Radius Problems For Functions Associated with a Nephroid Domai

Lemma 2.1. Let 1/3 < a < 5/3, and let $r_a$ be given by $$r_a = \begin{cases} a - 1/3 & \text{if } 1/3 < a \le 1, \\ 5/3 - a & \text{if } 1 \le a < 5/3. \end{cases}$$ Then $$\{w \in \mathbb{C} : |w - a| < r_a\} \subseteq \Omega_{Ne},$$ where $\Omega_{Ne} := \varphi_{Ne}(\mathbb{D})$ , the region bounded by the nephroid (1.4).

Lemma 3.1. ([18, Lemma 2.1, p. 267]). If $p \in \mathcal{P}[A, B]$ , then for |z| = r < 1, $$\left| p(z) - \frac{1 - ABr^2}{1 - B^2r^2} \right| \le \frac{(A - B)r}{1 - B^2r^2}.$$ (i) Let $0 \le B < A \le 1$ . Then the $\mathcal{S}_{Ne}^*$ -radius for $\mathcal{S}^*[A, B]$ is given by Theorem 3.1. $$R_{\mathcal{S}_{N_e}^*}\left(\mathcal{S}^*[A,B]\right) = \min\left\{1, \frac{2}{3A - B}\right\}.$$ In particular, if $1 - B \le 3(1 - A)$ , then $\mathcal{S}^*[A, B] \subset \mathcal{S}^*_{Ne}$ . (ii) Let $-1 \le B < A \le 1$ with $B \le 0$ . Then the $\mathcal{S}^*_{Ne}$ -radius for $\mathcal{S}^*[A, B]$ is given by $$R_{\mathcal{S}_{Ne}^*}(\mathcal{S}^*[A, B]) = \min\left\{1, \frac{2}{3A - 5B}\right\}.$$ In particular, if $3(1+A) \leq 5(1+B)$ , then $\mathcal{S}^*[A,B] \subset \mathcal{S}^*_{Ne}$ (i): Let $f \in \mathcal{S}^*[A, B]$ . Then $\mathcal{Q}_f \in \mathcal{P}[A, B]$ , and Lemma 3.1 yields $$\left| \mathcal{Q}_f(z) - \frac{1 - ABr^2}{1 - B^2 r^2} \right| \le \frac{(A - B)r}{1 - B^2 r^2}. \tag{3.1}$$ Also, The inequality (3.1) represents a disk with center at $(1 - ABr^2)/(1 - B^2r^2)$ and radius $(A - B)r/(1 - B^2r^2)$ . Since $B \ge 0$ , we have $(1 - ABr^2)/(1 - B^2r^2) \le 1$ . Thus, by Lemma 2.1, the disk (3.1) lies completely inside the region $\Omega_{Ne}$ bounded by the nephroid curve (1.4) if $$\frac{(A-B)r}{1-B^2r^2} \le \frac{1-ABr^2}{1-B^2r^2} - \frac{1}{3}$$ Simplifying the above inequality, we obtain $r \leq 2/(3A - B) =: \rho_J$ . For sharpness of the estimate, consider the function $$f_{J}(z) := \begin{cases} z \left(1 + Bz\right)^{\frac{A-B}{B}}, & \text{if } B \neq 0\\ ze^{Az}, & \text{if } B = 0. \end{cases}$$ The function $f_J(z)$ satisfies $\mathcal{Q}_{f_J}(z) = (1 + Az)/(1 + Bz)$ , implying that $f_J \in \mathcal{S}^*[A, B]$ . Also, for $z_0 = -2/(3A - B) \in \partial \mathbb{D}_{\rho_J}$ , we have $$Q_{f_J}(z_0) = 1/3 \in \partial \Omega_{Ne}$$ i.e., $$Q_{f_I}(\partial \mathbb{D}_{\rho_I}) \cap \partial \Omega_{Ne} = \{1/3\} \neq \emptyset.$$ This proves that the result is sharp and $f_J \in \mathcal{S}^*[A, B]$ is the extremal function. In particular, if $1-B \leq 3(1-A)$ , then $2/(3A-B) \geq 1$ and hence, $R_{\mathcal{S}_{N_e}^*}(\mathcal{S}^*[A, B]) = 1$ . That is, $\mathcal{S}^*[A, B] \subset \mathcal{S}_{N_e}^*$ . (ii): As afore, $f \in \mathcal{S}^*[A, B]$ implies the inequality (3.1). Now, $B \leq 0$ implies that $(1 - ABr^2)/(1 - B^2r^2) \geq 1$ . Therefore, in this case, the disk (3.1) lies in the interior of $\Omega_{Ne}$ if $$\frac{(A-B)r}{1-R^2r^2} \le \frac{5}{3} - \frac{1-ABr^2}{1-R^2r^2}.$$ Solving, we get $r \leq 2/(3A - 5B)$ . Choosing $z_0 = 2/(3A - 5B)$ and $f_J \in \mathcal{S}^*[A, B]$ defined above, it can be easily verified that $\mathcal{Q}_{f_J}(z_0) = 5/3 \in \partial \Omega_{Ne}$ . Hence, the result is sharp. Let $0 \leq B < A \leq 1$ . Using a result proved by Wani and Swaminathan [27], the following remark also shows that $1 - B \leq 3(1 - A)$ is indeed a sufficient condition for the inclusion $\mathcal{S}^*[A, B] \subset \mathcal{S}^*_{Ne}$ . Remark 1. Since, $1-B \le 3(1-A)$ implies $1-B^2 \le 3(1-AB+(B-A)) < 3(1-AB)$ and, $0 \le B < A \le 1$ gives $3(1-AB) \le 3(1-B^2)$ . Combining, we obtain $1-B^2 < 3(1-AB) \le 3(1-B^2)$ . Now the inclusion relation $\mathcal{S}^*[A,B] \subset \mathcal{S}^*_{Ne}$ follows from [27, Theorem 3.3]. Specializing the constants A and B in Theorem 3.1, the following important sharp radii results are obtained.

Notation repaired from the source PDF text layer (arxiv_1912_06328.pdf, p. 6, 7; pdftotext (poppler) 25.02.0). Show original OCR text

Lemma 3.1. ([18, Lemma 2.1, p. 267]). If  $p \in \mathcal{P}[A, B]$ , then for |z| = r < 1,

$$\left| p(z) - \frac{1 - ABr^2}{1 - B^2r^2} \right| \le \frac{(A - B)r}{1 - B^2r^2}.$$

(i) Let  $0 \le B < A \le 1$ . Then the  $\mathcal{S}_{Ne}^$ -radius for  $\mathcal{S}^[A, B]$  is given by Theorem 3.1.

$$R_{\mathcal{S}_{N_e}^}\left(\mathcal{S}^[A,B]\right) = \min\left\{1, \frac{2}{3A - B}\right\}.$$

In particular, if  $1 - B \le 3(1 - A)$ , then  $\mathcal{S}^[A, B] \subset \mathcal{S}^_{Ne}$ . (ii) Let  $-1 \le B < A \le 1$  with  $B \le 0$ . Then the  $\mathcal{S}^_{Ne}$ -radius for  $\mathcal{S}^[A, B]$  is given by

$$R_{\mathcal{S}_{Ne}^}(\mathcal{S}^[A, B]) = \min\left\{1, \frac{2}{3A - 5B}\right\}.$$

In particular, if  $3(1+A) \leq 5(1+B)$ , then  $\mathcal{S}^[A,B] \subset \mathcal{S}^_{Ne}$ 

(i): Let  $f \in \mathcal{S}^*[A, B]$ . Then  $\mathcal{Q}_f \in \mathcal{P}[A, B]$ , and Lemma 3.1 yields

$$\left| \mathcal{Q}_f(z) - \frac{1 - ABr^2}{1 - B^2 r^2} \right| \le \frac{(A - B)r}{1 - B^2 r^2}. \tag{3.1}$$

Also,

The inequality (3.1) represents a disk with center at  $(1 - ABr^2)/(1 - B^2r^2)$  and radius  $(A - B)r/(1 - B^2r^2)$ . Since  $B \ge 0$ , we have  $(1 - ABr^2)/(1 - B^2r^2) \le 1$ . Thus, by Lemma 2.1, the disk (3.1) lies completely inside the region  $\Omega_{Ne}$  bounded by the nephroid curve (1.4) if

$$\frac{(A-B)r}{1-B^2r^2} \le \frac{1-ABr^2}{1-B^2r^2} - \frac{1}{3}$$

Simplifying the above inequality, we obtain  $r \leq 2/(3A - B) =: \rho_J$ . For sharpness of the estimate, consider the function

$$f_{J}(z) := \begin{cases} z \left(1 + Bz\right)^{\frac{A-B}{B}}, & \text{if } B \neq 0\\ ze^{Az}, & \text{if } B = 0. \end{cases}$$

The function  $f_J(z)$  satisfies  $\mathcal{Q}_{f_J}(z) = (1 + Az)/(1 + Bz)$ , implying that  $f_J \in \mathcal{S}^*[A, B]$ . Also, for  $z_0 = -2/(3A - B) \in \partial \mathbb{D}_{\rho_J}$ , we have

$$Q_{f_J}(z_0) = 1/3 \in \partial \Omega_{Ne}$$

i.e.,

$$Q_{f_I}(\partial \mathbb{D}_{\rho_I}) \cap \partial \Omega_{Ne} = \{1/3\} \neq \emptyset.$$

This proves that the result is sharp and  $f_J \in \mathcal{S}^[A, B]$  is the extremal function. In particular, if  $1-B \leq 3(1-A)$ , then  $2/(3A-B) \geq 1$  and hence,  $R_{\mathcal{S}_{N_e}^}(\mathcal{S}^[A, B]) = 1$ . That is,  $\mathcal{S}^[A, B] \subset \mathcal{S}_{N_e}^*$ .

(ii): As afore,  $f \in \mathcal{S}^*[A, B]$  implies the inequality (3.1). Now,  $B \leq 0$  implies that  $(1 - ABr^2)/(1 - B^2r^2) \geq 1$ . Therefore, in this case, the disk (3.1) lies in the interior of  $\Omega_{Ne}$  if

$$\frac{(A-B)r}{1-R^2r^2} \le \frac{5}{3} - \frac{1-ABr^2}{1-R^2r^2}.$$

Solving, we get  $r \leq 2/(3A - 5B)$ . Choosing  $z_0 = 2/(3A - 5B)$  and  $f_J \in \mathcal{S}^*[A, B]$  defined above, it can be easily verified that  $\mathcal{Q}_{f_J}(z_0) = 5/3 \in \partial \Omega_{Ne}$ . Hence, the result is sharp.

Let  $0 \leq B < A \leq 1$ . Using a result proved by Wani and Swaminathan [27], the following remark also shows that  $1 - B \leq 3(1 - A)$  is indeed a sufficient condition for the inclusion  $\mathcal{S}^[A, B] \subset \mathcal{S}^_{Ne}$ .

Remark 1. Since,  $1-B \le 3(1-A)$  implies  $1-B^2 \le 3(1-AB+(B-A)) < 3(1-AB)$  and,  $0 \le B < A \le 1$  gives  $3(1-AB) \le 3(1-B^2)$ . Combining, we obtain  $1-B^2 < 3(1-AB) \le 3(1-B^2)$ . Now the inclusion relation  $\mathcal{S}^[A,B] \subset \mathcal{S}^_{Ne}$  follows from [27, Theorem 3.3].

Specializing the constants A and B in Theorem 3.1, the following important sharp radii results are obtained.

Corollary 3.1. The sharp $S_{Ne}$ -radius for $S^(\alpha) = S^*[1-2\alpha,-1]$ is $2/(3(1-2\alpha)+5)$ . The estimate is sharp for the function $k_{\alpha}(z) = z(1-z)^{2\alpha-2}$ , $0 \le \alpha < 1$ .

Corollary 3.2. The sharp $S_{Ne}$ -radius for the class of starlike functions $S$ is 1/4, and the sharpness holds for the well-known Koebe function $k(z) := z/(1-z)^2$ .

Corollary 3.3. The sharp $\mathcal{S}_{Ne}^*$ -radius for the convex class $\mathcal{C}$ is 2/5.

Theorem 3.2. For the function classes $\mathcal{BS}^(\alpha)$ , $\mathcal{S}_L^(\alpha)$ , and $\mathcal{S}_{\alpha,e}^*$ , the following radius results hold: - (i) For $\alpha \in [0, 1)$ , $R_{\mathcal{S}_{N_e}}(\mathcal{BS}^(\alpha)) = \rho_{\mathcal{B}}(\alpha) := 4/(3 + \sqrt{9 + 16\alpha})$ . (ii) For $\alpha \in [0, 1/3]$ , $R_{\mathcal{S}_{N_e}}(\mathcal{S}_L^(\alpha)) = \rho_L(\alpha) := 4(2 3\alpha)/9(1 \alpha)^2$ . In particular, $R_{\mathcal{S}_{Ne}}(\mathcal{S}_L^) = \rho_L := 8/9.$ - (iii) For $\alpha \in [0, (3e-5)/(3e-3)]$ , $R_{\mathcal{S}_{Ne}}\left(\mathcal{S}_{\alpha,e}^\right) = \rho_e(\alpha) := \log((5-3\alpha)/(3-3\alpha))$ . In particular, $R_{\mathcal{S}_{N_e}}(\mathcal{S}_e^) = \rho_e := \log(5/3) \approx 0.510826.$ Each of the estimates is best possible.

Theorem 3.3. For the function classes $\mathcal{S}_{RL}$ , $\mathcal{S}_C$ and $\mathcal{S}_R^*$ , we have: - (i) $R_{\mathcal{S}_{Ne}}(\mathcal{S}_{RL}^) = \rho_{RL} := 56/(122 41\sqrt{2}) \approx 0.874764$ , - (ii) $R_{\mathcal{S}_{Ne}}(\mathcal{S}_C^) = \rho_c := \sqrt{2} 1 \approx 0.414214$ , - (iii) $R_{\mathcal{S}_{N_e}}(\mathcal{S}_R^) = \rho_R := 1/(3\sqrt{2} 3) \approx 0.804738.$ The estimate in each part is sharp (see Figure 6).

Theorem 3.4. The $\mathcal{S}_{Ne}$ -radii for the classes $\mathcal{S}_{\mathcal{A}}$ and $\mathcal{S}_{S}^*$ are given by - (i) $R_{\mathcal{S}_{N_e}}(\mathcal{S}_{\mathcal{C}}^) = (\sqrt{17} 1)/6 \approx 0.520518,$ - (ii) $R_{\mathcal{S}_N}(\mathcal{S}_S^) = \sinh^{-1}(2/3) = \log((\sqrt{13} + 2)/3) \approx 0.625145.$ The estimates are not sharp.

Lemma 4.1. ([18],[20]). Let $p \in \mathcal{P}_n(\alpha)$ . Then, for |z| = r, the following results hold. (a) (cf. [18, Lemma 2.1]) $$\left| p(z) - \frac{1 + (1 - 2\alpha)r^{2n}}{1 - r^{2n}} \right| \le \frac{2(1 - \alpha)r^n}{1 - r^{2n}}.$$ (b) (cf. [20, Lemma 2]) $$\left| \frac{zp'(z)}{p(z)} \right| \le \frac{2nr^n(1-\alpha)}{(1-r^n)(1+(1-2\alpha)r^n)}.$$

Theorem 4.1. For the function classes $\mathcal{G}_1$ , $\mathcal{G}_2$ , $\mathcal{G}_3$ and $\mathcal{G}_4$ , we have the following radii (i) $$R_{\mathcal{S}_{Ne,n}^*}(\mathcal{G}_1) = \rho_1 := \left(3n + \sqrt{9n^2 + 1}\right)^{-1/n}$$ . (ii) $$R_{\mathcal{S}_{N_{e,n}}^*}(\mathcal{G}_2) = \rho_2 := \sqrt[n]{4} \times \left(9n + \sqrt{81n^2 + 24n + 16}\right)^{-1/n}$$ . (iii) $$R_{\mathcal{S}_{Ne,n}^*}(\mathcal{G}_3) = \rho_3 := \sqrt[n]{4} \times \left(9n + \sqrt{(4+9n)^2 - 48n}\right)^{-1/n}$$ . (iii) $$R_{\mathcal{S}_{Ne,n}^*}(\mathcal{G}_3) = \rho_3 := \sqrt[n]{4} \times \left(9n + \sqrt{(4+9n)^2 - 48n}\right)^{-1/n}$$ . (iv) $R_{\mathcal{S}_{Ne,n}^*}(\mathcal{G}_4) = \rho_4 := \sqrt[n]{4} \times \left(3(n+1) + \sqrt{(1+3n)^2 + 36n}\right)^{-1/n}$ . Each estimate is sharp Each estimate is sharp. (i): Let $f \in \mathcal{G}_1$ . Further, suppose that Proof. $$p(z) = g(z)/z$$ and $q(z) = f(z)/g(z)$ . Clearly, $p, q \in \mathcal{P}_n$ . Thus, for |z| = r, the following inequalities follow from Lemma 4.1(b) $$\left| \frac{zp'(z)}{p(z)} \right| \le \frac{2nr^n}{1 - r^{2n}} \quad \text{and} \quad \left| \frac{zq'(z)}{q(z)} \right| \le \frac{2nr^n}{1 - r^{2n}}. \tag{4.1}$$ Moreover, we have f(z) = g(z)q(z) = zp(z)q(z), which gives the identity $$Q_f(z) = \frac{zf'(z)}{f(z)} = 1 + \frac{zp'(z)}{p(z)} + \frac{zq'(z)}{q(z)}.$$ (4.2) Using the inequalities (4.1), the above identity yields the disk $$|Q_f(z) - 1| \le 4nr^n/(1 - r^{2n}).$$ In view of Lemma 2.1, the quantity $Q_f(z)$ takes values from the region bounded by the nephroid (1.4) provided $4nr^n/(1-r^{2n}) \le 2/3$ , or equivalently, $r^{2n}+6nr^n-1 \le 0$ . This further gives $r \leq \rho_1$ . To verify the sharpness of the bound $\rho_1$ , consider the functions $f_1, g_1 \in \mathcal{A}_n$ given by $$f_1(z) = z \left(\frac{1+z^n}{1-z^n}\right)^2$$ and $g_1(z) = z \left(\frac{1+z^n}{1-z^n}\right)$ . It is clear that $f_1(z)/g_1(z) = g_1(z)/z = (1+z^n)/(1-z^n) \in \mathcal{P}_n$ , and hence $f_1 \in \mathcal{G}_1$ . Also, $$Q_{f_1}(z) = 1 + 4nz^n/(1-z^{2n}),$$ and at the points $z = \pm \rho_1 \in \partial \mathbb{D}_{\rho_1}$ , we have $$Q_{f_1}(-\rho_1) = 1/3 \in \partial \Omega_{Ne} \text{ and } Q_{f_1}(\rho_1) = 5/3 \in \partial \Omega_{Ne}.$$ This proves sharpness of the estimate. (ii): Let $f \in \mathcal{G}_2$ . Define the functions $p, q : \mathbb{D} \to \mathbb{C}$ by p = g/z and q = f/g. Then f(z) = zp(z)q(z) with $p \in \mathcal{P}_n(1/2)$ and $q \in \mathcal{P}_n$ . In light of the identity (4.2), it follows from Lemma 4.1(b) that $$|\mathcal{Q}_f(z) - 1| \le \frac{2nr^n}{1 - r^{2n}} + \frac{nr^n}{1 - r^n} = \frac{3nr^n + nr^{2n}}{1 - r^{2n}}.$$ Now, from Lemma 2.1, $f \in \mathcal{S}_{N_e}^*$ if $$\frac{3nr^n + nr^n}{1 - r^{2n}} \le \frac{2}{3}.$$ That is, if $(3n+2)r^{2n} + 9nr^n - 2 \le 0$ , which gives the desired result $r \le \rho_2$ . To prove that $\rho_2$ is the sharp $\mathcal{S}_{Ne,n}^*$ -radius for the function family $\mathcal{G}_2$ , define $$f_2(z) = \frac{z(1+z^n)}{(1-z^n)^2}$$ and $g_2(z) = \frac{z}{1-z^n}$ . It is easy to see that $f_2/g_2 \in \mathcal{P}_n$ and $g_2/z \in \mathcal{P}_n(1/2)$ , which shows that $f_2$ is a member of $\mathcal{G}_2$ . Further, $$Q_{f_2}(z)\Big|_{z=\rho_2} = \frac{1+3nz^n+(n-1)z^{2n}}{1-z^{2n}}\Big|_{z=\rho_2} = \frac{5}{3}.$$ This shows that $$Q_{f_2}(\partial \mathbb{D}_{\rho_2}) \cap \partial \Omega_{Ne} = \{5/3\} \neq \emptyset,$$ and hence the estimate is sharp. (iii): Let $f \in \mathcal{G}_3$ . Further, suppose that p(z) = g(z)/z and q(z) = g(z)/f(z). Obviously $p \in \mathcal{P}_n$ , and $q \in \mathcal{P}_n(1/2)$ due to the fact that $$\left| \frac{f(z)}{g(z)} - 1 \right| < 1 \iff \frac{g}{f} \in \mathcal{P}_n\left(\frac{1}{2}\right).$$ Verify that f(z) = zp(z)/q(z) and $$Q_f(z) = 1 + \frac{zp'(z)}{p(z)} - \frac{zq'(z)}{q(z)}.$$ Applying Lemma 4.1(b) and simplifying, we obtain $$|\mathcal{Q}_f(z) - 1| \le (3nr^n + nr^{2n})/(1 - r^{2n}) \le 2/3$$ if $(3n+2)r^{2n}+9nr^n-2\leq 0$ , or, $r\leq \rho_3$ . Therefore, by Lemma 2.1, $f\in\mathcal{S}_{Ne}^*$ provided $r\leq \rho_3$ . For sharpness, consider $$f_3(z) = \frac{z(1+z^n)^2}{1-z^n}$$ and $g_3(z) = \frac{z(1+z^n)}{1-z^n}$ satisfying $$\left| \frac{f_3(z)}{g_3(z)} - 1 \right| = |z|^n < 1$$ and $\frac{g_3(z)}{z} = \frac{1 + z^n}{1 - z^n} \in \mathcal{P}_n$ , so that $f_3 \in \mathcal{G}_3$ . Moreover, it is easy to see that $\mathcal{Q}_{f_3}(z)$ assumes the value 1/3 at the point $z = \rho_3 e^{i\pi/n} \in \partial \mathbb{D}_{\rho_3}$ . This shows that the estimate is best possible. (iv): Let $f \in \mathcal{G}_4$ and let $q : \mathbb{D} \to \mathbb{C}$ be given by q = g/f, where $g \in \mathcal{A}_n$ is some convex function. As earlier, we have $q \in \mathcal{P}_n(1/2)$ and hence Lemma 4.1(b) gives $$\left| \frac{zq'(z)}{q(z)} \right| \le \frac{nr^n}{1 - r^n}.\tag{4.3}$$ Further, since convexity of g implies $zg'/g \in \mathcal{P}_n(1/2)$ , we have the following inequality from Lemma 4.1(a), $$\left| \frac{zg'(z)}{g(z)} - \frac{1}{1 - r^{2n}} \right| \le \frac{r^n}{1 - r^{2n}}.$$ (4.4) In view of the inequalities (4.3) and (4.4), and the identity $$Q_f(z) = \frac{zg'(z)}{g(z)} - \frac{zq'(z)}{q(z)},$$ we calculate that $$\left| \mathcal{Q}_f(z) - \frac{1}{1 - r^{2n}} \right| \le \frac{(n+1)r^n + nr^{2n}}{1 - r^{2n}}.$$ (4.5) As the number $1 - r^{2n} < 1$ for each $r \in (0, 1)$ , it follows from Lemma 2.1 that the disc (4.5) lies in the interior of the nephroid domain $\Omega_{Ne}$ if $$\frac{(n+1)r^n + nr^{2n}}{1 - r^{2n}} \le \frac{1}{1 - r^{2n}} - \frac{1}{3}.$$ Or equivalently, if $$(3n-1)r^{2n} + 3(n+1)r^n - 2 \le 0.$$ This further gives $r \leq \rho_4$ . For sharpness of the estimate, consider $$f_4(z) = \frac{z(1+z^n)}{(1-z^n)^{1/n}}$$ and $g_4(z) = \frac{z}{(1-z^n)^{1/n}}$ . Since $|f_4(z)/g_4(z) - 1| = |z|^n < 1$ and $g_4 \in C_n$ , we have $f_4 \in G_4$ . Moreover, $Q_{f_4}(-\rho_4) = 1/3$ and $Q_{f_4}(\rho_4) = 5/3$ , i.e., $$Q_{f_4}(\partial \mathbb{D}_{\rho_4}) \cap \partial \Omega_{Ne} = \{1/3, 5/3\} \neq \emptyset.$$ Therefore, the estimate is sharp. In continuation of our discussion, we provide radii results for two classes of functions that possess nice geometrical properties. These classes had less attention with respect to properties other than the radii results, after the celebrated de Branges's theorem. The close-to-starlike class $\mathcal{CS}_n^(\alpha)$ . For $\alpha \in [0,1)$ , the class of close-to-starlike functions $\mathcal{CS}_n^(\alpha)$ of type $\alpha$ was introduced by Reade [19] and is defined as $$\mathcal{CS}^(\alpha) := \left\{ f \in \mathcal{A}_n : \frac{f}{g} \in \mathcal{P}_n, g \in \mathcal{S}_n^(\alpha) \right\}.$$ The following theorem determines the $\mathcal{S}_{Ne,n}$ -radius for $\mathcal{CS}_n^(\alpha)$ .

Theorem 4.2. The sharp $\mathcal{S}_{Ne,n}$ -radius for $\mathcal{CS}_n^(\alpha)$ is given by $$R_{\mathcal{S}_{Ne,n}}\left(\mathcal{CS}_n^(\alpha)\right) = \rho_{cs} := \left(\frac{2}{3(1+n-\alpha) + \sqrt{9(1+n-\alpha)^2 + 4(4-3\alpha)}}\right)^{1/n}.$$

Theorem 4.3. The sharp $\mathcal{S}_{N_{e,n}}^*$ -radius for $\mathcal{W}_n$ is $$R_{\mathcal{S}_{Ne,n}^*}(\mathcal{W}_n) = \rho_{\mathcal{W}} := \left(2/(3n + \sqrt{9n^2 + 4})\right)^{1/n}$$

Theorem 4.4. The $S_{Ne,n}^*$ -radius for the class $\mathcal{M}_n(\beta)$ is given by $$R_{\mathcal{S}_{Ne,n}^*}(\mathcal{M}_n(\beta)) = (3\beta - 2)^{-1/n}.$$ The result is sharp for the function $f(z) = z(1-z^n)^{2(\beta-1)/n}$ .

Definition 1.1. Let $S^*(\varphi)$ denote the class of functions characterized as $$S^*(\varphi) := \{ f \in \mathcal{A} : \mathcal{Q}_f(z) \prec \varphi(z) \}, \tag{1.2}$$ where the analytic function $\varphi: \mathbb{D} \to \mathbb{C}$ is required to satisfy the following conditions: (i) $\varphi(z)$ is univalent with $\operatorname{Re}(\varphi) > 0$ , ²⁰¹⁰ Mathematics Subject Classification. 30C45, 30C80. Key words and phrases. Starlike functions, Subordination, Radius problem, Bernoulli and Booth lemniscates, Cardioid, Nephroid. - (ii) $\varphi(\mathbb{D})$ is starlike with respect to $\varphi(0) = 1$ , - (iii) $\varphi(\mathbb{D})$ is symmetric about the real line, and - (iv) $\varphi'(0) > 0$ . Evidently, for every such $\varphi$ , $\mathcal{S}^(\varphi)$ is always a subclass of the class of starlike functions $\mathcal{S}$ , and $\mathcal{S}^(\varphi) = \mathcal{S}$ for $\varphi(z) = (1+z)/(1-z)$ . We call $\mathcal{S}^(\varphi)$ as the Ma-Minda type starlike class associated with $\varphi$ . Specializing the function $\varphi$ in (1.2) yields several important function classes. For instance, $\mathcal{S}^(\alpha) := \mathcal{S}^\left(\frac{1+(1-2\alpha)z}{1-z}\right)$ , the Janowski class $\mathcal{S}^[A,B] :=$ $\mathcal{S}^*\left(\frac{1+Az}{1+Bz}\right)$ , where $-1 \leq B < A \leq 1$ , and many more. This interesting approach of constructing starlike classes has attracted many researchers, as a result, several Ma-Minda type starlike classes have been introduced and studied recently by modifying the right side function $\varphi$ in (1.2) appropriately. Below we mention a few of the recently introduced Ma-Minda type classes which will be later used in our discussion. - (1) $\mathcal{S}_L^ := \mathcal{S}^(\sqrt{1+z})$ . This class was first introduced by Sokół and Stankiewicz [23], and later discussed extensively by many authors, see [1, 3, 24, 25] and references therein. The function $\varphi_L(z) := \sqrt{1+z}$ maps $\mathbb D$ onto the region bounded by the right-loop of lemniscate of Bernoulli $|w^2 - 1| = 1$ . - (2) $\mathcal{S}_{RL}^ := \mathcal{S}^(\varphi_{RL})$ with $$\varphi_{RL}(z) := \sqrt{2} - (\sqrt{2} - 1)\sqrt{\frac{1 - z}{1 + 2(\sqrt{2} - 1)z}}, \qquad z \in \mathbb{D},$$ (1.3) was introduced by Mendiratta et al. [14]. The function $\varphi_{RL}(z)$ maps $\mathbb{D}$ univalently onto the interior of left-loop of shifted lemniscate of Bernoulli $|(w-\sqrt{2})^2-1|=1$ . - (3) $\mathcal{S}_{\mathcal{C}}^ := \mathcal{S}^(z + \sqrt{1+z^2})$ . Introduced by Raina and Sokół [17], the analytic function $\varphi_{\mathbb{C}}(z) := z + \sqrt{1+z^2}$ sends $\mathbb{D}$ onto a crescent shaped region. - (4) $S_e^ := S^(e^z)$ was considered by Mendiratta et al. [15]. - (5) $\mathcal{S}_C^{\varepsilon} := \mathcal{S}^(1+4z/3+2z^2/3)$ was introduced by Sharma et al. [21]. The function $\varphi_C(z) := 1 + 4z/3 + 2z^2/3$ sends $\mathbb{D}$ onto a region bounded by a cardioid. (6) $\mathcal{S}_R^ := \mathcal{S}^*(\varphi_0)$ , where $\varphi_0(z)$ is the rational function $$\varphi_0(z) := 1 + \frac{z}{k} \left( \frac{k+z}{k-z} \right) = 1 + \frac{1}{k} z + \frac{2}{k^2} z^2 + \cdots, \quad k = \sqrt{2} + 1.$$ - The class $\mathcal{S}_R$ was introduced by Kumar and Ravichandran [11]. (7) $\mathcal{BS}^(\alpha) := \mathcal{S}^(1+z/(1-\alpha z^2)), \ 0 \le \alpha < 1$ , is the class associated with the Booth* lemniscate which was introduced by Kargar et al. [9]. - (8) $\mathcal{S}_S^ := \mathcal{S}^(1 + \sin z)$ was introduced by Cho et al. [5]. The boundary of $\varphi_S(\mathbb{D})$ , where $\varphi_s(z) := 1 + \sin z$ , is an eight-shaped curve. - (9) For $0 < \alpha < 1$ , Khatter et al. [10] introduced $$\mathcal{S}_{\alpha,e}^ := \mathcal{S}^ \left( \alpha + (1-\alpha)e^z \right) \text{ and } \mathcal{S}_L^(\alpha) := \mathcal{S}^ \left( \alpha + (1-\alpha)\sqrt{1+z} \right).$$ For $\alpha = 0$ , the above classes reduce to $\mathcal{S}_e$ and $\mathcal{S}_L$ , respectively. (10) Very recently, Wani and Swaminathan [27] introduced the function class $$\mathcal{S}_{Ne}^ := \mathcal{S}^ \left( 1 + z - z^3 / 3 \right).$$ The authors proved that the analytic function $\varphi_{Ne}(z) := 1 + z - z^3/3$ maps $\mathbb{D}$ univalently onto the interior of a 2-cusped kidney-shaped curve called nephroid given by $$\left( (u-1)^2 + v^2 - \frac{4}{9} \right)^3 - \frac{4v^2}{3} = 0.$$ (1.4) Apart from studying several characteristic properties of the region bounded by the nephroid curve (1.4) (see Figure 1), the authors in [27] discussed certain inclusion relations, coefficient problems, and a few subordination results related to the function class $\mathcal{S}_{Ne}^*$ .  Figure 1. Nephroid: $((u-1)^2 + v^2 - 4/9)^3 - 4v^2/3 = 0$ .

Definition 2.1. Let $\mathbb{D}_r := \{z : |z| < r\}$ and $\Omega_{\varphi} := \varphi(\mathbb{D})$ . By $\mathcal{S}^(\varphi)$ -radius for the class $\mathcal{G} \subset \mathcal{A}$ , denoted by $R_{\mathcal{S}^_{\varphi}}(\mathcal{G})$ , we mean the largest number $\rho \in (0,1)$ such that each $f \in \mathcal{G}$ satisfies $$Q_f(\mathbb{D}_r) \subset \Omega_{\varphi}$$ for every $r \leq \rho$ , where $Q_f(z)$ is defined in (1.1). Moreover, the number $\rho$ is said to be sharp if we can find a function $f_0 \in \mathcal{G}$ such that $$Q_{f_0}(\partial \mathbb{D}_{\rho}) \cap \partial \Omega_{\varphi} \neq \emptyset.$$ For certain recent results in this direction, we provide the following table along with the respective references. Also, see [1, 2, 10, 18] and the references therein. | $R_{\mathcal{S}_{\varphi}^*}(\mathcal{G})$ | ρ | Approx. value | Extremal function | Refer | |----------------------------------------------------------------|-------------------------------------------------|---------------|----------------------------------------------------|-------| | $R_{\mathcal{S}_{\mathbb{Q}}}\left(\mathcal{S}^\right)$ | $\sqrt{2}-1$ | 0.41421 | $\frac{z}{(1-z)^2}$ | [6] | | $R_{\mathcal{S}_{RL}}(\mathcal{S}^)$ | $\sqrt{(\sqrt{5}-2)(5\sqrt{2}-7)}$ | 0.129525 | " | [14] | | $R_{\mathcal{S}_e}(\mathcal{S}^)$ | $\frac{e-1}{e+1}$ | 0.462117 | " | [15] | | $R_{\mathcal{S}_{\mathbb{Q}}^{*}}\left(\mathcal{C}\right)$ | $2-\sqrt{2}$ | 0.58578 | $\frac{z}{1-z}$ | [6] | | $R_{\mathcal{S}_e^*}(\mathcal{C})$ | $\frac{e-1}{e}$ | 0.632121 | " | [15] | | $R_{\mathcal{S}_S}(\mathcal{S}_L^)$ | $(2-\sin 1)\sin 1$ | 0.975 | $\frac{4z\exp(2\sqrt{1+z}-2)}{1+\sqrt{1+z}}$ | [5] | | $R_{\mathcal{S}_R}(\mathcal{S}_L^)$ | $8\sqrt{2} - 11$ | 0.313708 | " | [11] | | $R_{\mathcal{S}^(\alpha)}(\mathcal{S}_L^)$ | $1-\alpha^2$ | | " | [24] | | $R_{\mathcal{S}_{\mathbb{Q}}}\left(\mathcal{S}_{e}^\right)$ | $\log\left(\frac{\sqrt{2}+1}{2}\right)$ | 0.188226 | $z \exp\left(\int_0^z \frac{e^t - 1}{t} dt\right)$ | [6] | | $R_{\mathcal{S}_S}(\mathcal{S}_C^)$ | $\frac{1}{2}\left(\sqrt{2(2+3\sin 1)}-2\right)$ | 0.504 | $z\exp\left(\frac{4z}{3} + \frac{z^2}{3}\right)$ | [5] | | $R_{\mathcal{S}_L}(\mathcal{S}_C^)$ | $-1 + \sqrt{-\frac{1}{2} + \frac{3}{\sqrt{2}}}$ | 0.273311 | " | [21] | | $R_{\mathcal{BS}^(\alpha)}(\mathcal{S}_S^)$ | $\sinh^{-1}\left(\frac{1}{1+\alpha}\right)$ | | $z \exp\left(\int_0^z \frac{\sin t}{t} dt\right)$ | [4] | | $R_{\mathcal{S}_R}(\mathcal{S}_S^)$ | $\log(3 - 2\sqrt{2}) + \sqrt{6}(\sqrt{2} - 1)$ | 0.170742 | " | [11] | Table 2. Some recent radii results Motivated by the aforementioned works, in this work, we consider the function class $\mathcal{S}_{Ne}$ introduced by the authors in [27] and find sharp $\mathcal{S}_{Ne}$ -radii estimates for certain geometrically defined function classes available in the literature. More explicitly, in Section 3, the $\mathcal{S}_{Ne}$ -radius problem is discussed for some of the recently introduced Ma-Minda type starlike classes. In particular, the Janowski class $\mathcal{S}^[A, B]$ is considered, and consequently, the best possible $\mathcal{S}_{Ne}$ -radii for $\mathcal{S}$ and $\mathcal{C}$ are obtained as special cases. Finally, in Section 4, we consider several function families defined by the ratio of analytic functions and calculate $\mathcal{S}_{Ne}^*$ -radius for them, where $$\mathcal{S}_{Ne,n}^ := \mathcal{S}_{Ne}^ \cap \mathcal{A}_n,$$ and $\mathcal{A}_n$ consists of analytic functions of the form $f(z) = z + \sum_{k=n+1}^{\infty} a_k z^k$ . Apart from proving the sharpness of radii estimates analytically, certain graphical illustrations are also provided. The following lemma related to the largest disks contained in the region bounded by the nephroid curve (1.4) is instrumental in proving our results.