Coefficients of the Inverse Functions and Radius Estimates of Certain Starlike Functions

Theorem 2.1. Let the function $f \in \mathscr{S}^*(\varphi)$ and $f^{-1}(\omega) = \omega + \sum_{n=2}^{\infty} A_n \omega^n$ in some neighbourhood of the origin. Then $$|A_3 - \mu A_2^2| \le \frac{B_1}{2} \max\{1, |\nu - 1|\}$$ where $$\nu = \frac{1}{B_1}((3-2\mu)B_1^2 + B_1 - B_2).$$ In particular, we have $$|A_2| \le B_1$$ and $|A_3| \le \frac{B_1}{2} \max \left\{ 1, \frac{1}{B_1} |3B_1^2 - B_2| \right\}$ . The bounds obtained are sharp. Theorem 2.1 can be obtained as a direct application of the following lemma.

Lemma 2.2. [20] Let the function $p(z) = 1 + \sum_{n=1}^{\infty} p_n z^n \in \mathscr{P}$ , then for any complex number $\nu$ , we have $$\left| p_2 - \frac{\nu}{2} p_1^2 \right| \le 2 \max\{1, |\nu - 1|\} = \begin{cases} 2, & \text{if } 0 \le \nu \le 2\\ 2|\nu - 1|, & \text{elsewhere.} \end{cases}$$ Indeed, using equations (2.5) and (2.6), we have $$|A_3 - \mu A_2^2| = \frac{B_1}{4} \left| c_2 - \frac{1}{2B_1} ((3 - 2\mu)B_1^2 + B_1 - B_2)c_1^2 \right|.$$ Now the result follows from Lemma 2.2. Note that the equality holds for the bound on $A_2$ if and only if the function f is given by $zf'(z)/f(z) = \varphi(\epsilon z)$ where $|\epsilon| = 1$ . If $|3B_1^2 - B_2| < B_1$ , then equality holds for the bound on $A_3$ if and only if the function f is given by $zf'(z)/f(z) = \varphi(\epsilon z^2)$ and if $|3B_1^2 - B_2| > B_1$ , then equality holds if and only if the function f is given by $zf'(z)/f(z) = \varphi(\epsilon z)$ . If $|3B_1^2 - B_2| = B_1$ , then equality holds if and only if the f is given by $zf'(z)/f(z) = (\lambda \varphi(\epsilon z) + (1-\lambda)(\varphi(\epsilon z))^{-1})^{-1}$ where $0 \le \lambda \le 1$ . Remark 2.3. Letting $\varphi(z) = ((1+z)/(1-z))^{\alpha}$ for $0 < \alpha \le 1$ , then Theorem 2.1 reduces to [2, Theorem 1, p. 68] for the uppper bounds on $|A_2|$ and $|A_3|$ . Taking $\varphi(z) = (1+(1-2\alpha)z)/(1-z)$ where $0 < \alpha < 1$ , then Theorem 2.1 simplifies to [12, Theorem 1, p. 105]. If $\varphi(z) = \sqrt{1+z}$ , then $|A_2| \le 1/2$ and $|A_3| \le 7/16$ which are obtained in [37, Theorem 4.1, p. 90] and when $f \in \mathscr{S}^*[A,B]$ where $-1 \le B \le 1 < A$ , then $|A_2| \le A - B$ and $|A_3| \le (3A^2 - 5AB + 2B^2)/2$ which are proved in [30, Theorem 2.1, p. 3541]. Next, we estimate the first five initial coefficients of the inverse function of functions belonging to the functions belonging to $\mathscr{S}_e^*$ .

Theorem 2.4. Let the function $f \in \mathscr{S}_e^*$ and $f^{-1}(\omega) = \omega + \sum_{n=2}^{\infty} A_n \omega^n$ in some neighbourhood of the origin. Then the following sharp estimates hold: $$|A_2| \le 1$$ , $|A_3| \le 5/4$ , $|A_4| \le 31/18$ and $|A_5| \le 361/144$ . In order to prove the above result, we will using the following lemma in which the inequality (2.9) was given by [7] while the inequalities (2.10) and (2.11) were given by [15].

Lemma 2.5. [7, 15] If the function $p(z) = 1 + \sum_{n=1}^{\infty} p_n z^n \in \mathscr{P}$ , then $$|p_2 - \frac{1}{2}p_1^2| \le 2 - \frac{1}{2}|p_1|^2,\tag{2.9}$$ <span id="page-4-2"></span><span id="page-4-1"></span> $$|p_1^3 - 2p_1p_2 + p_3| \le 2 (2.10)$$ and <span id="page-4-3"></span> $$|p_1^4 + p_2^2 - 3p_1^2p_2 + 2p_1p_3 - p_4| \le 2. (2.11)$$ Proof of Theorem 2.4. Since the function $f \in \mathscr{S}_e^*$ , we have $B_1 = 1$ , $B_2 = 1/2$ , $B_3 = 1/6$ and $B_4 = 1/24$ . Using the values of $B_1$ and $B_2$ in Theorem 2.1, the bounds for $A_2$ and $A_3$ are obtained. To obtain the upper bound on the fourth inverse coefficient $A_4$ , we use the values of $B_i$ 's in equation (2.7) to obtain $$|A_4| = \frac{1}{24} \left| 4c_3 - 14c_1c_2 + \frac{67}{6}c_1^3 \right|$$ $$= \frac{1}{24} \left| 4(c_3 - 2c_1c_2 + c_1^3) - 6c_1c_2 + \frac{43}{6}c_1^3 \right|$$ $$\leq \frac{1}{24} \left[ 4|c_3 - 2c_1c_2 + c_1^3| + 6|c_1| \left| c_2 - \frac{43}{36}c_1^2 \right| \right]$$ where the last step follows from the triangle inequality. Using Lemma 2.5 and Lemma 2.2 with $\nu = 43/18$ , we have $$|A_4| \le \frac{1}{24} \left[ 8 + 12 \cdot 2 \left| \frac{43}{18} - 1 \right| \right] = \frac{31}{18}.$$ From equation (2.8), we get $$|A_5| = \frac{1}{2304} \left| 1261c_1^4 - 2496c_1^2c_2 + 432c_2^2 + 1104c_1c_3 - 288c_4 \right|$$ $$= \frac{1}{8} \left| \frac{1261}{288}c_1^4 - \frac{26}{3}c_1^2c_2 + \frac{3}{2}c_2^2 + \frac{23}{6}c_1c_3 - c_4 \right|$$ $$= \frac{1}{8} \left| B + \frac{11}{6}c_1C + \frac{5}{4}c_1^2D - \frac{1}{2}DE + \frac{13}{288}c_1^4 \right|$$ (2.12) where <span id="page-5-0"></span> $$B = c_1^4 + c_2^2 - 3c_1^2c_2 + 2c_1c_3 - c_4,$$ $$C = c_1^3 - 2c_1c_2 + c_3,$$ $$D = c_1^2 - c_2$$ and $$E = c_2 - \frac{1}{2}c_1^2.$$ Lemma 2.2 and Lemma 2.5 readily show that $$|B| \le 2$$ , $|C| \le 2$ , $|D| \le 2$ and $|E| \le 2 - \frac{1}{2}|c_1|^2$ . Using triangle inequality in (2.12) along with the above inequalities and the fact that $|c_1| \leq 2$ , we obtain $$|A_5| \le \frac{1}{8} \left[ |B| + \frac{11}{6} |c_1| |C| + \frac{5}{4} |c_1|^2 |D| + \frac{1}{2} |D| |E| + \frac{13}{288} |c_1|^4 \right]$$ $$\le \frac{1}{8} \left[ 4 + \frac{11}{3} |c_1| + 2|c_1|^2 + \frac{13}{288} |c_1|^4 \right] \le \frac{361}{144}.$$ Let the function $f_0 \colon \mathbb{D} \to \mathbb{C}$ be defined by $$f_0(z) = z \exp\left(\int_0^z \frac{e^{\epsilon t} - 1}{t} dt\right) = z + \epsilon z^2 + \frac{3\epsilon^2}{4} z^3 + \frac{17\epsilon^3}{36} z^4 + \frac{19\epsilon^4}{72} z^5 + \cdots\right)$$ where $|\epsilon| = 1$ . Then $f_0(0) = f_0'(0) - 1 = 0$ and $zf_0'(z)/f_0(z) = e^{\epsilon z}$ . Therefore the function $f_0 \in \mathscr{S}_e^*$ and we have $$f_0^{-1}(\omega) = \omega - \epsilon \omega^2 + \frac{5\epsilon^2}{4}\omega^3 - \frac{31\epsilon^3}{18}\omega^4 + \frac{361\epsilon^4}{144}\omega^5 + \cdots$$ Hence all the bounds estimated above are sharp for the function $f_0$ . In the last result of this section, we determine the first five initial coefficients of the inverse function of the functions belonging to the subclass $\mathscr{S}_{\mathcal{R}}^*$ .

Theorem 2.6. Let the function $f \in \mathscr{S}_{\mathcal{R}}^*$ and $f^{-1}(\omega) = \omega + \sum_{n=2}^{\infty} A_n \omega^n$ for all $\omega$ in some neighbourhood of the origin. Then $$|A_n| \le \frac{\sqrt{2} - 1}{n - 1}$$ (for $n = 2, 3, 4$ ) and $|A_5| \le \frac{69}{\sqrt{2}} - \frac{387}{8}$ . First three estimated bounds are sharp.

Theorem 3.1. Let the function $f \in \mathscr{S}^*(\varphi)$ and $f^{-1}(\omega) = \omega + \sum_{n=2}^{\infty} A_n \omega^n$ for all $\omega$ in some neighbourhood of the origin. 1. If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$|3B_1^2 - B_2| \le B_1,$$ $|5B_1^4 - 6B_1^2B_2 - 3B_2^2 + 4B_1B_3| - 3B_1^2 \le 0$ then $$|A_2A_4 - A_3^2| \le \frac{B_1^2}{4}.$$ 2. If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$|3B_1^2 - B_2| \ge B_1$$ , $|5B_1^4 - 6B_1^2B_2 - 3B_2^2 + 4B_1B_3| - B_1|3B_1^2 - B_2| - 2B_1^2 \ge 0$ or the conditions $$|3B_1^2 - B_2| \le B_1$$ , $|5B_1^4 - 6B_1^2B_2 - 3B_2^2 + 4B_1B_3| - 3B_1^2 \ge 0$ then $$|A_2A_4 - A_3^2| \le \frac{1}{12} |5B_1^4 - 6B_1^2B_2 - 3B_2^2 + 4B_1B_3|.$$ 3. If $B_1$ , $B_2$ and $B_3$ satisfy the conditions $$|3B_1^2 - B_2| > B_1$$ , $|5B_1^4 - 6B_1^2B_2 - 3B_2^2 + 4B_1B_3| - B_1|3B_1^2 - B_2| - 2B_1^2 \le 0$ then $$|A_2A_4 - A_3^2| \le \frac{B_1^2}{12} \left( \frac{3|5B_1^4 - 6B_1^2B_2 - 3B_2^2 + 4B_1B_3|}{-4B_1|3B_1^2 - B_2| - (3B_1^2 - B_2)^2 - 4B_1^2}{|5B_1^4 - 6B_1^2B_2 - 3B_2^2 + 4B_1B_3| - 2B_1|3B_1^2 - B_2| - B_1^2} \right).$$ To prove the above result, we shall use the following lemma which is given by Libera and Złotkiewicz [16].

Lemma 3.2. [16] Let the function $p \in \mathscr{P}$ and $p(z) = 1 + \sum_{n=1}^{\infty} p_n z^n$ , then $$2p_2 = p_1^2 + \gamma(4 - p_1^2)$$ $$4p_3 = p_1^3 + 2p_1(4 - p_1^2)\gamma - p_1(4 - p_1^2)\gamma^2 + 2(4 - p_1^2)(1 - |\gamma|^2)z$$ for some complex valued $\gamma$ and z satisfying $|\gamma| < 1$ and |z| < 1. Proof of Theorem 3.1. Using expressions (2.5), (2.6) and (2.7), we have $$A_2A_4 - A_3^2 = \frac{B_1}{192} \left[ \left( 5B_1^3 - \frac{3B_2^2}{B_1} + 4B_3 + 6B_1^2 - 2B_2 + B_1 - 6B_1B_2 \right) c_1^4 + 4(-3B_1^2 - B_1 + B_2)c_1^2c_2 - 12B_1c_2^2 + 16B_1c_1c_3 \right].$$ Let us suppose that <span id="page-8-1"></span> $$d_{1} = 16B_{1}, \quad d_{2} = 4(-3B_{1}^{2} - B_{1} + B_{2}),$$ $$d_{3} = -12B_{1}, \quad d_{4} = 5B_{1}^{3} - \frac{3B_{2}^{2}}{B_{1}} + 4B_{3} + 6B_{1}^{2} - 2B_{2} + B_{1} - 6B_{1}B_{2},$$ $$T = \frac{B_{1}}{192}.$$ (3.1) This gives $$|A_2A_4 - A_3^2| = T|d_1c_1c_3 + d_2c_1^2c_2 + d_3c_2^2 + d_4c_1^4|. (3.2)$$ Since $\mathscr{S}^*(\varphi)$ is rotationally invariant and if the function $p \in \mathscr{P}$ , then $p(e^{i\theta}z) \in \mathscr{P}$ (where $\theta$ is a real), we can always suppose that $c_1 > 0$ and since $|c_1| \leq 2$ , without loss of generality assume that $c_1 = c \in [0, 2]$ . In view of Lemma 3.2, we have $$|A_2A_4 - A_3^2| = \frac{T}{4} |(d_1 + 2d_2 + d_3 + 4d_4)c^4 + 2c^2(4 - c^2)(d_1 + d_2 + d_3)\gamma + (4 - c^2)\gamma^2(-d_1c^2 + d_3(4 - c^2)) + 2d_1c(4 - c^2)(1 - |\gamma|^2)z).$$ Applying the triangle inequality in above equation, replacing $|\gamma|$ by $\mu$ and substituting the values of $d_1$ , $d_2$ , $d_3$ and $d_4$ from (3.1), we have $$|A_2A_4 - A_3^2| \le \frac{T}{4} \left[ 4c^4 \left| 5B_1^3 - \frac{3B_2^2}{B_1} + 4B_3 - 6B_1B_2 \right| + 8c^2(4 - c^2)|3B_1^2 - B_2|\mu \right]$$ $$+ 4B_1(4 - c^2)(12 + c^2)\mu^2 + 32B_1c(4 - c^2)(1 - \mu^2)$$ $$= T \left[ c^4 \left| 5B_1^3 - \frac{3B_2^2}{B_1} + 4B_3 - 6B_1B_2 \right| + 2c^2(4 - c^2)|3B_1^2 - B_2|\mu \right]$$ $$+ 8B_1c(4 - c^2) + B_1(4 - c^2)(c - 2)(c - 6)\mu^2$$ $$= F(c, \mu).$$ For fixed c, since $\partial F/\partial \mu > 0$ in the region $\Omega = \{(c, \mu) : 0 \le c \le 2, 0 \le \mu \le 1\}$ , $F(c, \mu)$ is an increasing function of $\mu$ in the closed interval [0, 1] which implies the function $F(c, \mu)$ attains its maximum value at $\mu = 1$ for some fixed $c \in [0, 2]$ , that is, $$\max F(c, \mu) = F(c, 1) =: G(c)$$ where $$G(c) = \frac{B_1}{192} \left[ c^4 \left| 5B_1^3 - \frac{3B_2^2}{B_1} + 4B_3 - 6B_1B_2 \right| + 2c^2(4 - c^2)|3B_1^2 - B_2| + B_1(4 - c^2)(12 + c^2) \right]$$ $$= \frac{B_1}{192} \left[ c^4 \left( \left| 5B_1^3 - \frac{3B_2^2}{B_1} + 4B_3 - 6B_1B_2 \right| - 2|3B_1^2 - B_2| - B_1 \right) + 8c^2 \left( |3B_1^2 - B_2| - B_1 \right) + 48B_1 \right].$$ Let us set $$P = \left| 5B_1^3 - \frac{3B_2^2}{B_1} + 4B_3 - 6B_1B_2 \right| - 2|3B_1^2 - B_2| - B_1,$$ $$Q = 8\left( |3B_1^2 - B_2| - B_1 \right),$$ $$R = 48B_1.$$ (3.3) Since $$\max_{0 \leq t \leq 4}(Pt^2 + Qt + R) = \begin{cases} R, & Q \leq 0, P \leq -Q/4 \\ 16P + 4Q + R, & Q \geq 0, P \geq -Q/8 \text{ or } Q \leq 0, P \geq -Q/4 \\ (4PR - Q^2)/4P, & Q > 0, P \leq -Q/8 \end{cases}$$ we have $$|A_2A_4 - A_3^2| \le \frac{B_1}{192} \begin{cases} R, & Q \le 0, P \le -Q/4 \\ 16P + 4Q + R, & Q \ge 0, P \ge -Q/8 \text{ or } Q \le 0, P \ge -Q/4 \\ (4PR - Q^2)/4P, & Q > 0, P \le -Q/8 \end{cases}$$ <span id="page-9-0"></span> where P, Q are R are same is in (3.3). This completes the proof. As a consequence of Theorem 3.1, we have the following:

Corollary 3.3. 1. If the function $f \in \mathscr{S}^*$ , then $|A_2A_4 - A_3^2| \leq 3$ . - 2. If the function $f \in \mathscr{S}_{L}^{*}$ , then $|A_{2}A_{4} A_{3}^{2}| \leq 19/280$ . - 3. If the function $f \in \mathscr{S}_{e}^{}$ , then $|A_{2}A_{4} A_{3}^{2}| \leq 29/98$ . 4. If the function $f \in \mathscr{S}_{\mathcal{R}}^{}$ , then $|A_{2}A_{4} A_{3}^{2}| \leq 1/4k^{2}$ . Now let the function $f \in \mathscr{S}_{\mathcal{R}}^*$ , then $B_1 = 1/k$ , $B_2 = 2/k$ , $B_3 = 2/k^2$ . Using the values of $B_i$ 's in (2.1), (2.2) and (2.3), we get <span id="page-10-0"></span> $$a_2 = \frac{1}{2k}c_1 \tag{3.4}$$ <span id="page-10-1"></span> $$a_3 = \frac{1}{8k^2}(2kc_2 + (3-k)c_1^2) \tag{3.5}$$ <span id="page-10-2"></span> $$a_4 = \frac{1}{48k^3} \left( (11 - 11k + 2k^2)c_1^3 + 2(11 - 4k)kc_1c_2 + 8k^2c_3 \right). \tag{3.6}$$ The following estimate of the Fekete-Szegö functional for $f \in \mathscr{S}_{\mathcal{R}}^*$ is a direct consequence of Lemma 2.2: <span id="page-10-3"></span> $$|a_3 - \mu a_2^2| \le \frac{1}{2k} \max\left\{1, \frac{1}{k}|2\mu - 3|\right\}, \qquad k = \sqrt{2} + 1$$ (3.7) and in particular, we have $$|a_2| \le \frac{1}{k}$$ and $|a_3| \le \frac{3}{2k^2}$ . The extremal function h for the class $\mathscr{S}_{\mathcal{R}}^*$ is given by $$h(z) := \frac{k^2 z}{(k-z)^2} e^{-z/k}$$ $$= z + \frac{1}{k} z^2 + \frac{3}{2k^2} z^3 + \frac{11}{6k^3} z^4 + \cdots$$ (3.8) Hence we conclude the following: <span id="page-10-4"></span>Conjecture 3.4. Let $f \in \mathscr{S}_{\mathcal{R}}^*$ and $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ , then $$|a_n| \le \frac{1}{k^{n-1}} \left( \sum_{p=0}^{n-1} (-1)^p \frac{n-p}{p!} \right).$$ Following result can be easily obtained by putting the values of $B_i$ 's mentioned above in [14, Theorem 1. p. 3]:

Theorem 3.5. Let $f \in \mathscr{S}_{\mathcal{R}}^*$ and $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ , then $$|a_2a_4 - a_3^2| \le \frac{1}{4k^2} \approx 0.0428932.$$ The bound obtained is sharp for the function $f_2$ defined by (2.13).

Theorem 3.6. Let $f \in \mathscr{S}_{\mathcal{R}}^*$ and $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ , then $$|a_2a_3 - a_4| \le \frac{5220 + 3683\sqrt{2} + 359\sqrt{359 + 246\sqrt{2}} + 246\sqrt{718 + 492\sqrt{2}}}{1458(1 + \sqrt{2})^5} \approx 0.244395.$$

Theorem 4.1. The $\mathscr{S}^_{\mathcal{R}}$ -radii for the subclasses $\mathscr{CS}^(\alpha)$ , $\mathscr{S}^_q$ and $\mathscr{BS}^(\alpha)$ , $\mathscr{M}(\beta)$ -radius for the class $\mathscr{S}_{\mathcal{R}}$ and $\mathscr{S}_{\mathcal{L}}$ -radius for the class $\mathscr{S}_{\mathcal{R}}^*$ are given by: (a) $$\mathscr{R}_{\mathscr{S}_{\mathcal{P}}}(\mathscr{C}\mathscr{S}^(\alpha)) = \rho_0 := (2 - \alpha + \sqrt{7 - 6\alpha + \alpha^2})/(-3 + 2\alpha)$$ (b) $\mathscr{R}_{\mathscr{S}_{\mathcal{P}}}^(\mathscr{S}_q^) = \left(-2 + \sqrt{2} + \sqrt{-4 + 4\sqrt{2}}\right)/2 \approx 0.350701$ which is the smallest positive root of the equation $4r^4 - 4r^2 + (57 - 40\sqrt{2}) = 0$ $$(c) \,\,\mathcal{R}_{\mathcal{S}_{\mathcal{R}}}(\mathcal{B}\mathcal{S}^(\alpha)) = \left(-\left(3 + 2\sqrt{2}\right) + \sqrt{4\alpha + 17 + 12\sqrt{2}}\right)/2\alpha$$ (d) $$\mathscr{R}_{\mathscr{M}(\beta)}(\mathscr{S}_{\mathcal{R}}^*) = \begin{cases} 1 & \text{if } \beta \ge 2\\ k(-\beta + \sqrt{\beta^2 + 4\beta - 4})/2 & \text{if } \beta \le 2 \end{cases}$$ (e) $\mathscr{R}_{\mathscr{S}_{\mathcal{L}}}(\mathscr{S}_{\mathcal{R}}^) = (-1 + \sqrt{2})(-4 - 3\sqrt{2} + \sqrt{62 + 44\sqrt{2}})/2 \approx 0.601232$ (e) $$\mathcal{R}_{\mathcal{S}_L}(\mathcal{S}_{\mathcal{R}}^) = (-1 + \sqrt{2})(-4 - 3\sqrt{2} + \sqrt{62 + 44\sqrt{2}})/2 \approx 0.601232$$ respectively. The radii obtained are sharp. The subclass of $\mathscr{P}$ which satisfies $\operatorname{Re} p(z) > \alpha$ where $0 \le \alpha < 1$ is denoted by $\mathscr{P}(\alpha)$ . In general, for $|B| \leq 1$ and $A \neq B$ , the class $\mathscr{P}[A,B]$ consists of all those functions p with the normalization p(0) = 1 satisfying $p(z) \prec (1 + Az)/(1 + Bz)$ . The following lemmas will be used in our investigation:

Lemma 4.2. [33] If $p \in \mathscr{P}(\alpha)$ , then $$\left| \frac{zp'(z)}{p(z)} \right| \le \frac{2r(1-\alpha)}{(1-r)(1+(1-2\alpha)r)}, \quad |z| = r < 1.$$

Lemma 4.3. [29] If $p \in \mathscr{P}[A, B]$ , then $$\left| p(z) - \frac{1 - ABr^2}{1 - B^2r^2} \right| \le \frac{|A - B|r}{1 - B^2r^2}, \quad |z| = r < 1.$$

Lemma 4.4. [13] For $2(\sqrt{2}-1) < a < 2$ , let $r_a$ be defined by $$r_a = \begin{cases} a - 2(\sqrt{2} - 1), & \text{if } 2(\sqrt{2} - 1) < a \le \sqrt{2}; \\ 2 - a, & \text{if } \sqrt{2} \le a < 2. \end{cases}$$ Then $\{w \in \mathbb{C} : |w-a| < r_a\} \subset \varphi_{\mathcal{R}}(\mathbb{D})$ where $$\varphi_{\mathcal{R}}(\mathbb{D}) := \{ w \in \mathbb{C} \colon |w + (w^2 + 4w - 4)^{1/2}| < 2/k \}.$$ Proof of Theorem 4.1. (a) Let the function $f \in \mathscr{CS}^(\alpha)$ and the function $g \in \mathscr{S}^(\alpha)$ be such that $p(z) = f(z)/g(z) \in \mathcal{P}$ . Then $zg'(z)/g(z) \in \mathcal{P}(\alpha)$ and Lemma 4.3 gives $$\left| \frac{zg'(z)}{q(z)} - \frac{1 + (1 - 2\alpha)r^2}{1 - r^2} \right| \le \frac{2(1 - \alpha)r}{1 - r^2}.$$ Since the function $p \in \mathcal{P}$ , Lemma 4.2 yields $$\left| \frac{zp'(z)}{p(z)} \right| \le \frac{2r}{1 - r^2}.$$ Using the above estimates in the identity $$\frac{zf'(z)}{f(z)} = \frac{zg'(z)}{g(z)} + \frac{zp'(z)}{p(z)}$$ we can see that <span id="page-13-0"></span> $$\left| \frac{zf'(z)}{f(z)} - \frac{1 + (1 - 2\alpha)r^2}{1 - r^2} \right| \le \frac{2(2 - \alpha)r}{1 - r^2}.$$ (4.1) Let $0 \le r \le \rho_0$ . Then it can be easily seen that if $a := (1 + (1 - 2\alpha)r^2)/(1 - r^2)$ , then $a \le 2$ . Therefore from Lemma 4.4, we can see that the disk (4.1) lies inside the domain $\varphi_{\mathcal{R}}(\mathbb{D})$ if and only if $$\frac{2(2-\alpha)r}{1-r^2} \le 2 - \frac{1+(1-2\alpha)r^2}{1-r^2}.$$ The last inequality reduces to $-1 + 2(2 - \alpha)r + (3 - 2\alpha)r^2 \le 0$ , which holds if $r \le \rho_0$ and the result follows. Consider the functions $f, g \in \mathscr{A}$ defined by $$f(z) = \frac{z(1+z)}{(1-z)^{3-2\alpha}}$$ and $g(z) = \frac{z}{(1-z)^{2-2\alpha}}$ . Since $$\operatorname{Re} \frac{f(z)}{g(z)} = \operatorname{Re} \frac{1+z}{1-z} > 0$$ and $\operatorname{Re} \frac{zg'(z)}{g(z)} = \operatorname{Re} \frac{1+(1-2\alpha)z^2}{1-z^2} > \alpha$ then $g \in \mathscr{S}^(\alpha)$ and hence $f \in \mathscr{CS}^(\alpha)$ . Also at the point $z = \rho_0$ , we see that $$\frac{zf'(z)}{f(z)} = \frac{1 + 2(2 - \alpha)z + (1 - 2\alpha)z^2}{1 - z^2}$$ $$= \frac{1 + (1 - 2\alpha)\rho_0^2}{1 - \rho_0^2} - \frac{2(2 - \alpha)\rho_0}{1 - \rho_0^2} = 2.$$ This proves the sharpness of the result. (b) Since the function f is in the class $\mathscr{S}_q^*$ , for |z|=r, we have <span id="page-13-1"></span> $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le |z + \sqrt{1 + z^2} - 1| \le 1 - r - \sqrt{1 - r^2}. \tag{4.2}$$ In view of Lemma 4.4, the disk (4.2) lies in the domain $\varphi_{\mathcal{R}}(\mathbb{D})$ if $1-r-\sqrt{1-r^2} \leq 3-2\sqrt{2}$ or $4r^4-4r^2+(57-40\sqrt{2})\leq 0$ which gives the desired radius estimate and this estimate is best possible for the function <span id="page-13-3"></span> $$f_q(z) := z \exp(q(z) - \log(1 - z + q(z)) + \log 2 - 1). \tag{4.3}$$ (c) Let the function $f \in \mathscr{BS}^*(\alpha)$ and |z| = r. Then a simple calculation yields <span id="page-13-2"></span> $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \left| \frac{z}{1 - \alpha z^2} \right| \le \frac{r}{1 - \alpha r^2}. \tag{4.4}$$ Using Lemma 4.4, we see that the disk (4.4) is contained in the domain $\varphi_{\mathcal{R}}(\mathbb{D})$ if $$\frac{r}{1 - \alpha r^2} \le 3 - 2\sqrt{2}$$ <span id="page-14-3"></span>or $\alpha r^2 + (3 + 2\sqrt{2})r - 1 \le 0$ . This gives the required radius estimate. The function defined $$f_B(z) := z \exp\left(\frac{\tanh^{-1}(\sqrt{\alpha}z)}{\sqrt{\alpha}}\right)$$ (4.5) proves that the estimation is sharp. (d) Let the function $f \in \mathscr{S}_{\mathcal{R}}^*$ . Case 1. Let $\beta \geq 2$ . For |z| = r < 1, using the definition of subordination, it is easy to see $$\operatorname{Re} \frac{zf'(z)}{f(z)} \le \max_{|z|=r} \varphi_{\mathcal{R}}(z) \le 1 + \frac{r}{k} \left( \frac{k+r}{k-r} \right) < \frac{k^2+1}{k(k-1)} \le \beta.$$ Case 2. Let $\beta \leq 2$ . For $|z| = r < k(-\beta + \sqrt{\beta^2 + 4\beta - 4})/2$ , using the same technique as in Case 1, it follows that $$\operatorname{Re} \frac{zf'(z)}{f(z)} \le 1 + \frac{r}{k} \left(\frac{k+r}{k-r}\right) < \beta.$$ This proves the desired result. Sharpness follows for the function <span id="page-14-0"></span> $$f_r(z) = \frac{kz}{(k-z)^2} e^{-z/k}. (4.6)$$ (e) Since $f \in \mathscr{S}_{\mathcal{R}}^*$ , we have $$|\varphi_{\mathcal{R}}(z) - 1|^2 = \frac{r^2}{k^2} \left( \frac{k^2 + r^2 + 2kr\cos t}{k^2 + r^2 - 2kr\cos t} \right) < (\sqrt{2} - 1)^2$$ if $$\left| \frac{zf'(z)}{f(z)} - 1 \right| < \sqrt{2} - 1, \quad |z| = r < \frac{1}{2}(-1 + \sqrt{2})(-4 - 3\sqrt{2} + \sqrt{62 + 44\sqrt{2}}).$$ Therefore the result follows from [3, Lemma 2.2, p. 6559]. The radius estimate is sharp for the function $f_r$ defined by (4.6). Next result yields the sharp radius estimates related to the class $\mathscr{S}_e^*$ .

Theorem 4.5. The $\mathscr{S}_e$ -radii for the subclasses $\mathscr{S}_L$ , $\mathscr{S}_q$ , $\mathscr{S}_R$ , $\mathscr{S}_C$ and $\mathscr{BS}^(\alpha)$ are given - (a) $\mathscr{R}_{\mathscr{S}_e}(\mathscr{S}_L^) = (e^2 1)/e^2 \approx 0.864665$ (b) $\mathscr{R}_{\mathscr{S}_e}(\mathscr{S}_q^) = (-2e + \sqrt{-4e^2 + 8e^4})/(4e^2) \approx 0.498824$ which is the smallest positive root of the equation $4r^4 - 4r^2 + ((e^2 - 1)/e^2)^2 = 0$ - $(c) \,\,\mathcal{R}_{\mathcal{S}_{e}^{}}(\mathcal{S}_{\mathcal{R}}^{}) = (k 2ek + k\sqrt{1 8e + 8e^{2}})/(2e) \approx 0.780444$ $(d) \,\,\mathcal{R}_{\mathcal{S}_{e}^{}}(\mathcal{S}_{C}^{}) = (-2e + \sqrt{10e^{2} 4e})/2e \approx 0.395772$ $(e) \,\,\mathcal{R}_{\mathcal{S}_{e}^{}}(\mathcal{B}\mathcal{S}^{}(\alpha)) = (-e + \sqrt{e^{2} + 4(e 1)^{2}\alpha})/(2\alpha(e 1))$ respectively. The results are all sharp. To prove our estimations, we will make use of the following result.

Lemma 4.6. [21] For 1/e < a < e, let $r_a$ be defined by $$r_a = \begin{cases} a - e^{-1}, & \text{if } e^{-1} < a \le (e + e^{-1})/2; \\ e - a, & \text{if } (e + e^{-1})/2 \le a < e. \end{cases}$$ Then $\{w \in \mathbb{C} : |w - a| < r_a\} \subset \{w \in \mathbb{C} : |\log w| < 1\}.$ Proof of Theorem 4.5. (a) Since the function $f \in \mathscr{S}_L^*$ , we have $$\left| \frac{zf'(z)}{f(z)} - 1 \right| = |\sqrt{1+z} - 1| \le 1 - \sqrt{1-r}, \quad |z| \le r.$$ By applying Lemma 4.6, we note that the function $f \in \mathscr{S}_e^*$ if $1 - \sqrt{1 - r} \le 1 - 1/e$ which leads to the inequality $r \le e^2 - 1/e^2$ . The obtained radius estimate is sharp for the function $f_L$ defined as $$\frac{zf_L'(z)}{f_L'(z)} = \sqrt{1+z}.$$ (b) In view of Lemma 4.6, we see that the disk (4.2) is contained in the domain $e^z(\mathbb{D}) := \{w \in \mathbb{C} : |\log w| < 1\}$ provided $1 - r - \sqrt{1 - r^2} \le 1 - 1/e$ or $r + \sqrt{1 - r^2} \ge 1/e$ or equivalently $$4r^4 - 4r^2 + \left(\frac{e^2 - 1}{e^2}\right)^2 \le 0.$$ The previous inequality yields the desired estimate for the radius. The sharpness follows for the function $f_q$ defined by (4.3). (c) Let the function $f \in \mathscr{S}_{\mathcal{R}}^*$ . Then <span id="page-15-0"></span> $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{r}{k} \left( \frac{k+r}{k-r} \right), \qquad |z| \le r. \tag{4.7}$$ Using Lemma 4.6, we see that the disk (4.7) lies in the domain $\{w \in \mathbb{C} : |\log w| < 1\}$ if $$\frac{r}{k}\left(\frac{k+r}{k-r}\right) \le 1 - \frac{1}{e}.$$ The above inequality simplies to $$er^2 + k(2e-1)r - k^2(e-1) < 0$$ which gives the required radius estimate. Sharpness follows for the function $f_r$ defined by (4.6). (d) Let the function $f \in \mathscr{S}_{C}^{*}$ . Then a simple calculation gives <span id="page-15-1"></span> $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{1}{3}(4r + 2r^2), \qquad |z| \le r.$$ (4.8) Using Lemma 4.6, we note that the disk (4.8) is contained in the domain $\{w \in \mathbb{C} : |\log w| < 1\}$ if $(4r + 2r^2)/3 \le 1 - 1/e$ or $2er^2 + 4er - 3(e - 1) \le 0$ . The last inequality gives $$r \le \frac{-2e + \sqrt{10e^2 - 6e}}{2e}.$$ The result is sharp for the function $$f_C(z) := z \exp\left(\frac{4z}{3} + \frac{2z^2}{3}\right).$$ (e) Using Lemma 4.6, note that the disk (4.4) lies in the domain $e^z(\mathbb{D})$ provided $$\frac{r}{1 - \alpha r^2} \le 1 - \frac{1}{e}.$$ By a simple computation, the last inequality becomes $\alpha(e-1)r^2 + er - (e-1) \leq 0$ which gives $$r \le \frac{-e + \sqrt{e^2 + 4(e-1)^2 \alpha}}{2\alpha(e-1)}.$$ The function $f_B$ defined by (4.5) shows the sharpness of this radius estimate.