$\operatorname{Re} \frac{g(z)}{f(z)} > 1/2 \quad \text{and} \quad \operatorname{Re} \left(\frac{1+z}{z}g(z)\right) > 0 \quad (z \in \mathbb{D}).$$
 (2.22)

Define the functions  $p_1, p_2 : \mathbb{D} \longrightarrow \mathbb{C}$  by

$$p_1(z) = \frac{1+z}{z}g(z)$$
 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

$$\frac{zf'(z)}{f(z)} = \frac{zp'_1(z)}{p_1(z)} - \frac{zp'_2(z)}{p_2(z)} + \frac{1}{1+z}.$$
 (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

$$\left| \frac{zf'(z)}{f(z)} - \frac{1}{1 - r^2} \right| \le \frac{4r + r^2}{1 - r^2}.$$
 (2.25)

The classes we discuss here are all subclasses of starlike functions. By (2.25), we have

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

$$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \left| \frac{zf'(z)}{f(z)} - \frac{1}{1 - r^2} \right| + \frac{r^2}{1 - r^2} \le \frac{4r + 2r^2}{1 - r^2}. \tag{2.27}$$

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

$$\frac{4r + 2r^2}{1 - r^2} \le \sqrt{2} - 1. \tag{2.28}$$

Therefore, by (2.27) and (2.28), for  $0 < r \le R_{\mathcal{S}_L^*}$ , we have

$$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{4r + 2r^2}{1 - r^2} \le \sqrt{2} - 1. \tag{2.29}$$

For  $0 < r \le R_{\mathcal{S}_L^*}$ , using triangle inequality and (2.29), we have

$$\left| \frac{zf'(z)}{f(z)} + 1 \right| \le \sqrt{2} + 1$$
 (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

$$\frac{r^2 + 4r - 1}{1 - r^2} \le 2 - 2\sqrt{2}, \quad 0 < r \le R_{\mathcal{S}_R^*}.$$

![](_page_11_Figure_2.jpeg)

![](_page_11_Figure_3.jpeg)

(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 $$\operatorname{Re}\left(\frac{1+z}{z}f(z)\right) > 0 \quad (z \in \mathbb{D}).$$ (2.32) Define the function $h: \mathbb{D} \longrightarrow \mathbb{C}$ by $$h(z) = \frac{1+z}{z}f(z).$$ (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 $$\frac{zf'(z)}{f(z)} = \frac{zh'(z)}{h(z)} + \frac{1}{1+z}. (2.34)$$ Using (2.5) and (2.6), it follows from (2.34) that f maps the disk $|z| \le r$ onto the disk $$\left| \frac{zf'(z)}{f(z)} - \frac{1}{1 - r^2} \right| \le \frac{3r}{1 - r^2}.$$ (2.35) As the classes we discuss here are all subclasses of starlike functions. By (2.35), we have 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 $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \left| \frac{zf'(z)}{f(z)} - \frac{1}{1 - r^2} \right| + \frac{r^2}{1 - r^2} \le \frac{3r + r^2}{1 - r^2}.$$ (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 $$\frac{3r+r^2}{1-r^2} \le \sqrt{2} - 1. \tag{2.38}$$ Therefore, by (2.37) and (2.38), for $0 < r \le R_{\mathcal{S}_L^*}$ , we have $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{3r + r^2}{1 - r^2} \le \sqrt{2} - 1. \tag{2.39}$$ For $0 < r \le R_{\mathcal{S}_L^*}$ , using triangle inequality and (2.39), we have $$\left| \frac{zf'(z)}{f(z)} + 1 \right| \le \sqrt{2} + 1$$ (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 $$\frac{zf'(z)}{f(z)} = \frac{1-3\rho}{1-\rho^2} = \frac{3\rho-\rho^2}{1-\rho^2} = \left|\frac{\rho^2-3\rho}{1-\rho^2}\right| = \left|\frac{zf'(z)}{f(z)}-1\right|.$$    (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)).    (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

$$\operatorname{Re}\left(\frac{(1+z)^2}{z}f(z)\right) > 0 \quad (z \in \mathbb{D}).$$
 (2.42)

Define the function  $h: \mathbb{D} \longrightarrow \mathbb{C}$  by

$$h(z) = \frac{(1+z)^2}{z} f(z). \tag{2.43}$$

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

$$\frac{zf'(z)}{f(z)} = \frac{zh'(z)}{h(z)} + \frac{1-z}{1+z}. (2.44)$$

The bilinear transformation (1-z)/(1+z) maps the disk  $|z| \le r$  onto the disk

$$\left| \frac{1-z}{1+z} - \frac{1+r^2}{1-r^2} \right| \le \frac{2r}{1-r^2}.$$
 (2.45)

Using (2.45) and (2.6), it follows from (2.44) that f maps the disk  $|z| \le r$  onto the disk

$$\left| \frac{zf'(z)}{f(z)} - \frac{1+r^2}{1-r^2} \right| \le \frac{4r}{1-r^2}.$$
 (2.46)

The classes we discuss here are all subclasses of starlike functions. By (2.46), we have

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

$$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \left| \frac{zf'(z)}{f(z)} - \frac{1+r^2}{1-r^2} \right| + \frac{2r^2}{1-r^2} \le \frac{4r+2r^2}{1-r^2}. \tag{2.48}$$

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

$$\frac{4r + 2r^2}{1 - r^2} \le \sqrt{2} - 1. \tag{2.49}$$

Therefore, by (2.48) and (2.49), for  $0 < r \le R_{\mathcal{S}_r^*}$ , we have

$$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{4r + 2r^2}{1 - r^2} \le \sqrt{2} - 1. \tag{2.50}$$

For  $0 < r \le R_{\mathcal{S}_L^*}$ , using triangle inequality and (2.50), we have

$$\left| \frac{zf'(z)}{f(z)} + 1 \right| \le \sqrt{2} + 1$$
 (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

$$\left| \left( \frac{z f_4'(z)}{f_4(z)} \right)^2 - 1 \right| = \left| \left( \frac{1 - 4z + z^2}{1 - z^2} \right)^2 - 1 \right| = \left| \left( \frac{1 + 4\rho + \rho^2}{1 - \rho^2} \right)^2 - 1 \right| = 1.$$

![](_page_18_Figure_2.jpeg)

![](_page_18_Figure_3.jpeg)

![](_page_18_Figure_4.jpeg)

(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  $0 < r \le R_{\mathcal{S}_{sin}^*}$ 

$$\frac{4r}{1-r^2} \le \sin 1 - \frac{2r^2}{1-r^2}.$$

![](_page_19_Figure_2.jpeg)

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 $$f_2(z) = \frac{z(1-z)^2}{(1+z)^2}$$ 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 $$f_3(z) = \frac{z(1-z)}{(1+z)^2} \tag{2.31}$$ 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 $$f_4(z) = \frac{z(1-z)}{(1+z)^3}. (2.41)$$ 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 $2-\sqrt{3}\approx 0.276949$ .