Starlikeness of Analytic Functions with Subordinate Ratios

Lemma 1.2. Every $p(z) \prec e^z$ satisfies the sharp inequalities (1.6) $$e^{-r} \le |p(z)| \le e^r, \quad |z| \le r,$$ and (1.7) $$\left| \frac{zp'(z)}{p(z)} \right| \le \begin{cases} r, & |z| \le r \le \sqrt{2} - 1\\ \frac{(1+r^2)^2}{4(1-r^2)}, & |z| = r \ge \sqrt{2} - 1. \end{cases}$$ PROOF. Let $p(z) \prec e^z$ . Since $p(z) = e^{w(z)}$ for some Schwarz self-map w satisfying $|w(z)| \leq |z|$ , it follows that <span id="page-3-3"></span><span id="page-3-1"></span> $$|p(z)| = e^{\operatorname{Re} w(z)} \le e^{|w(z)|} \le e^{|z|}.$$ The function w also satisfy the sharp inequality (see [4, Corollary, p. 199]) (1.8) $$|w'(z)| \le \begin{cases} 1, & r = |z| \le \sqrt{2} - 1\\ \frac{(1+r^2)^2}{4r(1-r^2)}, & r \ge \sqrt{2} - 1. \end{cases}$$ From zp'(z)/p(z) = zw'(z), we conclude that <span id="page-3-0"></span> $$\left| \frac{zp'(z)}{p(z)} \right| \le \begin{cases} r, & r = |z| \le \sqrt{2} - 1\\ \frac{(1+r^2)^2}{4(1-r^2)}, & r \ge \sqrt{2} - 1. \end{cases}$$ This inequality is sharp for $p(z) = e^z$ and $r = |z| \le \sqrt{2} - 1$ . It is also sharp in the remaining interval for the function $p(z) = e^{w(z)}$ , where w is the extremal function for which equality holds in (1.8). For $f \in \mathcal{T}_2$ , let $p_1(z) = f(z)/g(z)$ and $p_2(z) = g(z)/zp(z)$ . Then $f(z) = zp(z)p_1(z)p_2(z)$ and <span id="page-3-2"></span> $$\left|\frac{zf'(z)}{f(z)} - 1\right| \le \left|\frac{zp'(z)}{p(z)}\right| + \left|\frac{zp'_1(z)}{p_1(z)}\right| + \left|\frac{zp'_2(z)}{p_2(z)}\right|.$$ Since $p, p_1, p_2 \prec e^z$ , estimates (1.7) and (1.9) show that (1.10) $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \begin{cases} 3r, & r = |z| \le \sqrt{2} - 1\\ \frac{3(1+r^2)^2}{4(1-r^2)}, & r \ge \sqrt{2} - 1. \end{cases}$$ for each function $f \in \mathcal{T}_2$ . It also follows from (1.6) that <span id="page-3-4"></span> $$re^{-3r} \le |f(z)| \le re^{3r}$$ holds for each function $f \in \mathcal{T}_2$ , and that these estimates are sharp. In this paper, we shall adopt the commonly used notations for subclasses of A. First, for $0 \le \alpha < 1$ , let $\mathcal{S}^*(\alpha)$ denote the class of starlike functions of order $\alpha$ consisting of functions $f \in \mathcal{A}$ satisfying the subordination $$\frac{zf'(z)}{f(z)} \prec \frac{1 + (1 - 2\alpha)z}{1 - z}.$$ Thus <span id="page-4-1"></span> $$\operatorname{Re} \frac{zf'(z)}{f(z)} > \alpha, \qquad z \in \mathbb{D}.$$ The case $\alpha = 0$ corresponds to the classical functions whose image domains are starlike with respect to the origin. Various other starlike subclasses of $\mathcal{A}$ occurring in the literature can be expressed in terms of the subordination $$\frac{zf'(z)}{f(z)} \prec \varphi(z)$$ for suitable choices of the superordinate function $\varphi$ . When $\varphi: \mathbb{D} \to \mathbb{C}$ is chosen to be $\varphi(z) := (1 + Az)/(1 + Bz), -1 \le B < A \le 1$ , the subclass derived is denoted by $\mathcal{S}^[A,B]$ . Functions $f \in \mathcal{S}^[A,B]$ are known as Janowski starlike. When $\varphi(z) :=$ $1+(2/\pi^2)((\log((1+\sqrt{z})/(1-\sqrt{z})))^2)$ , the subclass is denoted by $\mathcal{S}_p^*$ , and its functions are called parabolic starlike. In Section 2 of this paper, the radius of starlikeness, Janowski starlikeness, and parabolic starlikeness are found for the classes $\mathcal{T}_i$ , with i=1,2. Section 3 deals with the determination of the $\mathcal{G}$ -radius for the class $\mathcal{T}_i$ with i=1,2, for certain other subclasses $\mathcal{G}$ occurring in the literature. These classes are associated with particular choices of the superordinate function $\varphi$ in (1.11). As mentioned earlier, the $\mathcal{G}$ -radius for a given class $\mathcal{M}$ , denoted by $R_{\mathcal{G}}(\mathcal{M})$ , is the largest number R such that $r^{-1}f(rz) \in \mathcal{G}$ for every $0 < r \le R$ and $f \in \mathcal{M}$ . It will become apparent in the forthcoming proofs that there are common features in the methodology of finding the $\mathcal{G}$ -radius for each of these subclasses.

Theorem 2.1. Let $0 \le \alpha < 1$ . The radius of starlikeness of order $\alpha$ for $\mathcal{T}_1$ and $\mathcal{T}_2$ are - <span id="page-4-0"></span>(i) $R_{S^(\alpha)}(\mathcal{T}_1) = R_{S^_{\alpha}}(\mathcal{T}_1) = 2(1-\alpha)/(5-2\alpha),$ (ii) $R_{S^(\alpha)}(\mathcal{T}_2) = R_{S^_{\alpha}}(\mathcal{T}_2) = (1-\alpha)/3.$ PROOF. (i) The function $\sigma(r) = (2-5r)/(2-2r)$ is a decreasing function on [0,1). Further, the number $R_1 := 2(1-\alpha)/(5-2\alpha)$ is the root of the equation $\sigma(r) = \alpha$ . For $f \in \mathcal{T}_1$ and $0 < r \le R_1$ , the inequality (1.5) readily yields Re $$\frac{zf'(z)}{f(z)} \ge 1 - \frac{3r}{2(1-r)} = \frac{2-5r}{2-2r} = \sigma(r) \ge \sigma(R_1) = \alpha$$ and $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{3r}{2(1-r)} = 1 - \sigma(r) \le 1 - \sigma(R_1) = 1 - \alpha.$$ At $z = -R_1$ , the function $f_1 \in \mathcal{T}_1$ given by $f_1(z) = z(1+z)^{3/2}$ yields $$\frac{zf_1'(z)}{f_1(z)} = \frac{2+5z}{2+2z} = \frac{2-5R_1}{2-2R_1} = \alpha.$$ Thus $$\operatorname{Re} \frac{zf_1'(z)}{f_1(z)} = \alpha$$ and $\left| \frac{zf_1'(z)}{f_1(z)} - 1 \right| = 1 - \alpha$ . This proves that the $\mathcal{S}^(\alpha)$ and $\mathcal{S}^_{\alpha}$ radii for $\mathcal{T}_1$ are the same number $R_1$ . (ii) Consider $\omega(r) = 1 - 3r$ , $0 \le r < 1$ . The number $R_2 = (1 - \alpha)/3 < 1/3$ is clearly the root of the equation $\omega(r) = \alpha$ . Since $\omega$ is decreasing, then $\omega(r) \ge \omega(R_2) = \alpha$ for each $f \in \mathcal{T}_2$ and $0 < r \le R_2$ . It follows from (1.10) that Re $$\frac{zf'(z)}{f(z)} \ge 1 - 3r = \omega(r) \ge \alpha$$ , and $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le 3r = 1 - \omega(r) \le 1 - \alpha.$$ Evaluating the function $f_2(z) = ze^{3z}$ at $z = -R_2$ yields $$\frac{zf_2'(z)}{f_2(z)} = 1 - 3R = \alpha.$$ Hence $$\operatorname{Re} \frac{zf_2'(z)}{f_2(z)} = \alpha$$ and $\left| \frac{zf_2'(z)}{f_2(z)} - 1 \right| = 1 - \alpha$ . This proves that the $\mathcal{S}^(\alpha)$ and $\mathcal{S}^_{\alpha}$ radii for the class $\mathcal{T}_2$ are the same number $R_2$ . Next we find the $S^[A, B]$ -radius (Janowski starlikeness) for $\mathcal{T}_1$ and $\mathcal{T}_2$ . Recall that $S^[A, B]$ consists of analytic functions $f \in \mathcal{A}$ satisfying the subordination $zf'(z)/f(z) \prec (1 + Az)/(1 + Bz)$ , $-1 \leq B < A \leq 1$ .

Theorem 2.2. (i) Every $f \in \mathcal{T}_1$ is Janowski starlike in the disc $\mathbb{D}_r = \{z : |z| < r\}$ for $r \leq 2(A - B)/(3(1 + |B|) + 2(A - B))$ . If B < 0, then $R_{\mathcal{S}^*[A,B]}(\mathcal{T}_1) = 2(A - B)/(3 + 2A - 5B)$ . (ii) The radius of Janowski starlikeness for $\mathcal{T}_2$ is $R_{\mathcal{S}^*[A,B]}(\mathcal{T}_2) = (A-B)/(3(1-B))$ . PROOF. Since $S^[A, -1] = S^((1 - A)/2)$ , the results in the case B = -1 follow from Theorem 2.1. We now prove the results when $-1 < B < A \le 1$ . (i) Let $f \in \mathcal{T}_1$ and write w = zf'(z)/f(z). Then (1.5) shows that $|w-1| \le 3r/(2(1-r))$ for $|z| \le r$ . For $0 \le r \le R_1 := 2(A-B)/(3(1+|B|)+2(A-B))$ , then $3R_1/((2(1-R_1))=(A-B)/(1+|B|)$ . For $0 \le r \le R_1$ , we first show that the disc $$\left\{ w : |w - 1| \le \frac{3R_1}{2(1 - R_1)} = \frac{A - B}{1 + |B|} \right\}$$ is contained in the images of the unit disc under the mapping (1 + Az)/(1 + Bz). As $B \neq -1$ , the image is the disc given by $$\left\{ w : \left| w - \frac{1 - AB}{1 - B^2} \right| < \frac{A - B}{1 - B^2} \right\}.$$ Silverman [30, p. 50-51] has shown that the disc $$\{w : |w - c| < d\} \subset \{w : |w - a| < b\}$$ if and only if $|a-c| \le b-d$ . With the choices c = 1, d = (A-B)/(1+|B|), $a = (1-AB)/(1-B^2)$ and $b = (A-B)/(1-B^2)$ , then $|a-c| = |B|(A-B)/(1-B^2) = b-d$ . This proves that $\mathcal{S}^*[A, B]$ radius is at least $R_1$ . To prove sharpness, consider the function $f_1 \in \mathcal{T}_1$ given by $f_1(z) = z(1+z)^{3/2}$ . Evidently, $zf'_1(z)/f_1(z) = (2+5z)/(2+2z)$ . For B < 0, evaluating at $z = -R_1$ , then $zf'_1(z)/f_1(z) = 1 + 3z/(2+2z) = 1 - (A-B)/(1+|B|) = (1-A)/(1-B)$ . This shows that $$\left| \frac{zf_1'(z)}{f_1(z)} - \frac{1 - AB}{1 - B^2} \right| = \left| \frac{1 - A}{1 - B} - \frac{1 - AB}{1 - B^2} \right| = \frac{A - B}{1 - B^2},$$ proving sharpness in the case B < 0. (ii) Let $f \in \mathcal{T}_2$ and w := zf'(z)/f(z). It follows from (1.10) that $|w-1| \leq 3r$ for $|z| \leq r$ . For $0 \leq r \leq R_2 := (A-B)/(3(1+|B|))$ , we see that the disc $\{w : |w-1| \leq 3R_2 = (A-B)/(1+|B|)\}$ is contained in the disc $\{w : |w-(1-AB)/(1-B^2)| < (A-B)/(1-B^2)\}$ , as in the proof of (i). This proves that $\mathcal{S}^*[A, B]$ radius is at least $R_2$ . The result is sharp for the function $f_2 \in \mathcal{T}_2$ given by the function $f_2(z) = ze^{3z}$ . The function $\varphi_{PAR}: \mathbb{D} \to \mathbb{C}$ given by $$\varphi_{PAR}(z) := 1 + \frac{2}{\pi^2} \left( \log \frac{1 + \sqrt{z}}{1 - \sqrt{z}} \right)^2, \quad \operatorname{Im} \sqrt{z} \ge 0,$$ maps $\mathbb{D}$ into the parabolic region $$\varphi_{PAR}(\mathbb{D}) = \{ w = u + iv : v^2 < 2u - 1 \} = \{ w : \operatorname{Re} w > |w - 1| \}.$$ The class $C(\varphi_{PAR}) = \{ f \in \mathcal{A} : 1 + zf''(z)/f'(z) \prec \varphi_{PAR}(z) \}$ is the class of uniformly convex functions introduced by Goodman [7]. The corresponding class $\mathcal{S}_p^ := \mathcal{S}^(\varphi_{PAR}) = \{ f \in \mathcal{A} : zf'(z)/f(z) \prec \varphi_{PAR}(z) \}$ introduced by Rønning [24] is known as the class of parabolic starlike functions. The class $\mathcal{S}_p^*$ consists of functions $f \in \mathcal{A}$ satisfying $$\operatorname{Re}\left(\frac{zf'(z)}{f(z)}\right) > \left|\frac{zf'(z)}{f(z)} - 1\right|, \quad z \in \mathbb{D}.$$ Evidently, every parabolic starlike function is also starlike of order 1/2. The radius of parabolic starlikeness for the class $\mathcal{T}_1$ and $\mathcal{T}_2$ is given in the next result.

Corollary 2.3. The radius of parabolic starlikeness for $\mathcal{T}_1$ and $\mathcal{T}_2$ is respectively equal to its radius of starlikeness of order 1/2. Thus, - (i) $R_{\mathcal{S}_p^*}(\mathcal{T}_1) = 1/4$ , - (ii) $R_{\mathcal{S}_p^*}(\mathcal{T}_2) = 1/6$ . PROOF. Shanmugam and Ravichandran [25, p. 321] proved that $$\{w : |w - a| < a - 1/2\} \subseteq \{w : \operatorname{Re} w > |w - 1|\}$$ for $1/2 < a \le 3/2$ . Choosing a = 1, this implies that $\mathcal{S}_{1/2}^ \subset \mathcal{S}_p$ . Every parabolic starlike function is also starlike of order 1/2, whence the inclusion $\mathcal{S}_{1/2}^ \subset \mathcal{S}_p^ \subset \mathcal{S}^(1/2)$ . Therefore, for any class $\mathcal{F}$ , readily $R_{\mathcal{S}_{1/2}}(\mathcal{F}) \le R_{\mathcal{S}_p}(\mathcal{F}) \le R_{\mathcal{S}_p^(1/2)}(\mathcal{F})$ . When $\mathcal{F} = \mathcal{T}_i$ , i = 1, 2, Theorem 2.1 gives $R_{\mathcal{S}^(\alpha)}(\mathcal{T}_i) = R_{\mathcal{S}^_{\alpha}}(\mathcal{T}_i)$ . This shows that $R_{\mathcal{S}^_{1/2}}(\mathcal{T}_i) = R_{\mathcal{S}^_p}(\mathcal{T}_i) = R_{\mathcal{S}^(1/2)}(\mathcal{T}_i)$ . Since $R_{\mathcal{S}^(1/2)}(\mathcal{T}_1) = 1/4$ and $R_{\mathcal{S}^(1/2)}(\mathcal{T}_2) = 1/6$ from Theorem 2.1, it follows that $R_{\mathcal{S}^_p}(\mathcal{T}_1) = 1/4$ and $R_{\mathcal{S}^*_p}(\mathcal{T}_2) = 1/6$ .

Corollary 3.1. The $\mathcal{S}^*_{\exp}$ -radius for the class $\mathcal{T}_1$ is $$R_{\mathcal{S}_{\text{exp}}^*}(\mathcal{T}_1) = (2 - 2e)/(2 - 5e) \approx 0.296475,$$ while that of $\mathcal{T}_2$ is $$R_{\mathcal{S}_{\text{exp}}^*}(\mathcal{T}_2) = (e-1)/3e.$$ Proof. Mendiratta et al. [16, Lemma 2.2] proved that $$\{w : |w - a| < a - 1/e\} \subseteq \{w : |\log w| < 1\}$$ for $e^{-1} \leq a \leq (e+e^{-1})/2$ , and this inclusion with a=1 gives $\mathcal{S}_{1/e}^ \subset \mathcal{S}_{\exp}$ . It was also shown in [16, Theorem 2.1 (i)] that $\mathcal{S}_{\exp}^ \subset \mathcal{S}^(1/e)$ . Therefore, $\mathcal{S}_{1/e}^ \subset \mathcal{S}_{\exp}^ \subset \mathcal{S}^*(1/e)$ , which, as a consequence of Theorem 2.1, established the result. Corollary 3.2 investigates the radius of cardioid starlikeness for each class $\mathcal{T}_1$ and $\mathcal{T}_2$ . The class $S_C^ := \mathcal{S}^(\varphi_{CAR})$ , where $\varphi_{CAR}(z) = 1 + 4z/3 + 2z^2/3$ in (1.11), was introduced and studied in [21,26–28]. Descriptively, $f \in S_C^*$ provided zf'(z)/f(z) lies in the region bounded by the cardioid $\Omega_C := \{w = u + iv : (9u^2 + 9v^2 - 18u + 5)^2 - 16(9u^2 + 9v^2 - 6u + 1) = 0\}$ . <span id="page-7-0"></span>COROLLARY 3.2. The following are the $S_C^*$ -radius for the classes $\mathcal{T}_1$ and $\mathcal{T}_2$ : - (i) $R_{S_C^*}(\mathcal{T}_1) = 4/13$ , - (ii) $R_{S_C^*}(\mathcal{T}_2) = 2/9$ . PROOF. Sharma et al. [27] proved that $\{w : |w-a| < a-1/3\} \subseteq \Omega_C$ for $1/3 < a \le 5/3$ , and this inclusion with a = 1 gives $\mathcal{S}_{1/3}^ \subset \mathcal{S}_C$ . Thus $R_{\mathcal{S}_{1/3}}(\mathcal{T}_i) \le R_{\mathcal{S}_C}(\mathcal{T}_i)$ for i = 1, 2. To complete the proof, we demonstrate $R_{\mathcal{S}_C}(\mathcal{T}_i) \le R_{\mathcal{S}_{1/3}}(\mathcal{T}_i)$ for i = 1, 2. (i) Evaluating the function $f_1(z)=z(1+z)^{3/2}$ at $z=-R=-R_{S_{1/3}^*}(\mathcal{T}_1)=-4/13$ gives $$\frac{zf_1'(z)}{f_1(z)} = \frac{2+5z}{2+2z} = \frac{2-5R}{2-2R} = \frac{1}{3} = \varphi_{CAR}(-1).$$ Thus, $R_{S_C^*}(\mathcal{T}_1) \leq 4/13$ . (ii) Similarly, at $z = -R = -R_{S_{1/3}^*}(\mathcal{T}_2) = -2/9$ , the function $f_2(z) = ze^{3z}$ yields $$\frac{zf_2'(z)}{f_2(z)} = 1 + 3z = 1 - 3R = \frac{1}{3} = \varphi_{CAR}(-1).$$ This proves that $R_{S_c^*}(\mathcal{T}_2) \leq 2/9$ . In 2019, Cho et al. [3] studied the class $\mathcal{S}_{\sin}^ := \mathcal{S}^(1 + \sin z)$ consisting of functions $f \in \mathcal{A}$ satisfying the condition $zf'(z)/f(z) \prec 1 + \sin z$ . We find the $\mathcal{S}_{\sin}^*$ -radius for the classes $\mathcal{T}_1$ and $\mathcal{T}_2$ .

Corollary 3.3. The following are the $S_{sin}^*$ -radius for each class $\mathcal{T}_1$ and $\mathcal{T}_2$ : - (i) $R_{\mathcal{S}_{cin}^*}(\mathcal{T}_1) = 2(\sin 1)/(3 + 2\sin 1) \approx 0.35938.$ - (ii) $R_{S_{sin}^*}(\mathcal{T}_2) = (\sin 1)/3$ . PROOF. It was proved in [3] that $\{w: |w-a| < \sin 1 - |a-1|\} \subseteq q(\mathbb{D})$ for $|a-1| \le \sin 1$ , where $q(z) := 1 + \sin z$ . For a = 1, this implies that $\mathcal{S}^_{1-\sin 1} \subset \mathcal{S}^_{\sin}$ . Thus $R_{\mathcal{S}^_{1-\sin 1}}(\mathcal{T}_i) \le R_{\mathcal{S}^_{\sin}}(\mathcal{T}_i)$ for i = 1, 2. The proof is completed by demonstrating $R_{\mathcal{S}^_{\sin}}(\mathcal{T}_i) \le R_{\mathcal{S}^_{1-\sin 1}}(\mathcal{T}_i)$ for i = 1, 2. (i) Evaluating the function $f_1(z)=z(1+z)^{3/2}$ at $z=-R=-R_{S_{1-\sin 1}^*}(\mathcal{T}_1)=-2\sin 1/(3+2\sin 1)$ gives $$\frac{zf_1'(z)}{f_1(z)} = \frac{2+5z}{2+2z} = \frac{2-5R}{2-2R} = 1 - \sin 1 = q(-1).$$ Thus, $R_{S_{\text{sin}}^*}(\mathcal{T}_1) \leq 2 \sin 1/(3 + 2 \sin 1)$ . (ii) Similarly, at $z = \pm R = \pm R_{S_{1-\sin 1}^*}(\mathcal{T}_2) = \pm (\sin 1)/3$ , the function $f_2(z) = ze^{3z}$ yields $$\frac{zf_2'(z)}{f_2(z)} = 1 + 3z = 1 \pm 3R = 1 \pm \sin 1 = q(\pm 1).$$ This proves that $R_{\mathcal{S}_{\sin}^*}(\mathcal{T}_2) \leq (\sin 1)/3$ . Consider next the class $\mathcal{S}^_{\mathbb{Q}} := \mathcal{S}^(z + \sqrt{1+z^2})$ introduced by Raina and Sokół in [18]. Functions $f \in \mathcal{S}^*_{\mathbb{Q}}$ provided zf'(z)/f(z) lies in the region bounded by the lune $\Omega_l := \{w : |w^2 - 1| < 2|w|\}$ . The result below gives the radius of lune starlikeness for each class $\mathcal{T}_1$ and $\mathcal{T}_2$ .

Corollary 3.4. The following are the $\mathcal{S}_{\mathcal{A}}^*$ -radius for each class $\mathcal{T}_1$ and $\mathcal{T}_2$ : - (i) $R_{\mathcal{S}_{\mathcal{I}}^*}(\mathcal{T}_1) = 2(\sqrt{2} 2)/(2\sqrt{2} 7) \approx 0.280847.$ - (ii) $R_{\mathcal{S}_{\mathcal{J}}^*}(\mathcal{T}_2) = (2 \sqrt{2})/3.$ PROOF. It was shown by Gandhi and Ravichandran [5, Lemma 2.1] that $\{w: |w-a| < 1 - |\sqrt{2} - a|\} \subseteq \Omega_l$ for $\sqrt{2} - 1 < a \le \sqrt{2} + 1$ . Choosing a = 1, the inclusion gives $\mathcal{S}^_{\sqrt{2}-1} \subset \mathcal{S}^_{\mathbb{C}}$ . Thus $R_{\mathcal{S}^_{\sqrt{2}-1}}(\mathcal{T}_i) \le R_{\mathcal{S}^_{\mathbb{C}}}(\mathcal{T}_i)$ for i = 1, 2. We complete the proof by demonstrating $R_{\mathcal{S}^_{\mathbb{C}}}(\mathcal{T}_i) \le R_{\mathcal{S}^_{\sqrt{2}-1}}(\mathcal{T}_i)$ for i = 1, 2. (i) Evaluating the function $f_1(z) = z(1+z)^{3/2}$ at $z = -R = -R_{S_{\sqrt{2}-1}}(\mathcal{T}_1) = -2(\sqrt{2}-2)/(2\sqrt{2}-7)$ gives $$\left| \left( \frac{zf_1'(z)}{f_1(z)} \right)^2 - 1 \right| = \left| \left( \frac{2+5z}{2+2z} \right)^2 - 1 \right| = \left| \left( \frac{2-5R}{2-2R} \right)^2 - 1 \right| = 0.828 = 2 \left| \frac{zf_1'(z)}{f_1(z)} \right|.$$ Thus, $R_{S_{\mathcal{J}}^*}(\mathcal{T}_1) \leq 2(\sqrt{2}-2)/(2\sqrt{2}-7)$ . (ii) Similarly, at $z=-R=-R_{S_{\sqrt{2}-1}^*}(\mathcal{T}_2)=-(2-\sqrt{2})/3$ , the function $f_2(z)=ze^{3z}$ yields $$\left| \left( \frac{zf_2'(z)}{f_2(z)} \right)^2 - 1 \right| = \left| (1+3z)^2 - 1 \right| = \left| (1-3R)^2 - 1 \right| = 0.828 = 2 \left| \frac{zf_2'(z)}{f_2(z)} \right|.$$ This proves that $R_{S_{\mathcal{C}}^*}(\mathcal{T}_2) \leq (2 - \sqrt{2})/3$ . As a further example, consider next the class $\mathcal{S}_R^ := \mathcal{S}^(\eta(z))$ , where $\eta(z) = 1 + ((zk + z^2)/(k^2 - kz))$ , $k = \sqrt{2} + 1$ . This class associated with a rational function was introduced and studied by Kumar and Ravichandran in [10].

Corollary 3.5. The following are the $\mathcal{S}_R^*$ -radius for the classes $\mathcal{T}_1$ and $\mathcal{T}_2$ : - (i) $R_{\mathcal{S}_{R}^{*}}(\mathcal{T}_{1}) = 2(-3 + 2\sqrt{2})/(4\sqrt{2} 9) \approx 0.102642$ , - (ii) $R_{\mathcal{S}_R^*}(\mathcal{T}_2) = (3 2\sqrt{2})/3.$ PROOF. It was shown in [10] that $\{w: |w-a| < a-2(\sqrt{2}-1)\} \subseteq \eta(\mathbb{D}) \text{ for } 2(\sqrt{2}-1) < a \le \sqrt{2}$ . This inclusion with a=1 gives $\mathcal{S}^_{2(\sqrt{2}-1)} \subset \mathcal{S}^_R$ . Thus $R_{\mathcal{S}^_{2(\sqrt{2}-1)}}(\mathcal{T}_i) \le R_{\mathcal{S}^_R}(\mathcal{T}_i)$ for i=1,2. We next show that $R_{\mathcal{S}^_R}(\mathcal{T}_i) \le R_{\mathcal{S}^_{2(\sqrt{2}-1)}}(\mathcal{T}_i)$ for i=1,2. (i) At $z = -R = -R_{S_{2(\sqrt{2}-1)}^*}(\mathcal{T}_1) = -2(-3+2\sqrt{2})/(4\sqrt{2}-9)$ , the function $f_1(z) = z(1+z)^{3/2}$ yields $$\frac{zf_1'(z)}{f_1(z)} = \frac{2-5R}{2-2R} = 2(\sqrt{2}-1) = \eta(1).$$ Thus, $R_{\mathcal{S}_{R}^{*}}(\mathcal{T}_{1}) \leq 2(-3 + 2\sqrt{2})/(4\sqrt{2} - 9)$ . (ii) Evaluating $f_2(z) = ze^{3z}$ at $z = -R = -R_{S_{2(\sqrt{2}-1)}^*}(\mathcal{T}_2) = -(3-2\sqrt{2})/3$ gives $$\frac{zf_2'(z)}{f_2(z)} = 1 - 3R = 2(\sqrt{2} - 1) = \eta(1).$$ Thus $R_{\mathcal{S}_{R}^{*}}(\mathcal{T}_{2}) \leq (3 - 2\sqrt{2})/3$ . The class $\mathcal{S}_{N_e}^ := \mathcal{S}^(\psi(z))$ , where $\psi(z) = 1 + z - z^3/3$ , was introduced and studied by Wani and Swaminathan in [31]. Geometrically, $f \in \mathcal{S}_{N_e}^*$ provided zf'(z)/f(z) lies in the region bounded by the nephroid: a 2-cusped kidney shaped curve $\Omega_{N_e} := \{w = u + iv : ((u-1)^2 + v^2 - 4/9)^3 - 4v^2/3 = 0\}$ .

Corollary 3.6. The following are the $\mathcal{S}_{N_e}^*$ -radius for the classes $\mathcal{T}_1$ and $\mathcal{T}_2$ : - (i) $R_{\mathcal{S}_{N_e}^*}(\mathcal{T}_1) = 4/13$ , - (ii) $R_{\mathcal{S}_{N_e}^*}(\mathcal{T}_2) = 2/9.$ PROOF. It was shown in [31] that $\{w : |w-a| < a-1/3\} \subseteq \Omega_{N_e}$ for $1/3 < a \le 1$ . This inclusion with a = 1 gives $\mathcal{S}_{1/3}^ \subset \mathcal{S}_{N_e}$ . This shows that $R_{\mathcal{S}_{1/3}}(\mathcal{T}_i) \le R_{\mathcal{S}_{N_e}}(\mathcal{T}_i)$ for i = 1, 2. We next show that $R_{\mathcal{S}_{N_e}}(\mathcal{T}_i) \le R_{\mathcal{S}_{1/3}}(\mathcal{T}_i)$ for i = 1, 2. (i) Evaluating the function $f_1(z)=z(1+z)^{3/2}$ at $z=-R=-R_{S_{1/3}^*}(\mathcal{T}_1)=-4/13$ results in $$\frac{zf_1'(z)}{f_1(z)} = \frac{2-5R}{2-2R} = \frac{1}{3} = \psi(-1).$$ Thus, $R_{S_{N_e}^*}(\mathcal{T}_1) \leq 4/13$ . (ii) Similarly, evaluating $f_2(z) = ze^{3z}$ at $z = -R = -R_{S_{1/3}^*}(\mathcal{T}_2) = -2/9$ yields $$\frac{zf_2'(z)}{f_2(z)} = 1 - 3R = \frac{1}{3} = \psi(-1).$$ This proves that $R_{S_{N_c}^*}(\mathcal{T}_2) \leq 2/9$ . Finally, we consider the class $\mathcal{S}_{SG}^ := \mathcal{S}^(2/(1+e^{-z}))$ introduced by Goel and Kumar in [6]. Here $2/(1+e^{-z})$ is the modified sigmoid function that maps $\mathbb{D}$ onto the region $\Omega_{SG} := \{w = u + iv : |\log(w/(2-w))| < 1\}$ . Thus, $f \in \mathcal{S}_{SG}^*$ provided the function zf'(z)/f(z) maps $\mathbb{D}$ onto the region lying inside the domain $\Omega_{SG}$ .

Corollary 3.7. The $S_{SG}^*$ -radius for the class $\mathcal{T}_1$ is $$R_{\mathcal{S}_{SG}^*}(\mathcal{T}_1) = (2e-2)/(1+5e) \approx 0.23552,$$ while that of $\mathcal{T}_2$ is $$R_{\mathcal{S}_{SG}^*}(\mathcal{T}_2) = (e-1)/(3(1+e)).$$ PROOF. The inclusion $\{w: |w-a| < ((e-1)/(e+1)) - |a-1|\} \subseteq \Omega_{SG}$ holds for 2/(1+e) < a < 2e/(1+e) (see [6]). At a=1, the set inclusion shows that $\mathcal{S}^_{2/(e+1)} \subset \mathcal{S}^_{SG}$ . It was also shown in [6] that $\mathcal{S}^_{SG} \subset \mathcal{S}^(\alpha)$ for $0 \le \alpha \le 2/(e+1)$ . The desired result is now an immediate consequence of Theorem 2.1.