Starlike Functions associated with a Petal Shaped Domain

Theorem 2.1. The function $\rho(z) = 1 + \sinh^{-1}(z)$ is a convex univalent function. Proof. Let $h(z) = \sinh^{-1}(z)$ . Clearly, h(0) = 0. Since $h'(z) = 1/\sqrt{1+z^2}$ and $\sqrt{1+z^2} \prec \sqrt{1+z} \in \mathcal{P}$ , where $\mathcal{P}$ is the Carathéodory class. Therefore, $1/\sqrt{1+z^2} \in \mathcal{P}$ which implies that $\operatorname{Re} h'(z) > 0$ . Hence $\rho$ is univalent. Now a calculation yields $$1 + \frac{zh''(z)}{h'(z)} = \frac{1}{1+z^2}.$$ Since $$\frac{1}{1+z^2} \prec \frac{1}{1+z} \in \mathcal{P}.$$ Therefore, $\operatorname{Re}(1+zh''(z)/h'(z))>0$ which implies that h (and thus $\rho$ ) is a convex univalent function. Remark 2.1. Note that $\rho'(0) > 0$ and the function $\varphi(z) = z + \sqrt{1+z^2}$ satisfies $\varphi(\bar{z}) = \overline{\varphi(z)}$ . Therefore, $\rho(\bar{z}) = \overline{\rho(z)}$ and hence, the domain $\Omega_{\rho} = \rho(\mathbb{D})$ is symmetric about the real axis.

Theorem 2.2. The domain $\Omega_{\rho}$ is symmetric about the line Re(w) = 1. Proof. Since $\Omega_{\rho}$ is symmetric about the real axis, the condition $0 \leq \theta \leq \pi/2$ is sufficient to prove our result. As we know that symmetry along imaginary axis for $f \in \mathcal{A}$ holds if $\operatorname{Re}(f(\theta)) = -\operatorname{Re}(f(\pi-\theta))$ and $\operatorname{Im}(f(\theta)) = \operatorname{Im}(f(\pi-\theta))$ . Now let $h(z) = \sinh^{-1}(z) = \ln(z + \sqrt{1+z^2})$ . Then $\operatorname{Im}(h(z)) = \arg(z + \sqrt{1+z^2})$ . For $z = re^{it}$ , $t \in [0,\pi]$ and fixed $r \in (0,1)$ , we have the following expressions for $t \to \theta$ $$I_1 = \arg \left( r(\cos \theta + i \sin \theta) + \sqrt{1 + r^2(\cos(2\theta) + i \sin(2\theta))} \right)$$ = $\arg \left( z + \sqrt{1 + z^2} \right)$ , and for $t \to \pi - \theta$ $$I_{2} = \arg \left( r(\cos(\pi - \theta) + i\sin(\pi - \theta)) + \sqrt{1 + r^{2}(\cos(2(\pi - \theta)) + i\sin(2(\pi - \theta)))} \right)$$ $$= \arg \left( r(-\cos\theta + i\sin\theta) + \sqrt{1 + r^{2}(\cos(2\theta) - i\sin(2\theta))} \right)$$ $$= \arg \left( -\overline{z} + \sqrt{1 + \overline{z}^{2}} \right).$$ Now let us consider $(z + \sqrt{1+z^2})/(-\overline{z} + \sqrt{1+\overline{z}^2})$ . On rationalising the denominator, we get $$\frac{z + \sqrt{1 + z^2}}{-\overline{z} + \sqrt{1 + \overline{z}^2}} = \frac{(z + \sqrt{1 + z^2})(-z + \sqrt{1 + z^2})}{(-\overline{z} + \sqrt{1 + \overline{z}^2})(-z + \sqrt{1 + z^2})} = \frac{1}{|-z + \sqrt{1 + z^2}|^2} = k > 0,$$ where k is some real positive constant. Thus, $$\arg\left(\frac{z+\sqrt{1+z^2}}{-\overline{z}+\sqrt{1+\overline{z}^2}}\right) = \arg(k) = 0$$ $$\Rightarrow \arg\left(z+\sqrt{1+z^2}\right) = \arg\left(-\overline{z}+\sqrt{1+\overline{z}^2}\right)$$ $$\Rightarrow I_1 = I_2.$$ Similarly, $\operatorname{Re}(h(\theta)) = -\operatorname{Re}(h(\pi - \theta))$ for $0 \le \theta \le \pi/2$ . Hence, h(z) is symmetric about the imaginary axis and thus, by translation property, $\rho(z)$ is symmetric about the line $\operatorname{Re}(w) = 1$ . <span id="page-3-0"></span> FIGURE 1. $\rho(\mathbb{D})$ lies in the annular region bounded between the circles C1 and C2. Now using Theorem 2.2, we obtain the next result: <span id="page-3-2"></span>Corollary 2.1. The disk $\{w: |w-1| \leq \sinh^{-1}(r)\}\$ is contained in $\rho(|z| \leq r)$ and is maximal. Proof. Since $\min_{|z|=r} |\sinh^{-1}(z)| = |\sinh^{-1}(-r)| = \sinh^{-1}(r)$ and hence the conclusion can be drawn at once.

Theorem 2.3. We find that the following properties hold for $\rho(z) = 1 + \sinh^{-1}(z)$ : $$\begin{array}{l} (i) \ \rho(-r) \leq \operatorname{Re} \rho(z) \leq \rho(r) \quad (|z| \leq r < 1); \\ (ii) \ |\operatorname{Im} \rho(z)| \leq \pi/2 \quad (|z| \leq 1); \\ (iii) \ \rho(-r) \leq |\rho(z)| \leq \rho(r) \quad (|z| \leq r < 1); \\ (iv) \ |\operatorname{arg} \rho(z)| \leq \tan^{-1}(1/t) \ where \ t = \frac{4}{\pi} \sqrt{\sinh^{-1}(1)(1-\sinh^{-1}(1))}. \end{array}$$ Proof. (i) Since $\rho(z)$ is convex and typically real, the value of Re $\rho(z)$ falls between $\lim_{\theta\to 0} \rho(re^{\theta})$ and $\lim_{\theta\to \pi} \rho(re^{\theta})$ , thus the result follows. (ii) Using Theorem 2.2, it suffices to take $\theta \in [0, \pi/2]$ . Then the inequality follows by letting r tending to 1^- and observing that the function $$\operatorname{Im} \rho(z) = \operatorname{arg} \left( r \cos(\theta) + \sqrt{1 + r^2(\cos(2\theta) + i \sin(2\theta))} + ir \sin(\theta) \right)$$ is strictly increasing in the interval $[0, \pi/2]$ and hence the result follows at once. (iii) The radially farthest and nearest points in $\rho(\mathbb{D})$ from origin are respectively B and A (see Figure 1) and therefore the result obviously holds. Moreover we observe that these points A and B lie on the real line and hence the bounds of $|\rho(z)|$ and $\operatorname{Re} \rho(z)$ coincide. The proof of (iv) is evident from Theorem 3.1(iii) so skipped here. Next we have the following important result:

Lemma 2.1. For $1 - \sinh^{-1}(1) < a < 1 + \sinh^{-1}(1)$ , let $r_a$ be given by $$r_a = \begin{cases} a - (1 - \sinh^{-1}(1)), & 1 - \sinh^{-1}(1) < a \le 1; \\ 1 + \sinh^{-1}(1) - a, & 1 \le a < 1 + \sinh^{-1}(1). \end{cases}$$ Then $\{w : |w - a| < r_a\} \subset \Omega_{\rho}$ . We omit the proof of Lemma 2.1 as it directly follows from Theorem 2.2 and Corollary 2.1. Remark 2.2. Evidently the domain $\Omega_{\rho}$ is contained inside the disk $\{w: |w-1| < \pi/2\}$ .

Theorem 3.1. The class $S_{\rho}^*$ satisfies the following relationships: - (i) $S_{\rho}^ \subset S_{\alpha}^ \subset S^*$ for $0 \le \alpha \le 1 \sinh^{-1}(1)$ ; - (ii) $\mathcal{S}_{\varrho}^* \subset M(\beta)$ for $\beta \geq 1 + \sinh^{-1}(1)$ ; (iii) $$\mathcal{S}_{\rho}^ \subset \mathcal{SS}^(\gamma)$$ for $(2/\pi) \tan^{-1}(1/t) \leq \gamma \leq 1$ where $t = \frac{4}{\pi} \sqrt{\sinh^{-1}(1)(1-\sinh^{-1}(1))}$ ; (iv) $$k - \mathcal{ST} \subset \mathcal{S}_o^*$$ for $k \ge 1 + 1/\sinh^{-1}(1)$ . Proof. Consider $f \in \mathcal{S}_{\rho}^*$ which implies $zf'(z)/f(z) \prec 1 + \sinh^{-1}(z)$ . By Theorem 2.3, it is evident that for $z \in \mathbb{D}$ , $$1 - \sinh^{-1}(1) = \min_{|z|=1} \operatorname{Re}(1 + \sinh^{-1}(z)) \le \operatorname{Re} \frac{zf'(z)}{f(z)}$$ and $$\operatorname{Re} \frac{zf'(z)}{f(z)} \le \max_{|z|=1} \operatorname{Re}(1+\sinh^{-1}(z)) = 1+\sinh^{-1}(1).$$ This proves (i) and (ii). For (iii), let $w \in \mathbb{C}$ , X = Re(w), Y = Im(w), and $b = 1-\sinh^{-1}(1)$ . Now consider the parabolic domain $\Gamma_P$ with the boundary curve $\partial \Gamma_P = \gamma_p : Y^2 = 4a(X - b)$ . Then the focus a of the smallest parabola $\gamma_p$ which contains $\Omega_\rho$ will touch the peak points $1 \pm i\pi/2$ of $\mathcal{S}_\rho^*$ is $\pi^2/(16\sinh^{-1}(1))$ . Let P be any point on the parabola $\gamma_P$ with parametric coordinates $(b+at^2,2at)$ such that the tangent OE at P passes through origin for some parameter t. Let the equation of the tangent OE be y = mx, where m = dy/dx = (dy/dt)/(dx/dt) = 1/t. Therefore at P, we have $$m = \frac{y}{x} \Rightarrow \frac{1}{t} = \frac{2at}{b + at^2},$$ which yields <span id="page-4-1"></span> $$t = \sqrt{\frac{b}{a}} = \frac{4}{\pi} \sqrt{\sinh^{-1}(1)(1 - \sinh^{-1}(1))}$$ (3.1) and the argument of the tangent at P of $\gamma_p$ is $\tan^{-1}(1/t)$ . Since $\Omega_\rho \subset \Gamma_p$ , it gives $$\left|\arg\frac{zf'(z)}{f(z)}\right| \le \max_{|z|=1} \arg(\rho(z)) = \max \arg(\gamma_p) = \tan^{-1}(1/t),$$ which demonstrates $f \in \mathcal{SS}^*((2/\pi)\tan^{-1}(1/t))$ , where t is given by (3.1).  <span id="page-5-0"></span>FIGURE 2. Boundary curves, depicting some inclusion relations for $w = 1 + \sinh^{-1}(z)$ . To show (iv), consider $f \in k - \mathcal{ST}$ along with the conic domain $\Gamma_k = \{w \in \mathbb{C} : \operatorname{Re} w > k|w-1|\}$ . For k > 1, let $\partial \Gamma_k$ represent the horizontal ellipse $\gamma_k : x^2 = k^2(x-1)^2 + k^2y^2$ which may be rewritten as $$\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1,$$ where $x_0 = k^2/(k^2 - 1)$ , $y_0 = 0$ , $a = k/(k^2 - 1)$ and $b = 1/\sqrt{k^2 - 1}$ . For $\gamma_k \subset \Omega_\rho$ , the condition $x_0 + a \leq 1 + \sinh^{-1}(1)$ must hold, or equivalently $k \geq 1 + 1/\sinh^{-1}(1)$ . Since $\Gamma_{k_1} \subseteq \Gamma_{k_2}$ for $k_1 \geq k_2$ , it follows that for $k \geq 1 + 1/\sinh^{-1}(1)$ , $k - \mathcal{ST} \subset \mathcal{S}_\rho^*$ . Figure 2 clearly depicts these relations. For our next result, we consider $\mathcal{P}_n[C,D]$ , the class of functions p(z) of the form $1+\sum_{k=n}^{\infty}c_kz^k$ , satisfying $p(z) \prec (1+Cz)/(1+Dz)$ , where $-1 \leq D < C \leq 1$ . Denote by $\mathcal{P}_n(\alpha) := \mathcal{P}_n[1-2\alpha,-1]$ and $\mathcal{P}_n := \mathcal{P}_n(0)$ . For n=1, $\mathcal{P} = \mathcal{P}_1$ is the Carathéodory class. We need the following lemmas:

Lemma 3.1. [18] For $p \in \mathcal{P}_n(\alpha)$ , we have $$\left| \frac{zp'(z)}{p(z)} \right| \le \frac{2(1-\alpha)nr^n}{(1-r^n)(1+(1-2\alpha)r^n)}, \ (|z|=r).$$

Lemma 3.2. [16] For $p \in \mathcal{P}_n[C, D]$ , we have $$\left| p(z) - \frac{1 - CDr^{2n}}{1 - D^2r^{2n}} \right| \le \frac{(C - D)r^n}{1 - D^2r^{2n}}, \ (|z| = r).$$ Especially, for $p \in \mathcal{P}_n(\alpha)$ , we have $$\left| p(z) - \frac{1 + (1 - 2\alpha)r^{2n}}{1 - r^{2n}} \right| \le \frac{2(1 - \alpha)r^n}{1 - r^{2n}}, \ (|z| = r).$$

Theorem 3.2. Let $-1 < D < C \le 1$ . If either of the following two conditions holds: (i) $$(1 - \sinh^{-1}(1))(1 - D^2) < 1 - CD \le 1 - D^2$$ and $C - D \le (1 - D)\sinh^{-1}(1)$ ; (ii) $$1 - D^2 \le 1 - CD < (1 + \sinh^{-1}(1))(1 - D^2)$$ and $C - D \le (1 + D)\sinh^{-1}(1)$ . Then $\mathcal{S}^[C,D] \subset \mathcal{S}^_{\rho}$ . Proof. Let $f \in \mathcal{S}^*[C,D]$ which implies $zf'(z)/f(z) \in \mathcal{P}[C,D]$ . Using Lemma 3.2 we have <span id="page-6-0"></span> $$\left| \frac{zf'(z)}{f(z)} - \frac{1 - CD}{1 - D^2} \right| \le \frac{(C - D)}{1 - D^2}.$$ (3.2) Let $a=(1-CD)/(1-D^2)$ and assume that (i) holds. Now multiplying 1+D and dividing by $(1-D^2)$ on either sides of the inequality $(C-D) \leq (1-D)\sinh^{-1}(1)$ gives $(C-D)/(1-D^2) \leq a-(1-\sinh^{-1}(1))$ on simplification. Also, the inequality $(1-\sinh^{-1}(1))(1-D^2) < 1-CD \leq 1-D^2$ is equivalent to $1-\sinh^{-1}(1) < (1-CD)/(1-D^2) \leq 1$ . Therefore, from (3.2) we find w=zf'(z)/f(z) is contained inside the disk $|w-a| < r_a$ , where $r_a=a-(1-\sinh^{-1}(1))$ and $1-\sinh^{-1}(1) < a \leq 1$ . Hence $f \in \mathcal{S}_{\rho}^*$ by Lemma 2.1. A similar proof can be shown when (ii) holds.

Theorem 4.1. If $f \in \mathcal{S}_{\rho}^*$ , then the following results hold: - (i) For $1 \sinh^{-1}(1) \le \alpha < 1$ , we have $f \in \mathcal{S}_{\alpha}^*$ in $|z| \le \sinh(1 \alpha)$ . - (ii) For $1 < \beta \le 1 + \sinh^{-1}(1)$ , we have $f \in \mathcal{M}(\beta)$ in $|z| \le \sinh(\beta 1)$ . - (iii) For k > 0, we have $f \in k \mathcal{ST}$ in $|z| \le \sinh(1/(k+1))$ . The results are sharp. Proof. Since $f \in \mathcal{S}_{\rho}^*$ , $zf'(z)/f(z) \prec 1 + \sinh^{-1}(z)$ and hence for |z| = r < 1 Theorem 2.3 gives $$1 - \sinh^{-1}(r) \le \operatorname{Re} \frac{zf'(z)}{f(z)} \le 1 + \sinh^{-1}(r),$$ thereby validating the first two parts. Also, the constants $\sinh(1 - \alpha)$ and $\sinh(\beta - 1)$ are optimal for the function $f_0$ given by (1.6). Now to prove (iii), note that $f \in k - \mathcal{ST}$ in |z| < r, if $$\operatorname{Re}(1+\sinh^{-1}(w(z))) \ge k|1+\sinh^{-1}(w(z))-1| = k|\sinh^{-1}(w(z))|.$$ Here w denotes the Schwarz function. Since $\operatorname{Re}(1+\sinh^{-1}(w(z))) \geq 1-\sinh^{-1}(r)$ and $|\sinh^{-1}(w(z))| \leq \sinh^{-1}(r)$ , the inequality $\operatorname{Re}(1+\sinh^{-1}(w(z))) \geq k|\sinh^{-1}(w(z))|$ holds whenever $1-\sinh^{-1}(r) \geq k\sinh^{-1}(r)$ , which implies $r \leq \sinh(1/(1+k))$ . For the function $f_0$ given by (1.6) and for $z_0 = -\sinh(1/(1+k))$ , we have $$\operatorname{Re} \frac{z_0 f_0'(z_0)}{f_0(z_0)} = \operatorname{Re} (1 + \sinh^{-1}(z_0)) = \frac{k}{k+1} = k |\sinh^{-1}(z_0)| = k \left| \frac{z_0 f_0'(z_0)}{f_0(z_0)} - 1 \right|.$$ This concludes the proof. Corollary 4.1. Substituting k = 1 in part (iii) above, we find that $f \in \mathcal{S}_{\rho}^*$ is parabolic starlike [17] in $|z| \leq \sinh(1/2)$ . In the next result, we find the $\mathcal{K}_{\alpha}$ -radius for the class $\mathcal{S}_{\rho}^*$ .

Theorem 4.2. Let $f \in \mathcal{S}_{\rho}^*$ . Then $f \in \mathcal{K}_{\alpha}$ in $|z| < r_{\alpha}$ , where $r_{\alpha}$ is the least positive root of <span id="page-7-0"></span> $$(1 - r^2)\sqrt{1 + r^2} \left(1 - \sinh^{-1}(r)\right) \left(1 - \alpha - \sinh^{-1}(r)\right) - r = 0 \quad (0 \le \alpha < 1). \tag{4.1}$$ Proof. Let $f \in \mathcal{S}_{\rho}^*$ and w be a Schwarz function. Then $zf'(z)/f(z) = 1 + \sinh^{-1}(w(z))$ such that $$1 + \frac{zf''(z)}{f'(z)} = 1 + \sinh^{-1}(w(z)) + \frac{zw'(z)}{(1 + \sinh^{-1}(w(z)))\sqrt{1 + w^2(z)}}$$ which yields $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) \ge \operatorname{Re}\left(1 + \sinh^{-1}(w(z))\right) - \left|\frac{zw'(z)}{(1 + \sinh^{-1}(w(z)))\sqrt{1 + w^2(z)}}\right|.$$ We know for the Schwarz function w, the inequality $|w'(z)| \leq (1 - |w(z)|^2)/(1 - |z|^2)$ holds. Thus we observe that $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) \ge 1 - \sinh^{-1}(|z|) - \frac{|z|(1 - |w(z)|^2)}{(1 - \sinh^{-1}(|z|))(1 - |z|^2)\sqrt{1 + |z|^2}}$$ $$\ge 1 - \sinh^{-1}(|z|) - \frac{|z|}{(1 - \sinh^{-1}(|z|))(1 - |z|^2)\sqrt{1 + |z|^2}}.$$ Now consider the function $q(r) := 1 - \sinh^{-1}(r) - r / \left( (1 - \sinh^{-1}(r))(1 - r^2) \sqrt{1 + r^2} \right)$ . This is a decreasing function in [0,1) with q(0) = 1. Therefore $\text{Re}(1 + zf''(z)/f'(z)) > \alpha$ in $|z| < r_{\alpha} < 1$ , where $r_{\alpha}$ is given as the least positive root of the equation $q(r) = \alpha$ , which is same as (4.1) and hence the result. Remark 4.1. Note for $\alpha = 0$ , $r_0 \approx 0.37198$ which is not sharp, so the result can be further improved. The sharp $\mathcal{K}_0$ -radius for the class $\mathcal{S}_{\rho}^*$ is $r_0 \approx 0.400435$ , which we can guess graphically but a mathematical proof is yet to derive. For our next theorems 4.3 - 4.5, the following subclasses are required: Let $\mathcal{S}_n^[C,D] := \{ f \in \mathcal{A}_n : zf'(z)/f(z) \in \mathcal{P}_n[C,D] \}$ . Also, let $\mathcal{S}_n^(\alpha) := \mathcal{S}_n^[1-2\alpha,-1] = \mathcal{A}_n \cap \mathcal{S}_{\alpha}$ and $\mathcal{S}_{\rho,n}^ := \mathcal{A}_n \cap \mathcal{S}_{\rho}$ . Further, Ali et al. [2] studied the three classes $\mathcal{S}_n := \{ f \in \mathcal{A}_n : f(z)/z \in \mathcal{P}_n \}$ , $\mathcal{S}_n^*[C,D]$ and $$\mathcal{CS}_n(\alpha) := \left\{ f \in \mathcal{A}_n : \frac{f(z)}{g(z)} \in \mathcal{P}_n, \ g \in \mathcal{S}_n^*(\alpha) \right\}.$$ Now we obtain the $\mathcal{S}_{\rho,n}^*$ -radii for the classes defined above.

Theorem 4.3. For the class $S_n$ , the sharp $S_{n,n}^*$ -radius is given by: $$R_{\mathcal{S}_{\rho,n}^*}(\mathcal{S}_n) = \left(\frac{\sinh^{-1}(1)}{n + \sqrt{n^2 + \left(\sinh^{-1}(1)\right)^2}}\right)^{1/n}.$$ Proof. Let $f \in \mathcal{S}_n$ . Define $s : \mathbb{D} \to \mathbb{C}$ by s(z) = f(z)/z. Then $s \in \mathcal{P}_n$ and we can obtain zf'(z)/f(z) - 1 = zs'(z)/s(z) from the above definition of s. Using Lemma 2.1 and Lemma 3.1, the following holds $$\left| \frac{zf'(z)}{f(z)} - 1 \right| = \frac{zs'(z)}{s(z)} \le \frac{2nr^n}{1 - r^{2n}} \le \sinh^{-1}(1),$$ or equivalently $(\sinh^{-1}(1))r^{2n} + 2nr^n - \sinh^{-1}(1) \leq 0$ . Therefore, the $\mathcal{S}_{\rho,n}^*$ -radius of $\mathcal{S}_n$ is the least positive root of $(\sinh^{-1}(1))r^{2n} + 2nr^n - \sinh^{-1}(1) = 0$ for $r \in (0,1)$ . We can verify $\operatorname{Re}(f_0(z)/z) > 0$ holds in $\mathbb{D}$ where $f_0(z) = z(1+z^n)/(1-z^n)$ . Thus $f_0 \in \mathcal{S}_n$ and $zf'_0(z)/f_0(z) = 1 + 2nz^n/(1-z^{2n})$ . Moreover, the result is sharp since at $z = R_{\mathcal{S}_{\rho,n}^*}(\mathcal{S}_n)$ , we obtain $$\frac{zf_0'(z)}{f_0(z)} - 1 = \frac{2nz^n}{1 - z^{2n}} = \sinh^{-1}(1).$$ The proof is complete. Let $\mathcal{F}$ define the class of functions $f \in \mathcal{A}$ satisfying $f(z)/z \in \mathcal{P}$ . The radius of univalence and starlikeness of the class $\mathcal{F}$ is $\sqrt{2}-1$ , as shown in [12]. Corollary 4.2. For the class $\mathcal{F}$ , the $\mathcal{S}_o^*$ -radius is stated as $$R_{\mathcal{S}_{a}^{*}}(\mathcal{F}) = -e + \sqrt{1 + e^{2}} \approx 0.178105.$$

Theorem 4.4. For the class $CS_n(\alpha)$ , the sharp $S_{n,n}^*$ -radius is given by $$R_{\mathcal{S}_{\rho,n}^*}(\mathcal{CS}_n(\alpha)) = \left(\frac{\sinh^{-1}(1)}{n - \alpha + 1 + \sqrt{(n - \alpha + 1)^2 + (\sinh^{-1}(1) + 2(1 - \alpha))\sinh^{-1}(1)}}\right)^{1/n}.$$ Proof. Let $f \in \mathcal{CS}_n(\alpha)$ and $g \in \mathcal{S}_n^*(\alpha)$ . Considering s(z) = f(z)/g(z), clearly indicates $s \in \mathcal{P}_n$ . Also, it gives $$\frac{zf'(z)}{f(z)} = \frac{zs'(z)}{s(z)} + \frac{zg'(z)}{g(z)}.$$ The use of Lemmas (3.1 - 3.2) gives us <span id="page-8-0"></span> $$\left| \frac{zf'(z)}{f(z)} - \frac{1 + (1 - 2\alpha)r^{2n}}{1 - r^{2n}} \right| \le \frac{2(n - \alpha + 1)r^n}{1 - r^{2n}}.$$ (4.2) Considering $(1 + (1 - 2\alpha)r^{2n})/(1 - r^{2n}) \ge 1$ , the relation $f \in \mathcal{S}_{\rho,n}^*$ follows from (4.2) and Lemma 2.1 if the subsequent inequality is true: $$\frac{1 + 2(n - \alpha + 1)r^n + (1 - 2\alpha)r^{2n}}{1 - r^{2n}} \le 1 + \sinh^{-1}(1)$$ or equivalently, $(2-2\alpha+\sinh^{-1}(1))r^{2n}+2(n-\alpha+1)r^n-\sinh^{-1}(1)\leq 0$ holds. Thus, the least positive root of $$(2 - 2\alpha + \sinh^{-1}(1))r^{2n} + 2(n - \alpha + 1)r^{n} - \sinh^{-1}(1) = 0$$ gives the $\mathcal{S}_{a,n}^*$ -radius for the class $\mathcal{CS}_n(\alpha)$ . Next examine the following functions <span id="page-8-1"></span> $$f_0(z) = \frac{z(1+z^n)}{(1-z^n)^{(n+2-2\alpha)/n}}$$ and $g_0(z) = \frac{z}{(1-z^n)^{2(1-\alpha)/n}}$ , (4.3) which implies $f_0(z)/g_0(z) = (1+z^n)/(1-z^n)$ and $zg'_0(z)/g_0(z) = (1+(1-2\alpha)z^n)/(1-z^n)$ . Moreover, it is obvious that $\operatorname{Re}(f_0(z)/g_0(z)) > 0$ and $\operatorname{Re}(zg'_0(z)/g_0(z)) > \alpha$ in the unit disk $\mathbb{D}$ . Hence $f_0 \in \mathcal{CS}_n(\alpha)$ . At $z = R_{\mathcal{S}_{\rho,n}^*}(\mathcal{CS}_n(\alpha))$ , the function $f_0$ defined in (4.3) satisfies $$\frac{zf_0'(z)}{f_0(z)} = \frac{1 + 2(n - \alpha + 1)z^n + (1 - 2\alpha)z^{2n}}{1 - z^{2n}} = 1 + \sinh^{-1}(1),$$ which accomplish sharpness of the result.

Theorem 4.5. For the class $\mathcal{S}_n^[C,D]$ , the $\mathcal{S}_{\rho,n}$ -radius is given by $$R_{\mathcal{S}_{\rho,n}}(\mathcal{S}_n^[C,D]) = \left\{ \begin{array}{ll} \min\{1;R_1\}, & -1 \leq D < 0 < C \leq 1; \\ \min\{1;R_2\}, & 0 < D < C \leq 1, \end{array} \right.$$ where $$R_1 := \left(\frac{2\sinh^{-1}(1)}{C - D + \sqrt{(C - D)^2 + 4(D^2(1 + \sinh^{-1}(1)) - CD)\sinh^{-1}(1)}}\right)^{1/n}$$ and $$R_2 := \left(\frac{2\sinh^{-1}(1)}{C - D + \sqrt{(C - D)^2 + 4(D^2(\sinh^{-1}(1) - 1) + CD)\sinh^{-1}(1)}}\right)^{1/n}.$$ Proof. Let $f \in \mathcal{S}_n^*[C,D]$ . From Lemma 3.2, we have <span id="page-9-1"></span> $$\left| \frac{zf'(z)}{f(z)} - b \right| \le \frac{(C - D)r^n}{1 - D^2 r^{2n}},\tag{4.4}$$ where $b = (1 - CDr^{2n})/(1 - D^2r^{2n})$ , |z| = r, represents the center of the disk. We infer $b \ge 1$ for $-1 \le D < 0 < C \le 1$ . From Lemma 2.1, $f \in \mathcal{S}_{\rho,n}^*$ depends on whether following condition is true: $$\frac{1 + (C - D)r^n - CDr^{2n}}{1 - D^2r^{2n}} \le 1 + \sinh^{-1}(1),$$ which reduces to $$r \le \left(\frac{2\sinh^{-1}(1)}{C - D + \sqrt{(C - D)^2 + 4(D^2(1 + \sinh^{-1}(1)) - CD)\sinh^{-1}(1)}}\right)^{1/n} = R_1.$$ Further, taking D=0, we get b=1. Then (4.4) yields $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le Cr^n, \ (0 < C \le 1).$$ Now applying Lemma 2.1 with a=1 gives $f\in \mathcal{S}_{\rho,n}^*$ , if $r\leq ((\sinh^{-1}(1))/C)^{1/n}$ For $0 < D < C \le 1$ , we have b < 1. Thus, using Lemma 2.1 and (4.4), we have $f \in \mathcal{S}_{\rho,n}^*$ if the following holds: $$\frac{CDr^{2n} + (C-D)r^n - 1}{1 - D^2r^{2n}} \le \sinh^{-1}(1) - 1,$$ or equivalently, if $$r \le \left(\frac{2\sinh^{-1}(1)}{C - D + \sqrt{(C - D)^2 + 4(D^2(\sinh^{-1}(1) - 1) + CD)\sinh^{-1}(1)}}\right)^{1/n} = R_2.$$ This concludes the proof. The next theorem establishes radius results for some well-known classes mentioned earlier.

Theorem 4.6. The sharp $S_{\rho}$ -radii for the classes $S_L$ , $S_{RL}$ , $S_C$ , $S_e$ , $\Delta$ and $\mathcal{BS}^*(\alpha)$ are: (i) $$R_{\mathcal{S}_{o}^{}}(\mathcal{S}_{L}^{}) = \sinh^{-1}(1)(2 - \sinh^{-1}(1)) \approx 0.985928.$$ (ii) $$R_{\mathcal{S}^_{\rho}}(\mathcal{S}^_{RL}) = \frac{\left(2 + (1 + \sqrt{2})\sinh^{-1}(1)\right)\sinh^{-1}(1)}{5 - 3\sqrt{2} + \left(4(\sqrt{2} - 1) + 2\sinh^{-1}(1)\right)\sinh^{-1}(1)} \approx 0.964694.$$ (iii) $$R_{\mathcal{S}_{\rho}}(\mathcal{S}_C^) = \frac{1}{2} \left( \sqrt{2 \left( 2 + 3 \sinh^{-1}(1) \right)} - 2 \right) \approx 0.523831.$$ (iv) $$R_{\mathcal{S}_{\rho}}(\mathcal{S}_e^) = \ln(1 + \sinh^{-1}(1)) \approx 0.632002.$$ $$(v) \ R_{\mathcal{S}_{\rho}^{}}(\Delta^{}) = \frac{\sinh^{-1}(1)(2 + \sinh^{-1}(1))}{2(1 + \sinh^{-1}(1))} \approx 0.674924.$$ (vi) $$R_{\mathcal{S}^_{\rho}}(\mathcal{B}\mathcal{S}^(\alpha)) = \frac{-1 + \sqrt{1 + \alpha \left(2\sinh^{-1}(1)\right)^2}}{2\alpha\sinh^{-1}(1)}, \ \alpha \in [0, 1].$$

Theorem 4.7. For functions in the classes $\mathcal{F}_1$ , $\mathcal{F}_2$ and $\mathcal{F}_3$ , the sharp $\mathcal{S}_{\rho,n}^*$ -radii respectively, are: (i) $$R_{\mathcal{S}_{\rho,n}^*}(\mathcal{F}_1) = \left(\frac{\sqrt{4n^2 + (\sinh^{-1}(1))^2} - 2n}{\sinh^{-1}(1)}\right)^{1/n}$$ . (ii) $R_{\mathcal{S}_{\rho,n}^*}(\mathcal{F}_2) = \left(\frac{\sqrt{9n^2 + 4\sinh^{-1}(1)(n + \sinh^{-1}(1))} - 3n}{2(n + \sinh^{-1}(1))}\right)^{1/n}$ . (iii) $R_{\mathcal{S}_{\rho,n}}(\mathcal{F}_3) = R_{\mathcal{S}_{\rho,n}}(\mathcal{F}_2)$ .