Ahmad Sulaiman Ahmad El-Faqeer, Maisarah Haji Mohd, V. Ravichandran, Shamani Supramaniam · 2020 📄 arXiv →
Let $f$ and $g$ be analytic functions on the open unit disk of the complex plane with $f/g$ belonging to the class $\mathcal{P} $ of functions with positive real part consisting of functions $p$ with $p(0)=1$ and $\operatorname{Re} p(z)>0$ or to its subclass consisting of functions $p$ with $|p(z)-1|<1$. We obtain the sharp radius constants for the function $f$ to be starlike of order $α$, parabolic starlike, etc. when $g/k\in\mathcal{P}$ where $k$ denotes the Koebe function defined by $k(z)=z/(
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Lemma 2.7. [8, Lemma 2.2, p. 202] For $2(\sqrt{2}-1) < a < 2$ , let $r_a$ be given by
$$r_a = \begin{cases} a - 2(\sqrt{2} - 1), & 2(\sqrt{2} - 1) < a \le \sqrt{2} \\ 2 - a, & \sqrt{2} \le a \le 2. \end{cases}$$
Then $\{w: |w-a| < r_a\}$ , where $\Omega_r$ is the image of the disk $\mathbb{D}$ under the function $1 + (zk + z^2)/(k^2 - kz)$ , $k = \sqrt{2} + 1$ .
Proof of Theorem 1.1. Let the function $f \in \Pi_1$ . Then there is a function $g : \mathbb{D} \to \mathbb{C}$ satisfying
$$\operatorname{Re}\left(\frac{f(z)}{g(z)}\right) > 0 \quad \text{and} \quad \operatorname{Re}\left(\frac{(1-z)^2 g(z)}{z}\right) \quad \forall z \in \mathbb{D}.$$
(2.1)
Define functions $p_1, p_2 : \mathbb{D} \to \mathbb{C}$ as the following.
$$p_1(z) = \frac{(1-z)^2 g(z)}{z}$$
and $p_2(z) = \frac{f(z)}{g(z)}$ . (2.2)
By using (2.1) and (2.2), we have $p_1, p_2 \in \mathcal{P}$ , and $f(z) = zp_1(z)p_2(z)/(1-z)^2$ . Then it follows that
$$\frac{zf'(z)}{f(z)} = \frac{zp'_1(z)}{p_2(z)} + \frac{zp'_2(z)}{p_2(z)} + \frac{1+z}{1-z}.$$
(2.3)
The bilinear transformation (1+z)/(1-z) maps the disk $|z| \le r$ onto the disk
$$\left| \frac{1+z}{1-z} - \frac{1-r^2}{1+r^2} \right| \le \frac{2r}{1-r^2}.$$
(2.4)
For $p \in \mathcal{P}(\alpha)$ , we have
$$\left| \frac{zp'(z)}{p(z)} \right| \le \frac{2(1-\alpha)r}{(1-r)(1+(1-2\alpha)r)}, \quad |z| \le r.$$
(2.5)
By using (2.3), (2.4) and (2.5), function f maps disk |z| < r onto disk
$$\left| \frac{zf'(z)}{f(z)} - \frac{1+r^2}{1-r^2} \right| \le \frac{6r}{1-r^2}.$$
(2.6)
From (2.6), it follows that
$$\operatorname{Re} \frac{zf'(z)}{f(z)} \ge \frac{1 - 6r + r^2}{1 - r^2} \ge 0,$$
(2.7)
for all $0 \le r \le 3 - 2\sqrt{2}$ . Therefore, the function $f \in \Pi_1$ is starlike in $|z| \le 3 - 2\sqrt{2} \approx 0.171573$ . Hence, all our radii found here must be less than $3 - 2\sqrt{2}$ .
(1) The number $\rho = R_{\mathcal{S}^}(\alpha)$ , is the smallest positive root of the equation $(1+\alpha)r^2 - 6r + 1 - \alpha = 0$ in [0,1]. For $0 < r \le R_{\mathcal{S}^(\alpha)}$ , from (2.7), it follows that
Re
$$\frac{zf'(z)}{f(z)} \ge \frac{1 - 6r + r^2}{1 - r^2} \ge \frac{1 - 6\rho + \rho^2}{1 - \rho^2} = \alpha$$
.
This shows that the radius of starlikeness of order $\alpha$ is at least $R_{\mathcal{S}^*}(\alpha)$ . To show that it is sharp, consider the function $f_0 \in \Pi_1$ given in (1.1). For this function $f_0$ , we have
$$\frac{zf_0'(z)}{f_0(z)} = \frac{1+6z+z^2}{1-z^2}.$$
At $z = -\rho$ , we have
Re
$$\frac{zf_0'(z)}{f_0(z)} = \frac{1 - 6\rho + \rho^2}{1 - \rho^2} = \alpha$$
,
proving the sharpness of the radius.
(2) We can give a proof using Lemma 2.1 but we give a different proof here. The number $\rho := R_{\mathcal{S}_L}$ is the smallest positive root of the equation $(1 + \sqrt{2})r^2 + 6r + 1 - \sqrt{2} = 0$ in interval (0, 1), and, from (2.6), it is clear that, for $0 \le r \le \rho$ ,
$$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \left| \frac{zf'(z)}{f(z)} - \frac{1+r^2}{1-r^2} \right| + \frac{2r^2}{1-r^2} \le \frac{6r + 2r^2}{1-r^2} \le \frac{6\rho + \rho^2}{1-\rho^2} = \sqrt{2} - 1 \quad (2.8)$$
and
$$\left| \frac{zf'(z)}{f(z)} + 1 \right| \le 2 + \left| \frac{zf'(z)}{f(z)} - 1 \right| \le \sqrt{2} + 1.$$
(2.9)
Thus, from (2.8) and (2.9), it follows that, for $0 \le r \le \rho$ ,
$$\left| \left( \frac{zf'(z)}{f(z)} \right)^2 - 1 \right| = \left| \frac{zf'(z)}{f(z)} - 1 \right| \left| \frac{zf'(z)}{f(z)} + 1 \right| \le (\sqrt{2} + 1)(\sqrt{2} - 1) = 1.$$
For the function $f_0 \in \Pi_1$ given in (1.1), we have, at $z = \rho$ ,
$$\frac{zf_0'(z)}{f_0(z)} = 1 + \frac{6\rho + 2\rho^2}{1 - \rho^2} = \sqrt{2}$$
and so, at $z = \rho$ ,
$$\left| \left( \frac{z f_0'(z)}{f_0(z)} \right)^2 - 1 \right| = 1.$$
This proves the sharpness.
(3) For $\rho := R_{S_P} = (6 - \sqrt{33})/3$ , we have
$$\frac{1}{2} < 1 \le a = \frac{1+r^2}{1-r^2} \le \frac{1+\rho^2}{1-\rho^2} = \frac{3\sqrt{33}-1}{16} \approx 1.0146 < 3/2.$$
Also, for $\rho = R_{\mathcal{S}_P}$ , we have
$$\frac{6\rho}{(1-\rho^2)} \le \frac{1+\rho^2}{1-\rho^2} - \frac{1}{2}$$
and the disk in (2.6) for $r = \rho$ becomes
$$\left| \frac{zf'(z)}{f(z)} - a \right| = \left| \frac{zf'(z)}{f(z)} - \frac{1+\rho^2}{1-\rho^2} \right| \le \frac{1+\rho^2}{1-\rho^2} - \frac{1}{2} = a - \frac{1}{2}.$$
By Lemma 2.2, it follows that the disk in (2.6) lies inside region $\Omega_{PAR}$ . This proves that the radius of parabolic starlikeness is at least $R_{S_P}$ .
The radius is sharp for the function $f_0 \in \Pi_1$ . At the point $z = -\rho = -R_{\mathcal{S}_P}$ , we have
$$\operatorname{Re}\left(\frac{zf_0'(z)}{f_0(z)}\right) = \frac{1 - 6\rho + \rho^2}{1 - \rho^2} = \frac{1}{2} = \frac{6\rho - 2\rho^2}{1 - \rho^2} = \left|\frac{zf_0'(z)}{f_0(z)} - 1\right|.$$
(4) For $e^{-1} < a \le \frac{e + e^{-1}}{2}$ , Lemma 2.3 gives
$$\{w \in \mathbb{C} : |w - a| < a - e^{-1}\} \subseteq \{w \in \mathbb{C} : |\log w| < 1\} =: \Omega_e,$$
(2.10)
For $\rho = R_{\mathcal{S}_e^*}$ , we have
$$e^{-1} < a := \frac{1 + \rho^2}{1 - \rho^2} = \frac{1 + 9e^2}{e(1 + 3\sqrt{1 + 8e^2})} \approx 1.0236 \le \frac{e + e^{-1}}{2} \approx 1.5430$$
and, $\rho$ being smallest positive root of the equation $(1+e)r^2 - 6er + e - 1 = 0$ ,
$$\frac{6\rho}{1-\rho^2} \le \frac{1+\rho^2}{1-\rho^2} - \frac{1}{e} = a - e^{-1}.$$
Consequently, the disk in (2.6) for $r = \rho$ becomes
$$\left| \frac{zf'(z)}{f(z)} - a \right| = \left| \frac{zf'(z)}{f(z)} - \frac{1+\rho^2}{1-\rho^2} \right| \le \frac{1+\rho^2}{1-\rho^2} - \frac{1}{e} = a - e^{-1}.$$
By (2.10) the above disk is inside $\Omega_e$ proving that the $\mathcal{S}_e^$ radius for the class $\Pi_1$ is at least $R_{\mathcal{S}_{e}^{}}$ . The result is sharp for the function $f_{0}$ given in (1.1). Indeed, at $z = -\rho$ where $\rho = R_{\mathcal{S}_{e}^{*}}$ , we have
$$\left| \log \left( \frac{z f_0'(z)}{f_0(z)} \right) \right| = \left| \log \left( \frac{1 - 6\rho + \rho^2}{1 - \rho^2} \right) \right| = 1.$$
(5) For $\frac{1}{3} < a \le \frac{5}{3}$ , by an application of Lemma 2.4, it follows that
$$\left\{ w \in \mathbb{C} : |w - a| < a - \frac{1}{3} \right\} \subseteq \Omega_c, \tag{2.11}$$
where $\Omega_c$ is the domain bounded by the cardioid $\{x+iy: (9x^2+9y^2-18x+5)^2\}$ $-16(9x^2 + 9y^2 - 6x + 1) = 0$ . For $\rho = R_{\mathcal{S}_c^*}$ , we have
$$\frac{1}{3} < a := \frac{1+\rho^2}{1-\rho^2} = \frac{3\sqrt{73}-1}{24} \approx 1.0263 \le \frac{5}{3}$$
and, $\rho$ being the samllest positive root of the equation $2r^2 - 9r + 1 = 0$ ,
$$\frac{6\rho}{1-\rho^2} = \frac{1+\rho^2}{1-\rho^2} - \frac{1}{3}.$$
Therefore, the disk in (2.6) becomes
$$\left| \frac{zf'(z)}{f(z)} - a \right| = \left| \frac{zf'(z)}{f(z)} - \frac{1 + \rho^2}{1 - \rho^2} \right| \le \frac{1 + \rho^2}{1 - \rho^2} - \frac{1}{3} = a - \frac{1}{3}$$
and this disk is inside $\Omega_c$ . This shows that $\mathcal{S}_c^$ radius is at least $R_{\mathcal{S}_c^}$ .
For the function $f_0$ given in (1.1), at $z = \rho = R_{\mathcal{S}_c^*}$ , we have
$$\frac{zf_0'(z)}{f_0(z)} = \frac{1 - 6\rho + \rho^2}{1 - \rho^2} = \frac{1}{3} = \varphi_c(-1) \in \partial \varphi_c(\mathbb{D})$$
where $\varphi_c(z) = 1 + 4z/3 + 2z^2/3$ . (6) For $\rho = R_{\mathcal{S}_{\sin}^*}$ , and $a := (1 + r^2)/(1 - r^2)$ , we have
$$|a-1| = \frac{2\rho^2}{1-\rho^2} \approx 0.13199 < \sin 1 \approx 0.8414.$$
and
$$\frac{6\rho}{1-\rho^2} \le \sin 1 - \frac{2\rho^2}{1-\rho^2}.$$
The disk in (2.6) for $r = \rho$ becomes
$$\left| \frac{zf'(z)}{f(z)} - a \right| = \left| \frac{zf'(z)}{f(z)} - \frac{1 + \rho^2}{1 - \rho^2} \right| \le \sin 1 - \frac{2\rho^2}{1 - \rho^2} = \sin 1 - |1 - a|.$$
Lemma 2.5 shows that the disk in (2.6) is inside $\Omega_s$ where $\Omega_s =: \varphi_s(\mathbb{D})$ is the image of the unit disk $\mathbb{D}$ under the mapping $\varphi_s(z) = 1 + \sin z$ . This proves that the $\mathcal{S}_{\sin}^$ radius is at least $R_{\mathcal{S}_{\sin}^}$ . For the function $f_0$ given in (1.1), with $\rho = R_{\mathcal{S}_{\sin}^*}$ , we have
$$\left(\frac{zf'(z)}{f(z)}\right) = \frac{1 + 6\rho + \rho^2}{1 - \rho^2} = 1 + \sin 1 \in \varphi_s(1) \in \partial \varphi_s(\mathbb{D}).$$
(7) For $\rho = R_{\mathcal{S}_{\mathcal{O}}^*}$ , we have
$$a := \frac{1+\rho^2}{1-\rho^2} \approx 1.0202 \in (\sqrt{2}-1, \sqrt{2}+1)$$
and
$$\frac{1 - 6\rho + \rho^2}{1 - \rho^2} = \sqrt{2} - 1.$$
The disk in (2.6) becomes
$$\left| \frac{zf'(z)}{f(z)} - a \right| \le 1 - |\sqrt{2} - a|$$
and by Lemma 2.6 it lies inside $\{w: |w^2-1| < 2|w|\}$ . This shows that $\mathcal{S}^_{\mathbb{Q}}$ radius is at least $R_{\mathcal{S}^_{\mathbb{Q}}}$ . The sharpness follows as the function $f_0$ defined in (1.1) satisfies, at $z = \rho = R_{\mathcal{S}^*_{\mathbb{Q}}}$ ,
$$\left| \left( \frac{z f_0'(z)}{f_0(z)} \right)^2 - 1 \right| = \left| \left( \frac{1 - 6\rho + \rho^2}{1 - \rho^2} \right)^2 - 1 \right| = 2(\sqrt{2} - 1)$$
$$= 2 \frac{1 - 6\rho + \rho^2}{1 - \rho^2} = 2 \left| \frac{z f_0'(z)}{f_0(z)} \right|.$$
(8) For $\rho = R_{\mathcal{S}_{R}^{*}}$ , we have
$$2(\sqrt{2}-1) < a := \frac{1+\rho^2}{1-\rho^2} \approx 1.00167 \le \sqrt{2} < 2,$$
and
$$\frac{1 - 6\rho + \rho^2}{1 - \rho^2} = 2 - 2\sqrt{2}.$$
The disk (2.6) becomes
$$\left| \frac{zf'(z)}{f(z)} - a \right| < a - 2(\sqrt{2} - 1).$$
(2.12)
By Lemma 2.7, this disk lies inside the domain $\Omega_r$ . This proves that $\mathcal{S}_R^$ radius is at least $R_{\mathcal{S}_R^}$ .
To prove sharpness, consider the function $f_0 \in \Pi_1$ given in (1.1). At $z = -\rho = -R_{\mathcal{S}_R^*}$ , we have
$$\frac{zf'(z)}{f(z)} = \frac{1 - 6\rho + \rho^2}{1 - \rho^2} = 2(\sqrt{2} - 1) = \varphi_r(-1) \in \partial \varphi_r(\mathbb{D})$$
where
$$\varphi_r(z) = 1 + (kz + z^2)/(k^2 - kz), k = \sqrt{2} + 1.$$
Proof of Theorem 1.2. Since |w-1| < 1 is equivalent to Re(1/w) > 1/2, the condition |f(z)/g(z) - 1| < 1 is the same as the condition Re(g(z)/f(z)) > 1/2. Let the function $f \in \Pi_2$ . Let $g : \mathbb{D} \to \mathbb{C}$ be chosen such that
$$\operatorname{Re}\left(\frac{g(z)}{f(z)}\right) > \frac{1}{2} \quad \text{and} \quad \operatorname{Re}\left(\frac{(1-z)^2}{z}g(z)\right).$$
(2.13)
Define $p_1, p_2 : \mathbb{D} \to \mathbb{C}$ as
$$p_1(z) = \frac{(1-z)^2}{z}g(z), \quad p_2(z) = \frac{g(z)}{f(z)}.$$
(2.14)
From (2.13) and (2.14) it follows that $p_1 \in \mathcal{P}$ , and $p_2 \in \mathcal{P}(1/2)$ , and $f(z) = z/(1-z)^2 p_1(z)/p_2(z)$ . A calculation shows that
$$\frac{zf'(z)}{f(z)} = \frac{zp'_1(z)}{p_1(z)} - \frac{zp'_2(z)}{p_2(z)} + \frac{1+z}{1-z}.$$
(2.15)
The bilinear transformation $\omega = (1+z)/(1-z)$ maps the disk $|z| \leq r$ onto disk
$$\left| \frac{1+z}{1-z} - \frac{1+r^2}{1-r^2} \right| \le \frac{2r}{1-r^2}.$$
(2.16)
Recall that for $p \in \mathcal{P}(\alpha)$ , we have
$$\left| \frac{zp'(z)}{p(z)} \right| \le \frac{2(1-\alpha)r}{(1-r)(1+(1-2\alpha)r)}, \quad |z| \le r.$$
(2.17)
Using (2.16) and (2.17) in (2.15), we get
$$\left| \frac{zf'(z)}{f(z)} - \frac{1+r^2}{1-r^2} \right| \le \frac{5r+r^2}{1-r^2}.$$
(2.18)
From (2.18), it follows that
$$\operatorname{Re}\left(\frac{zf'(z)}{f(z)}\right) \ge \frac{1-5r}{1-r^2} \ge 0 \tag{2.19}$$
for $0 \le r \le 1/5$ . For the function $f_3$ given in (1.3), we have
$$\frac{zf_3'(z)}{f_3(z)} = \frac{1+5z}{1-z^2} = 0$$
for z = -1/5. Thus, the radius of starlikeness of the class $\Pi_2$ is 1/5. All radius values to be computed here will be less than 1/5.
(1) The number $\rho := R_{\mathcal{S}^}(\alpha)$ is the smallest positive root of the equation $\alpha r^2 - 5r + 1 - \alpha = 0$ . For $0 < r \le R_{\mathcal{S}^}(\alpha)$ , from (2.19), we have
$$\operatorname{Re}\left(\frac{zf'(z)}{f(z)}\right) \ge \frac{1-5r}{1-r^2} \ge \frac{1-5\rho}{1-\rho^2} = \alpha.$$
For the function $f_2 \in \Pi_2$ given in (1.3), we have, at $z = -\rho = -R_{\mathcal{S}^*}(\alpha)$ ,
$$\frac{zf_3'(z)}{f_3(z)} = \frac{1 - 5\rho}{1 - \rho^2} = \alpha.$$
This proves that the radius of starlikeness of order $\alpha$ is $R_{\mathcal{S}^*}(\alpha)$ .
(2) From (2.18), it follows that
$$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \left| \frac{zf'(z)}{f(z)} - \frac{1+r^2}{1-r^2} \right| + \frac{2r^2}{1-r^2} \le \frac{5r+3r^2}{1-r^2}. \tag{2.20}$$
The number $\rho = R_{\mathcal{S}_L^}$ , is the positive root of the equation $5r + 3r^2 - (1 - r^2)(\sqrt{2} - 1) = 0$ . For $0 < r \le \rho = R_{\mathcal{S}_I^}$ , we have
$$\frac{5r+3r^2}{1-r^2} \le \frac{5\rho+3\rho^2}{1-\rho^2} = \sqrt{2}-1. \tag{2.21}$$
Therefore, by (2.20), (2.21), and for $0 < r \le \rho = R_{\mathcal{S}_L^*}$ , it follows that
$$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \sqrt{2} - 1,$$
(2.22)
and
$$\left| \frac{zf'(z)}{f(z)} + 1 \right| \le \sqrt{2} + 1.$$
(2.23)
The last two inequalities immediately yields
$$\left| \left( \frac{zf'(z)}{f(z)} \right)^2 - 1 \right| \le \left| \frac{zf'(z)}{f(z)} + 1 \right| \left| \frac{zf'(z)}{f(z)} - 1 \right| \le (\sqrt{2} + 1)(\sqrt{2} - 1) = 1.$$
This proves that $\mathcal{S}_L^$ is at least $R_{\mathcal{S}_L^}$ .
(3) For $0 \le r \le \rho := R_{\mathcal{S}_P} = 5 - 2\sqrt{6}$ , we have for
$$\frac{1}{2} < 1 \le a = \frac{1 + \rho^2}{1 - \rho^2} = \frac{5\sqrt{6}}{12} < 3/2$$
and, $\rho$ being the smallest positive root of the equation $r^2 - 10r + 1 = 0$ ,
$$\frac{5\rho + \rho^2}{(1 - \rho^2)} \le \frac{1 + \rho^2}{1 - \rho^2} - \frac{1}{2}.$$
The disk in (2.18) becomes
$$\left| \frac{zf'(z)}{f(z)} - \frac{1+\rho^2}{1-\rho^2} \right| \le \frac{1+\rho^2}{1-\rho^2} - \frac{1}{2}.$$
By Lemma 2.2, the disk in (2.18) is inside the region $\Omega_{PAR}$ . Thus, the radius of parabolic starlikeness of the class $\Pi_2$ is at least $R_{\mathcal{S}_p}$ .
For the function $f_3$ given in (1.3) at $z = -\rho$ where $\rho = R_{\mathcal{S}_p}$ , we have
$$\operatorname{Re}\left(\frac{zf_3'(z)}{f_3(z)}\right) = \frac{1 - 5\rho}{1 - \rho^2} = \frac{5\rho - \rho^2}{1 - \rho^2} = \left|\frac{zf_3'(z)}{f_3(z)} - 1\right|.$$
(4) For $\rho = R_{\mathcal{S}_e^*}$ , we have $1/e < a := (1 + \rho^2)/(1 - \rho^2) \approx 1.0331 \le (e + e^{-1})/2$ and
$$\frac{5\rho + \rho^2}{1 - \rho^2} = \frac{1 + \rho^2}{1 - \rho^2} - \frac{1}{e}.$$
The disk in (2.18) becomes
$$\left| \frac{zf'(z)}{f(z)} - \frac{1+\rho^2}{1-\rho^2} \right| \le \frac{1+\rho^2}{1-\rho^2} - \frac{1}{e}.$$
By Lemma 2.3, this disk is inside the region $\Omega_e$ , proving that $\mathcal{S}_e^$ radius is at least $R_{\mathcal{S}_e^}$ .
The result is sharp for the function $f_3$ given in (1.3). For this function, we have, at $z = -\rho$ where $\rho = R_{\mathcal{S}_e^*}$ ,
$$\left| \log \left( \frac{z f_3'(z)}{f_3(z)} \right) \right| = \left| \log \left( \frac{1 - 5\rho}{1 - \rho^2} \right) \right| = \left| \log(e^{-1}) \right| = 1.$$
(5) For $\rho = R_{\mathcal{S}_c^*}$ , we have $1/3 < a := (1+\rho^2)/(1-\rho^2) = \frac{1}{72}(1+5\sqrt{217}) \approx 1.03686 \le 5/2$ and, $\rho$ being the smallest positive root of $r^2 - 15r + 2 = 0$ ,
$$\frac{5\rho + \rho^2}{1 - \rho^2} = \frac{1 + \rho^2}{1 - \rho^2} - \frac{1}{3}.$$
The disk in (2.18) becomes
$$\left| \frac{zf'(z)}{f(z)} - \frac{1+\rho^2}{1-\rho^2} \right| \le \frac{1+\rho^2}{1-\rho^2} - \frac{1}{3}.$$
By Lemma 2.3, this disk is inside the region $\Omega_c$ , proving that $\mathcal{S}_c^$ radius is at least $R_{\mathcal{S}_c^}$ .
The radius is sharp for the function $f_3$ given in (1.3). At $z = -\rho$ where $\rho = R_{\mathcal{S}_c^*}$ , we have
$$\frac{zf_3'(z)}{f_3(z)} = \frac{1-5\rho}{1-\rho^2} = \frac{1}{3} = \varphi_c(-1) \in \partial \varphi_c(\mathbb{D})$$
where $\varphi_c(z) = 1 + 4z/3 + 2z^2/3$ .
(6) For $\rho = R_{\mathcal{S}_{\sin}^*}$ , and $a := (1 + \rho^2)/(1 - \rho^2)$ , we have
$$|a-1| = \frac{2\rho^2}{1-\rho^2} \approx 0.0465396 < \sin 1 \approx 0.8414.$$
and
$$\frac{5\rho + \rho^2}{1 - \rho^2} \le \sin 1 - \frac{2\rho^2}{1 - \rho^2}.$$
The disk in (2.6) for $r = \rho$ becomes
$$\left| \frac{zf'(z)}{f(z)} - a \right| = \left| \frac{zf'(z)}{f(z)} - \frac{1 + \rho^2}{1 - \rho^2} \right| \le \sin 1 - \frac{2\rho^2}{1 - \rho^2} = \sin 1 - |1 - a|.$$
Lemma 2.5 shows that the disk in (2.18) is inside $\Omega_s$ where $\Omega_s =: \varphi_s(\mathbb{D})$ is the image of the unit disk $\mathbb{D}$ under the mapping $\varphi_s(z) = 1 + \sin z$ . This proves that the $\mathcal{S}_{\sin}^$ radius is at least $R_{\mathcal{S}_{\sin}^}$ .
(7) For $\rho = R_{\mathcal{S}_{\mathcal{O}}^*}$ , we have
$$a := \frac{1+\rho^2}{1-\rho^2} \approx 1.02839 \in (\sqrt{2}-1, \sqrt{2}+1)$$
and
$$\frac{5\rho + \rho^2}{1 - \rho^2} = \frac{1 + \rho^2}{1 - \rho^2} + 1 - \sqrt{2}.$$
The disk in (2.18) becomes
$$\left| \frac{zf'(z)}{f(z)} - a \right| \le 1 - |\sqrt{2} - a|$$
and by Lemma 2.6 it lies inside $\{w: |w^2-1|<2|w|\}$ . This shows that $\mathcal{S}^_{\mathbb{Q}}$ radius is at least $R_{\mathcal{S}^_{\mathbb{Q}}}$ . The sharpness follows as the function $f_3$ defined in (1.1) satisfies, at $z=\rho=R_{\mathcal{S}^*_{\mathbb{Q}}}$ ,
$$\left| \left( \frac{zf_3'(z)}{f_3(z)} \right)^2 - 1 \right| = \left| \left( \frac{1 - 5\rho}{1 - \rho^2} \right)^2 - 1 \right| = 2(\sqrt{2} - 1)$$
$$= 2\frac{1 - 5\rho}{1 - \rho^2} = 2 \left| \frac{zf_3'(z)}{f_3(z)} \right|.$$
(8) For $\rho = R_{\mathcal{S}_{\mathcal{B}}^*}$ , we have
$$2(\sqrt{2}-1) < a := \frac{1+\rho^2}{1-\rho^2} \approx 1.00238 \le \sqrt{2} < 2,$$
and
$$\frac{5\rho + \rho^2}{1 - \rho^2} = \frac{1 + \rho^2}{1 - \rho^2} - 2(\sqrt{2} - 1).$$
The disk (2.18) becomes
$$\left| \frac{zf'(z)}{f(z)} - a \right| < a - 2(\sqrt{2} - 1).$$
By Lemma 2.7, this disk lies inside the domain $\Omega_r$ . This proves that $\mathcal{S}_R^$ radius is at least $R_{\mathcal{S}_R^}$ .
To prove sharpness, consider the function $f_3 \in \Pi_1$ given in (1.1). At $z = -\rho = -R_{\mathcal{S}_{\mathcal{P}}^*}$ , we have
$$\frac{zf_3'(z)}{f_3(z)} = \frac{1 - 5\rho}{1 - \rho^2} = 2(\sqrt{2} - 1) = \varphi_r(-1) \in \partial \varphi_r(\mathbb{D})$$
where $\varphi_r(z) = 1 + (kz + z^2)/(k^2 - kz), k = \sqrt{2} + 1.$
We have only obtained lower bounds for the $\mathcal{S}_L^$ and $\mathcal{S}_{sin}^$ radii for the class $\Pi_2$ and we believe the bounds are sharp but unable to prove it.We use first-party site metrics to improve navigation: page visits, internal clicks, search result counts (not search terms), and approximate active time. A random tab ID groups events for seven days; GPC/DNT turns this off. The separate visitor count uses a self-hosted pseudonymous fingerprint and also respects GPC/DNT.