Starlikeness of Certain Non-Univalent Functions

Theorem 3.2. The following results for the class $\mathcal{S}_L^*$ are sharp. (i) $$R_{\mathcal{S}_L^*}(\mathcal{G}_1) = (3 - 2\sqrt{2})/(\sqrt{2} - 1)\left(3 + \sqrt{12 - 2\sqrt{2}}\right) \approx 0.0687.$$ (ii) $R_{\mathcal{S}_L^*}(\mathcal{G}_2) = 2(\sqrt{2} - 1)/\left(5 + \sqrt{33 - 4\sqrt{2}}\right) \approx 0.0809.$ (ii) $$R_{\mathcal{S}_L^*}(\mathcal{G}_2) = 2(\sqrt{2} - 1)/(5 + \sqrt{33 - 4\sqrt{2}}) \approx 0.0809.$$ (iii) $$R_{\mathcal{S}_L^*}(\mathcal{G}_3) = (3 - 2\sqrt{2})/(\sqrt{2} - 1)\left(2 + \sqrt{7 - 2\sqrt{2}}\right) \approx 0.1025.$$ PROOF. (i) The function defined by $m(r) = 6r(1-r^2)^{-1}+1$ , $0 \le r < 1$ is an increasing function. Let $\rho = R_{\mathcal{S}_L}(\mathcal{G}_1)$ is the root of the equation $m(r) = \sqrt{2}$ . For $0 < r \le R_{\mathcal{S}_L}(\mathcal{G}_1)$ , we have $m(r) \leq \sqrt{2}$ . That is, $$\frac{6r}{1 - r^2} + 1 \le \sqrt{2} = m(\rho).$$ For the class $\mathcal{G}_1$ , the centre of the disc is 1, therefore the disc obtained in (2.7) is contained in the region bounded by lemniscate, by Lemma 3.3. For the function $f_1$ defined in (2.3), at $z = R_{\mathcal{S}_{\tau}^*}(\mathcal{G}_1) = \rho$ , $$\left| \left( \frac{zf'(z)}{f(z)} \right)^2 - 1 \right| = \left| \left( \frac{1 + 6\rho - \rho^2}{1 - \rho^2} \right)^2 - 1 \right| = 1.$$ (ii) The function $n(r): [0,1) \longrightarrow \mathbb{R}$ defined by $n(r) = (5r+r^2)(1-r^2)^{-1}+1$ is an increasing function. Let $\rho = R_{\mathcal{S}_L}(\mathcal{G}_2)$ is the root of the equation $n(r) = \sqrt{2}$ . For $0 < r \le R_{\mathcal{S}_L}(\mathcal{G}_2)$ , we have $n(r) \le \sqrt{2}$ . That is, $$\frac{5r+r^2}{1-r^2}+1 \le \sqrt{2} = n(\rho).$$ For the class $\mathcal{G}_2$ , the centre of the disc is 1, therefore the disc obtained in (2.12) is contained in the region bounded by lemniscate, by Lemma 3.3. For the function $f_2$ defined in (2.8), at $z = R_{\mathcal{S}_L^*}(\mathcal{G}_2) = \rho$ , $$\left| \left( \frac{zf'(z)}{f(z)} \right)^2 - 1 \right| = \left| \left( \frac{1+5\rho}{1-\rho^2} \right)^2 - 1 \right| = 1.$$ (iii) The function defined by $s(r) = 4r(1-r^2)^{-1} + 1$ , $0 \le r < 1$ is an increasing function. Let $\rho = R_{\mathcal{S}_L}(\mathcal{G}_3)$ is the root of the equation $s(r) = \sqrt{2}$ For $0 < r \le R_{\mathcal{S}_L}(\mathcal{G}_3)$ , we have $s(r) \le \sqrt{2}$ . That is, $$\frac{4r}{1 - r^2} + 1 \le \sqrt{2} = s(\rho).$$ For the class $\mathcal{G}_3$ , the centre of the disc is 1, therefore the disc obtained in (2.15) is contained in the region bounded by lemniscate, by Lemma 3.3. For the function $g_1$ defined in (2.3), at $z = R_{\mathcal{S}_{L}^*}(\mathcal{G}_3) = \rho$ , $$\left| \left( \frac{zf'(z)}{f(z)} \right)^2 - 1 \right| = \left| \left( \frac{1 + 4\rho - \rho^2}{1 - \rho^2} \right)^2 - 1 \right| = \left| (\sqrt{2})^2 - 1 \right| = 1.$$ Let $\varphi_{PAR}(z) := 1 + \left(2/\pi^2 \left(\log(1+\sqrt{z})/(1-\sqrt{z})\right)^2\right)$ . Since $\varphi_{PAR}(\mathbb{D}) = \{w : \operatorname{Re} w > |w-1|\}$ is a parabolic region, the functions in the class $\mathcal{S}_p := \mathcal{S}^*(\varphi_{PAR})$ are known as parabolic starlike functions. These functions are studied by authors in $[\mathbf{9}, \mathbf{21}, \mathbf{32}]$ . Shanmugam and Ravichandran $[\mathbf{36}, \operatorname{pp.321}]$ had proved that for 1/2 < a < 3/2, then <span id="page-6-0"></span> $$\{w : |w - a| < a - 1/2\} \subset \{w : \operatorname{Re} w > |w - 1|\}.$$ (3.4) The following theorem gives the radius of parabolic starlikeness of the three classes $\mathcal{G}_1$ , $\mathcal{G}_2$ and $\mathcal{G}_3$ .

Theorem 3.3. The following results hold for the class $S_p$ : - (i) $R_{S_n}(\mathcal{G}_1) = \sqrt{37} 6 \approx 0.0827$ . - (ii) $R_{S_n}(\mathcal{G}_2) \ge (2\sqrt{7} 5)/3 \approx 0.0972$ . - (iii) $R_{S_p}(\mathcal{G}_3) = \sqrt{17} 4 \approx 0.1231.$ PROOF. (i) The function defined by $m(r) = (1 - 6r - r^2)(1 - r^2)^{-1}$ , $0 \le r < 1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_p}(\mathcal{G}_1)$ is the root of the equation m(r) = 1/2. For $0 < r \le R_{\mathcal{S}_p}(\mathcal{G}_1)$ , we have $m(r) \ge 1/2$ . That is, $$\frac{6r}{1-r^2} \le \frac{1}{2} = m(\rho).$$ For the class $\mathcal{G}_1$ , the centre of the disc is 1, therefore the disc obtained in (2.7) is contained in the region bounded by parabola, by Lemma 3.4. For the function $f_1$ defined in (2.3), at $z = R_{\mathcal{S}_p}(\mathcal{G}_1) = \rho$ , Re $$\frac{zf_1'(z)}{f_1(z)} = \frac{1+6\rho-\rho^2}{1-\rho^2} = \frac{1}{2} = \left| \frac{zf_1'(z)}{f_1(z)} - 1 \right|.$$ (ii) The function $n(r): [0,1) \longrightarrow \mathbb{R}$ defined by $n(r) = (1-5r-2r^2)(1-r^2)^{-1}+1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_p}(\mathcal{G}_2)$ is the root of the equation n(r) = 1/2. For $0 < r \le R_{\mathcal{S}_p}(\mathcal{G}_2)$ , we have $n(r) \ge 1/2$ . That is, $$\frac{r(r+5)}{1-r^2} \le \frac{1}{2} = n(\rho).$$ For the class $\mathcal{G}_2$ , the centre of the disc is 1, therefore the disc obtained in (2.12) is contained in the region bounded by parabola, by Lemma 3.4. This shows that $R_{\mathcal{S}_p}(\mathcal{G}_2)$ is at least $\rho$ . (iii) The function defined by $s(r) = 1 - (4r(1-r^2)^{-1})$ , $0 \le r < 1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_p}(\mathcal{G}_3)$ is the root of the equation s(r) = 1/2. For $0 < r \le R_{\mathcal{S}_p}(\mathcal{G}_3)$ , we have $s(r) \ge 1/2$ . That is, $$\frac{4r}{1 - r^2} \le \frac{1}{2} = s(\rho).$$ For the class $\mathcal{G}_3$ , the centre of the disc is 1, therefore the disc obtained in (2.15) is contained in the region bounded by parabola, by Lemma 3.4. For the function $g_1$ defined in (2.3), at $z = -R_{\mathcal{S}_p}(\mathcal{G}_3) = -\rho$ , $$\operatorname{Re} \frac{zg_1'(z)}{g_1(z)} = \frac{1 - 4\rho - \rho^2}{1 - \rho^2} = \frac{1}{2} = \left| \frac{zg_1'(z)}{g_1(z)} - 1 \right|.$$ In 2015, Mendiratta et al. [27] introduced the class of starlike functions associated with the exponential function as $S_e^ = S^(e^z)$ and it satisfies the condition $|\log z f'(z)/f(z)| < 1$ . They had also proved that, for $e^{-1} \le a \le (e + e^{-1})/2$ , <span id="page-7-0"></span> $$\{w \in \mathbb{C} : |w - a| < a - e^{-1}\} \subseteq \{w \in \mathbb{C} : |\log w| < 1\}.$$ (3.5)

Theorem 3.4. The following results hold for the class $\mathcal{S}_e^*$ : - (i) $R_{S_{\epsilon}^*}(\mathcal{G}_1) = (e-1)/(3e+\sqrt{10e^2-2e+1}) \approx 0.1042.$ - (ii) $R_{\mathcal{S}_{e}^{*}}(\mathcal{G}_{2}) \ge 2(e^{2} + e 2)/(2 + e) \left(5e + \sqrt{8 + 4e + 29e^{2}}\right) \approx 0.1213.$ - (iii) $R_{\mathcal{S}_e^*}(\mathcal{G}_3) = 2(e-1)^2/(e+1)\left(4e+\sqrt{4-8e+20e^2}\right) \approx 0.1543.$ PROOF. (i) The function defined by $m(r) = (1 - 6r - r^2)(1 - r^2)^{-1}$ , $0 \le r < 1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_e}(\mathcal{G}_1)$ is the root of the equation m(r) = 1/e. For $0 < r \le R_{\mathcal{S}_e}(\mathcal{G}_1)$ , we have $m(r) \ge 1/e$ . That is, $$\frac{6r}{1-r^2} \le 1 - \frac{1}{e}.$$ For the class $\mathcal{G}_1$ , the centre of the disc is 1, therefore the disc obtained in (2.7) is contained in the region bounded by exponential function, by Lemma 3.5. For the function $f_1$ defined in (2.3), at $z = R_{\mathcal{S}_s^*}(\mathcal{G}_1) = \rho$ , $$\left| \log \frac{z f_1'(z)}{f_1(z)} \right| = \left| \log \frac{1 + 6\rho - \rho^2}{1 - \rho^2} \right| = 1.$$ (ii) The function $n(r): [0,1) \longrightarrow \mathbb{R}$ defined by $n(r) = (1 - 5r - 2r^2)(1 - r^2)^{-1} + 1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_e}(\mathcal{G}_2)$ is the root of the equation n(r) = 1/e. For $0 < r \le R_{\mathcal{S}_e}(\mathcal{G}_2)$ , we have $n(r) \ge 1/e$ . That is, $$\frac{r(r+5)}{1-r^2} \le 1 - \frac{1}{e}.$$ For the class $\mathcal{G}_2$ , the centre of the disc is 1, therefore the disc obtained in (2.12) is contained in the region bounded by the exponential function, by Lemma 3.5. This shows that $R_{\mathcal{S}_{\varepsilon}^*}(\mathcal{G}_2)$ is at least $\rho$ . (iii) The function defined by $s(r) = 1 - (4r(1-r^2)^{-1})$ , $0 \le r < 1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_e}(\mathcal{G}_3)$ is the root of the equation s(r) = 1/e. For $0 < r \le R_{\mathcal{S}_e}(\mathcal{G}_3)$ , we have $s(r) \ge 1/e$ . That is, $$\frac{4r}{1-r^2} \le \frac{e-1}{e}.$$ For the class $\mathcal{G}_3$ , the centre of the disc is 1, therefore the disc obtained in (2.15) is contained in the region bounded by the exponential function, by Lemma 3.5. For the function $g_1$ defined in (2.3), at $z = R_{\mathcal{S}_s^*}(\mathcal{G}_3) = \rho$ , $$\left| \log \frac{zg_1'(z)}{g_1(z)} \right| = \left| \log \frac{1 + 4\rho - \rho^2}{1 - \rho^2} \right| = 1.$$ Theorem 3.5 provides radii results for starlike functions associated with a cardioid. Sharma et al. [37] studied various properties of the class $S_c^ = S^(1 + (4/3)z + (2/3)z^2)$ . Geometrically, if a function $f \in S_c^*$ then zf'(z)/f(z) lies in the region bounded by the cardioid $\Omega_c = \{u + iv : (9u^2 + 9v^2 - 18u + 5)^2 - 16(9u^2 + 9v^2 - 6u + 1) = 0\}$ . They had also proved that, for $1/3 < a \le 5/3$ , <span id="page-8-1"></span> $$\{w \in \mathbb{C} : |w - a| < (3a - 1)/3\} \subseteq \Omega_c. \tag{3.6}$$ <span id="page-8-0"></span>Theorem 3.5. The following results hold for the class $\mathcal{S}_c^*$ : - (i) $R_{\mathcal{S}_c^*}(\mathcal{G}_1) = (\sqrt{85} 9)/2 \approx 0.1097.$ - (ii) $R_{\mathcal{S}_c^*}(\mathcal{G}_2) \ge (\sqrt{265} 15)/10 \approx 0.1279$ . - (iii) $R_{\mathcal{S}_c^*}(\mathcal{G}_3) = \sqrt{10} 3 \approx 0.1623.$ PROOF. (i) The function defined by $m(r) = (1 - 6r - r^2)(1 - r^2)^{-1}$ , $0 \le r < 1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_c}(\mathcal{G}_1)$ is the root of the equation m(r) = 1/3. For $0 < r \le R_{\mathcal{S}_c}(\mathcal{G}_1)$ , we have $m(r) \ge 1/3$ . That is, $$\frac{6r}{1 - r^2} \le 1 - \frac{1}{3}.$$ For the class $\mathcal{G}_1$ , the centre of the disc is 1, therefore the disc obtained in (2.7) is contained in the region bounded by the cardioid, by Lemma 3.6. For the function $f_1$ defined in (2.3), at $z = R_{\mathcal{S}_c^*}(\mathcal{G}_1) = \rho$ , $$\left| \frac{z f_1'(z)}{f_1(z)} \right| = \left| \frac{1 + 6\rho - \rho^2}{1 - \rho^2} \right| = \frac{1}{3} = \Omega_c(-1).$$ (ii) The function $n(r): [0,1) \to \mathbb{R}$ defined by $n(r) = (1-5r-2r^2)(1-r^2)^{-1}+1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_c}(\mathcal{G}_2)$ is the root of the equation n(r) = 1/3. For $0 < r \le R_{\mathcal{S}_c}(\mathcal{G}_2)$ , we have $n(r) \ge 1/3$ . That is, $$\frac{r(r+5)}{1-r^2} \le 1 - \frac{1}{3}.$$ For the class $\mathcal{G}_2$ , the centre of the disc is 1, therefore the disc obtained in (2.12) is contained in the region bounded by the cardioid, by Lemma 3.6. This shows that $R_{\mathcal{S}_{z}^{*}}(\mathcal{G}_2)$ is at least $\rho$ . (iii) The function defined by $s(r) = 1 - (4r(1-r^2)^{-1})$ , $0 \le r < 1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_c}(\mathcal{G}_3)$ is the root of the equation s(r) = 1/3. For $0 < r \le R_{\mathcal{S}_c}(\mathcal{G}_3)$ , we have $s(r) \ge 1/3$ . That is, $$\frac{4r}{1-r^2} \le \frac{2}{3}.$$ For the class $\mathcal{G}_3$ , the centre of the disc is 1, therefore the disc obtained in (2.15) is contained in the region bounded by the cardioid, by Lemma 3.6. For the function $g_1$ defined in (2.3), at $z = R_{\mathcal{S}_c^*}(\mathcal{G}_3) = \rho$ , $$\left| \frac{z f_1'(z)}{f_1(z)} \right| = \left| \frac{1 + 4\rho - \rho^2}{1 - \rho^2} \right| = \frac{1}{3} = \Omega_c(-1).$$ In 2019, Cho et al. [5] considered the class of starlike functions associated with sine function where the class $S_{sin}$ is defined as $S_{sin}^ = \{f \in \mathcal{A} : zf'(z)/f(z) \prec 1 + \sin z := q_0(z)\}$ for $z \in \mathbb{D}$ . For $|a-1| \leq \sin 1$ , they had established the following inclusion: <span id="page-9-0"></span> $$\{w \in \mathbb{C} : |w - a| < \sin 1 - |a - 1|\} \subseteq \Omega_s. \tag{3.7}$$ Here $\Omega_s := q_0(\mathbb{D})$ is the image of the unit disk $\mathbb{D}$ under the mappings $q_0(z) = 1 + \sin z$ .

Theorem 3.6. The following results are sharp for the class $S_{sin}^*$ . - (i) $R_{\mathcal{S}_{sin}^*}(\mathcal{G}_1) = \sin 1/\left(3 + \sqrt{9 + \sin^2 1}\right) \approx 0.1375.$ - (ii) $R_{\mathcal{S}_{sin}^*}(\mathcal{G}_2) = 2\sin 1/\left(5 + \sqrt{25 + 4\sin 1 + 4\sin^2 1}\right) \approx 0.1589.$ - (iii) $R_{\mathcal{S}_{sin}^*}(\mathcal{G}_3) = \sin 1/\left(2 + \sqrt{4 + \sin^2 1}\right) \approx 0.2018.$ PROOF. (i) The function defined by $m(r) = (1 - 6r - r^2)(1 - r^2)^{-1}$ , $0 \le r < 1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_{sin}}(\mathcal{G}_1)$ is the root of the equation $m(r) = 1 - \sin 1$ . For $0 < r \le R_{\mathcal{S}_{sin}}(\mathcal{G}_1)$ , we have $m(r) \ge 1 - \sin 1$ . That is, $$\frac{6r}{1-r^2} \le \sin 1.$$ For the class $\mathcal{G}_1$ , the centre of the disc is 1, therefore the disc obtained in (2.7) is contained in the region $\Omega_s$ bounded by the sine function, by Lemma 3.7. For the function $f_1$ defined in (2.3), at $z = -R_{\mathcal{S}_{sin}^*}(\mathcal{G}_1) = -\rho$ , $$\left| \frac{zf_1'(z)}{f_1(z)} \right| = \left| \frac{1 - 6\rho - \rho^2}{1 - \rho^2} \right| = 1 + \sin 1 = q_0(1).$$ (ii) The function $n(r): [0,1) \longrightarrow \mathbb{R}$ defined by $n(r) = (1-5r-2r^2)(1-r^2)^{-1}+1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_{sin}}(\mathcal{G}_2)$ is the root of the equation $n(r) = 1 - \sin 1$ . For $0 < r \le R_{\mathcal{S}_{sin}}(\mathcal{G}_2)$ , we have $n(r) \ge 1 - \sin 1$ . That is, $$\frac{r(r+5)}{1-r^2} \le \sin 1.$$ For the class $\mathcal{G}_2$ , the centre of the disc is 1, therefore the disc obtained in (2.12) is contained in the region bounded by the sine function, by Lemma 3.7. For the function $f_2$ defined in (2.8), at $z = R_{\mathcal{S}_{sin}^*}(\mathcal{G}_1) = \rho$ , $$\left| \frac{zf_2'(z)}{f_2(z)} \right| = \left| \frac{1+5\rho}{1-\rho^2} \right| = 1 + \sin 1 = q_0(1).$$ (iii) The function defined by $s(r) = 1 - (4r(1-r^2)^{-1})$ , $0 \le r < 1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_{sin}}(\mathcal{G}_3)$ is the root of the equation $s(r) = 1 - \sin 1$ . For $0 < r \le R_{\mathcal{S}_{sin}}(\mathcal{G}_3)$ , we have $s(r) \ge 1 - \sin 1$ . That is, $$\frac{4r}{1-r^2} \le \sin 1.$$ For the class $\mathcal{G}_3$ , the centre of the disc is 1, therefore the disc obtained in (2.15) is contained in the region bounded by the sine function, by Lemma 3.7. For the function $g_1$ defined in (2.3), at $z = -R_{\mathcal{S}_{sin}^*}(\mathcal{G}_3) = -\rho$ , $$\left| \frac{zf_1'(z)}{f_1(z)} \right| = \left| \frac{1 - 4\rho - \rho^2}{1 - \rho^2} \right| = 1 + \sin 1 = q_0(1).$$ In 2015, Raina and Sokół[29] introduced the class $\mathcal{S}^_{\mathbb{Q}} = \mathcal{S}^(z + \sqrt{1+z^2})$ . They showed that a function $f \in \mathcal{S}^*_{\mathbb{Q}}$ if and only if zf'(z)/f(z) belongs to a lune shaped region $\mathcal{L} := \{w \in \mathbb{C} : |w^2 - 1| < 2|w|\}$ . Gandhi and Ravichandran [8, Lemma 2.1] proved that <span id="page-10-0"></span> $$\{w \in \mathbb{C} : |w - a| < 1 - |\sqrt{2} - a|\} \subseteq \{w \in \mathbb{C} : |w^2 - 1| < 2|w|\}.$$ (3.8)

Theorem 3.7. The following results hold for the class $\mathcal{S}_{\mathbb{Q}}^*$ : (i) $$R_{\mathcal{S}_{0}^{*}}(\mathcal{G}_{1}) = (6 - 4\sqrt{2})/(2 - \sqrt{2})\left(3 + \sqrt{15 - 4\sqrt{2}}\right) \approx 0.0967.$$ (ii) $$R_{\mathcal{S}_{0}^{*}}(\mathcal{G}_{2}) \ge (16 - 10\sqrt{2})/(3 - \sqrt{2})\left(5 + \sqrt{57 - 20\sqrt{2}}\right) \approx 0.1131.$$ (iii) $$R_{\mathcal{S}_{0}^{*}}(\mathcal{G}_{3}) = (6 - 4\sqrt{2})/(2 - \sqrt{2})\left(2 + \sqrt{10 - 4\sqrt{2}}\right) \approx 0.1434.$$ PROOF. (i) The function defined by $m(r) = (1 - 6r - r^2)(1 - r^2)^{-1}$ , $0 \le r < 1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_{\mathbb{Q}}}(\mathcal{G}_1)$ is the root of the equation $m(r) = \sqrt{2} - 1$ . For $0 < r \le R_{\mathcal{S}_{\mathbb{Q}}}(\mathcal{G}_1)$ , we have $m(r) \ge \sqrt{2} - 1$ . That is, $$\frac{6r}{1-r^2} \le 2 - \sqrt{2}.$$ For the class $\mathcal{G}_1$ , the centre of the disc is 1, therefore the disc obtained in (2.7) is contained in the region bounded by the intersection of disks $\{w : |w-1| < \sqrt{2}\}$ and $\{w: |w+1| < \sqrt{2}\}$ , by Lemma 3.8. For the function $f_1$ defined in (2.3), at $z = -R_{\mathcal{S}_{\mathcal{J}}^*}(\mathcal{G}_1) = -\rho$ , $$\left| \left( \frac{z f_1'(z)}{f_1(z)} \right)^2 - 1 \right| = \left| \left( \frac{1 - 6\rho - \rho^2}{1 - \rho^2} \right)^2 - 1 \right| = 2 \left| \frac{1 - 6\rho - \rho^2}{1 - \rho^2} \right|.$$ (ii) The function $n(r): [0,1) \longrightarrow \mathbb{R}$ defined by $n(r) = (1-5r-2r^2)(1-r^2)^{-1}+1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_{\mathbb{Q}}}(\mathcal{G}_2)$ is the root of the equation $n(r) = \sqrt{2} - 1$ . For $0 < r \le R_{\mathcal{S}_{\mathbb{Q}}}(\mathcal{G}_2)$ , we have $n(r) \ge \sqrt{2} - 1$ . That is, $$\frac{r(r+5)}{1-r^2} \le 2 - \sqrt{2}.$$ For the class $\mathcal{G}_2$ , the centre of the disc is 1, therefore the disc obtained in (2.12) is contained in the region bounded by the lune, by Lemma 3.8. This shows that $R_{\mathcal{S}_{\mathfrak{C}}^*}(\mathcal{G}_2)$ is at least $\rho$ . (iii) The function $s(r): [0,1) \longrightarrow \mathbb{R}$ defined by $s(r) = 1 - (4r(1-r^2)^{-1})$ is a decreasing function. Let $\rho = R_{\mathcal{S}_{\mathcal{J}}}(\mathcal{G}_1)$ is the root of the equation $s(r) = \sqrt{2} - 1$ . For $0 < r \le R_{\mathcal{S}_{\mathcal{J}}}(\mathcal{G}_3)$ , we have $s(r) \ge \sqrt{2} - 1$ . That is, $$\frac{4r}{1-r^2} \le 2 - \sqrt{2}.$$ For the class $\mathcal{G}_3$ , the centre of the disc is 1, therefore the disc obtained in (2.15) is contained in the region bounded by the lune, by Lemma 3.8. For the function $g_1$ defined in (2.3), at $z = -R_{\mathcal{S}_{\mathcal{J}}^*}(\mathcal{G}_3) = -\rho$ , $$\left| \left( \frac{zg_1'(z)}{g_1(z)} \right)^2 - 1 \right| = \left| \left( \frac{1 - 4\rho - \rho^2}{1 - \rho^2} \right)^2 - 1 \right| = 2 \left| \frac{1 - 4\rho - \rho^2}{1 - \rho^2} \right|.$$ In the next theorem, we provide radii for starlike functions associated with a rational function. Kumar and Ravichandran [19] introduced the class of starlike functions associated with a rational function, $\psi(z) = 1 + (z^2k + z^2/(k^2 - kz))$ where $k = \sqrt{2} + 1$ , defined by $\mathcal{S}_R^ = \mathcal{S}^(\psi(z))$ . For $2(\sqrt{2} - 1) < a \le \sqrt{2}$ , they had proved that <span id="page-11-0"></span> $$\{w \in \mathbb{C} : |w - a| < a - 2(\sqrt{2} - 1)\} \subseteq \psi(\mathbb{D}). \tag{3.9}$$

Theorem 3.8. The following results hold for the class $\mathcal{S}_R^*$ : (i) $$R_{\mathcal{S}_R^*}(\mathcal{G}_1) = (3 - 2\sqrt{2}) / \left(3 + \sqrt{26 - 12\sqrt{2}}\right) \approx 0.0285.$$ (ii) $$R_{\mathcal{S}_R^*}(\mathcal{G}_2) \ge (20 - 14\sqrt{2})/(2 - \sqrt{2}) \left(5 + \sqrt{105 - 56\sqrt{2}}\right) \approx 0.0340.$$ (iii) $$R_{\mathcal{S}_R^*}(\mathcal{G}_3) = (3 - 2\sqrt{2})/\left(2 + \sqrt{21 - 12\sqrt{2}}\right) \approx 0.0428.$$ PROOF. (i) The function defined by $m(r) = (1 - 6r - r^2)(1 - r^2)^{-1}$ , $0 \le r < 1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_R^*}(\mathcal{G}_1)$ is the root of the equation $m(r) = 2(\sqrt{2} - 1)$ . For $0 < r \le R_{\mathcal{S}_R^*}(\mathcal{G}_1)$ , we have $m(r) \ge 2(\sqrt{2} - 1)$ . That is, $$\frac{6r}{1-r^2} \le 1 - 2(\sqrt{2} - 1).$$ For the class $\mathcal{G}_1$ , the centre of the disc is 1, therefore the disc obtained in (2.7) is contained in the region bounded by the rational function, by Lemma 3.9. For the function $f_1$ defined in (2.3), at $z = -R_{\mathcal{S}_{\mathcal{D}}^*}(\mathcal{G}_1) = -\rho$ , $$\left| \frac{zf_1'(z)}{f_1(z)} \right| = \left| \frac{1 - 6\rho - \rho^2}{1 - \rho^2} \right| = 2(\sqrt{2} - 1) = \psi(1).$$ (ii) The function $n(r):[0,1) \longrightarrow \mathbb{R}$ defined by $n(r)=(1-5r-2r^2)(1-r^2)^{-1}+1$ is a decreasing function. Let $\rho=R_{\mathcal{S}_R}^*(\mathcal{G}_2)$ is the root of the equation $n(r)=2(\sqrt{2}-1)$ . The function defined by $$n(r) = \frac{1 - 5r - 2r^2}{1 - r^2} + 1$$ is a decreasing function. For $0 < r \le R_{\mathcal{S}_R^*}(\mathcal{G}_2)$ , we have $n(r) \ge 2(\sqrt{2} - 1)$ . That is, $$\frac{r(r+5)}{1-r^2} \le 3 - 2\sqrt{2}.$$ For the class $\mathcal{G}_2$ , the centre of the disc is 1, therefore the disc obtained in (2.12) is contained in the region bounded by the rational function, by Lemma 3.9. This shows that $R_{\mathcal{S}_R^*}(\mathcal{G}_2)$ is at least $\rho$ . (iii) The function defined by $s(r) = 1 - (4r(1-r^2)^{-1}), \ 0 \le r < 1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_R}(\mathcal{G}_3)$ is the root of the equation $s(r) = 2(\sqrt{2}-1)$ . For $0 < r \le R_{\mathcal{S}_R}(\mathcal{G}_3)$ , we have $s(r) \ge 2(\sqrt{2}-1)$ . That is, $$\frac{4r}{1-r^2} \le 2 - \sqrt{2}.$$ For the class $\mathcal{G}_3$ , the centre of the disc is 1, therefore the disc obtained in (2.15) is contained in the region bounded by the rational function, by Lemma 3.9. For the function $g_1$ defined in (2.3), at $z = -R_{\mathcal{S}_{\mathcal{B}}^*}(\mathcal{G}_3) = -\rho$ , $$\left| \frac{zg_1'(z)}{g_1(z)} \right| = \left| \frac{1 - 4\rho - \rho^2}{1 - \rho^2} \right| = 2(\sqrt{2} - 1) = \psi(1).$$ Mendiratta et al. [26] studied the subclass of starlike function associated with left half of shifted lemniscate of Bernoulli, given by $|(w-\sqrt{2})^2-1|<1$ . The class $\mathcal{S}_{RL}^*$ is defined as $$S_{RL}^ = S^ \left( \sqrt{2} - (\sqrt{2} - 1) \sqrt{\frac{1 - z}{1 + 2(\sqrt{2} - 1)z}} \right).$$ For $\sqrt{2}/3 \le a < \sqrt{2}$ , they had proved the following inclusion: <span id="page-12-0"></span> $$\{w \in \mathbb{C} : |w - a| < r_{RL}\} \subseteq \{w \in \mathbb{C} : |(w - \sqrt{2})^2 - 1| < 1\},$$ (3.10) where $r_{RL} = \left(\left(1 - \left(\sqrt{2} - a\right)^2\right)^{1/2} - \left(1 - \left(\sqrt{2} - a\right)^2\right)\right)^{1/2}$ . Using this result, we obtain $\mathcal{S}_{RL}^*$ -radii of the classes $\mathcal{G}_1$ , $\mathcal{G}_2$ , $\mathcal{G}_3$ in the following theorem.

Theorem 3.9. Let $\eta = \sqrt{2(\sqrt{2}-1)} - 2(\sqrt{2}-1)$ . Then the following sharp results hold for the class $S_{RL}^*$ . - (i) $R_{\mathcal{S}_{RL}^*}(\mathcal{G}_1)$ is the smallest positive root ( $\approx 0.0475$ ) in (0,1) of the equation $(36+2\eta)r^2 \eta r^2 \eta = 0$ . - (ii) $R_{S_{RL}^*}(G_2)$ is the smallest positive root ( $\approx 0.0567$ ) in (0,1) of the equation $(1-\eta)r^4 + 10r^3 + (25-2\eta)r^2 \eta = 0$ . - (iii) $R_{\mathcal{S}_{RL}^*}(\mathcal{G}_3)$ is the smallest positive root ( $\approx 0.0711$ ) in (0,1) of the equation $\eta r^4 (16 + 2\eta)r^2 + \eta = 0$ . PROOF. (i) The function defined by $m(r) = (6r(1-r^2)^{-1}) + 1$ , $0 \le r < 1$ is an increasing function. Let $\rho = R_{\mathcal{S}_{RL}}(\mathcal{G}_3)$ is the root of the equation $m(r) = 1 + \sqrt{\eta}$ . For $0 < r \le R_{\mathcal{S}_{RL}}(\mathcal{G}_1)$ , we have $m(r) \le \sqrt{2}$ . That is, $$\left(\frac{6r}{1-r^2}\right)^2 \le \eta = (m(\rho) - 1)^2.$$ For the class $\mathcal{G}_1$ , the centre of the disc is 1, therefore the disc obtained in (2.7) is contained in the region bounded by the reverse lemniscate, by Lemma 3.10. This shows that $R_{\mathcal{S}_{RL}^*}(\mathcal{G}_1)$ is at least $\rho$ . For the function $f_1$ defined in (2.3), the radius is sharp. (ii) The function defined by $n(r) = (5r + r^2)(1 - r^2)^{-1} + 1$ , $0 \le r < 1$ is an increasing function. Let $R_{\mathcal{S}_{RL}}(\mathcal{G}_2)$ is the root of the equation $n(r) = 1 + \sqrt{\eta}$ . For $0 < r \le R_{\mathcal{S}_{RL}}(\mathcal{G}_2)$ , we have $n(r) \le \sqrt{2}$ . That is, $$\left(\frac{5r+r^2}{1-r^2}\right)^2 \le \eta = (n(\rho)-1)^2.$$ For the class $\mathcal{G}_2$ , the centre of the disc is 1, therefore the disc obtained in (2.12) is contained in the region bounded by reverse lemniscate, by Lemma 3.10. This shows that $R_{\mathcal{S}_{RL}^*}(\mathcal{G}_2)$ is at least $\rho$ . The obtained radius is sharp for the function $f_2$ defined in (2.8). (iii) The function defined by $s(r) = 4r(1-r^2)^{-1} + 1$ , $0 \le r < 1$ is an increasing function. Let $R_{\mathcal{S}_{RL}}(\mathcal{G}_3)$ is the root of the equation $s(r) = 1 + \sqrt{\eta}$ . For $0 < r \le R_{\mathcal{S}_{RL}}(\mathcal{G}_3)$ , we have $s(r) \le \sqrt{2}$ . That is, $$\left(\frac{4r}{1-r^2}\right)^2 \le \eta = (s(\rho)-1)^2.$$ For the class $\mathcal{G}_3$ , the centre of the disc is 1, therefore the disc obtained in (2.15) is contained in the region bounded by reverse lemniscate, by Lemma 3.10. This shows that $R_{\mathcal{S}_{RL}^*}(\mathcal{G}_3)$ is at least $\rho$ . The obtained radius is sharp for the function $g_1$ defined in (2.3). The sharpness of the results can be shown using the software Wolfram Mathematica. In 2020, Wani and Swaminathan [40, Lemma 2.2] introduced the class $\mathcal{S}_{Ne}^ = \mathcal{S}^(1 + z - (z^3/3))$ that maps the open unit disc $\mathbb{D}$ onto the interior of a two cusped kidney shaped curve $\Omega_{Ne} := \{u + iv : ((u-1)^2 + v^2 - 4/9)^3 - 4v^2/3 < 0\}$ . For $1/3 < a \le 1$ , they had proved that <span id="page-14-0"></span> $$\{w \in \mathbb{C} : |w - a| < a - 1/3\} \subseteq \Omega_{Ne}. \tag{3.11}$$ Our next theorem determines the $\mathcal{S}_{Ne}^*$ -radii results for the classes $\mathcal{G}_1$ , $\mathcal{G}_2$ and $\mathcal{G}_3$ .

Theorem 3.10. The following sharp results hold for the class $\mathcal{S}_{Ne}^*$ . - (i) $R_{\mathcal{S}_{N_e}^*}(\mathcal{G}_1) = (\sqrt{85} 9)/2 \approx 0.1097$ - (ii) $R_{S_{**}}(\mathcal{G}_2) = (\sqrt{265} 15)/10 \approx 0.1278$ - (iii) $R_{\mathcal{S}_{N_e}^*}(\mathcal{G}_3) = \sqrt{10} 3 \approx 0.1622$ PROOF. (i) The function defined by $m(r) = (1 - 6r - r^2)(1 - r^2)^{-1}$ , $0 \le r < 1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_{N_e}}(\mathcal{G}_1)$ is the root of the equation m(r) = 1/3. For $0 < r \le R_{\mathcal{S}_{N_e}}(\mathcal{G}_1)$ , we have $m(r) \ge 1/3$ . That is, $$\frac{6r}{1-r^2} \le 1 - \frac{1}{3}.$$ For the class $\mathcal{G}_1$ , the centre of the disc is 1, therefore the disc obtained in (2.7) is contained in the region bounded by the nephroid, by Lemma 3.11. For the function $f_1$ defined in (2.3), at $z = R_{\mathcal{S}_{N_e}^*}(\mathcal{G}_1) = \rho$ , $$\left| \frac{zf_1'(z)}{f_1(z)} \right| = \left| \frac{1 + 6\rho - \rho^2}{1 - \rho^2} \right| = \frac{1}{3} \in \partial\Omega_{Ne}$$ where $\partial\Omega_{Ne}$ denotes the boundary of nephroid domain. (ii) The function $n(r): [0,1) \longrightarrow \mathbb{R}$ defined by $n(r) = (1-5r-2r^2)(1-r^2)^{-1}+1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_{N_e}}(\mathcal{G}_2)$ is the root of the equation n(r) = 1/3. For $0 < r \le R_{\mathcal{S}_{N_e}}(\mathcal{G}_2)$ , we have $n(r) \ge 1/3$ . That is, $$\frac{r(r+5)}{1-r^2} \le \frac{2}{3}.$$ For the class $\mathcal{G}_2$ , the centre of the disc is 1, therefore the disc obtained in (2.12) is contained in the region bounded by the nephroid, by Lemma 3.11. For the function $f_2$ defined in (2.8), at $z = R_{\mathcal{S}_{N_c}^*}(\mathcal{G}_2) = \rho$ , $$\left| \frac{zf_2'(z)}{f_2(z)} \right| = \left| \frac{1+5\rho}{1-\rho^2} \right| = \frac{5}{3} \in \partial\Omega_{Ne}.$$ (iii) The function defined by $s(r) = 1 - (4r(1-r^2)^{-1})$ , $0 \le r < 1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_{N_e}}(\mathcal{G}_3)$ is the root of the equation s(r) = 1/3. For $0 < r \le R_{\mathcal{S}_{N_e}}(\mathcal{G}_3)$ , we have $s(r) \ge 1/3$ . That is, $$\frac{4r}{1-r^2} \le \frac{2}{3}.$$ For the class $\mathcal{G}_3$ , the centre of the disc is 1, therefore the disc obtained in (2.15) is contained in the region bounded by the nephroid, by Lemma 3.11. For the function $g_1$ defined in (2.3), at $z = R_{\mathcal{S}_{N_e}^*}(\mathcal{G}_3) = \rho$ , $$\left| \frac{zg_1'(z)}{g_1(z)} \right| = \left| \frac{1 + 4\rho - \rho^2}{1 - \rho^2} \right| = \frac{1}{3} \in \partial \Omega_{Ne}.$$ In 2020, Goel and Kumar [10] introduced the class $\mathcal{S}_{SG}$ that maps the open unit disc $\mathbb{D}$ onto a domain $\Delta_{SG} := \{ w \in \mathbb{C} : |\log w/(2-w)| < 1 \}$ and $\mathcal{S}_{SG}^ = \mathcal{S}^*(2/(1+e^{-z}))$ . For 2/(1+e) < a < 2e/(1+e), they had proved the following inclusion: <span id="page-15-1"></span> $$\{w \in \mathbb{C} : |w - a| < r_{SG}\} \subset \Delta_{SG},\tag{3.12}$$ provided $r_{SG} = ((e-1)/(e+1)) - |a-1|$ . Theorem 3.11 provides $\mathcal{S}_{SG}^*$ -radii of the classes $\mathcal{G}_1, \ \mathcal{G}_2, \ \mathcal{G}_3.$ <span id="page-15-0"></span>Theorem 3.11. The following sharp results hold for the class $\mathcal{S}_{SG}^*$ . - (i) $R_{\mathcal{S}_{SG}}(\mathcal{G}_1) = 2(e-1)/\left((6+6e) + \sqrt{40+64e+40e^2}\right) \approx 0.0766$ (ii) $R_{\mathcal{S}_{SG}}(\mathcal{G}_2) = 2(e-1)/\left((5+5e) + \sqrt{25+42e+33e^2}\right) \approx 0.0901$ (iii) $R_{\mathcal{S}_{SG}^*}(\mathcal{G}_3) = 2(e-1)/\left((4+4e) + \sqrt{20+24e+20e^2}\right) \approx 0.1140$ PROOF. (i) The function defined by $m(r) = (1 - 6r - r^2)(1 - r^2)^{-1}$ , $0 \le r < 1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_{SG}}(\mathcal{G}_1)$ is the root of the equation m(r) = 2/(1+e). For $0 < r \le R_{\mathcal{S}_{SG}}(\mathcal{G}_1)$ , we have $m(r) \ge 2/(1+e)$ . That is, $$\frac{6r}{1-r^2} \le \frac{e-1}{e+1}.$$ For the class $\mathcal{G}_1$ , the centre of the disc is 1, therefore the disc obtained in (2.7) is contained in the region bounded by the modified sigmoid, by Lemma 3.12. For the function $f_1$ defined in (2.3), at $z = R_{\mathcal{S}_{SG}^*}(\mathcal{G}_1) = \rho$ , $$\left|\log \frac{zf_1'(z)/f_1(z)}{2 - (zf_1'(z)/f_1(z))}\right| = \left|\log \frac{(1 + 6\rho - \rho^2)/(1 - \rho^2)}{2 - ((1 + 6\rho - \rho^2)/(1 - \rho^2))}\right| = 1.$$ (ii) The function $n(r):[0,1) \to \mathbb{R}$ defined by $n(r)=(1-5r-2r^2)(1-r^2)^{-1}+1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_{SG}}(\mathcal{G}_2)$ is the root of the equation n(r) = 2/(1+e). For $0 < r \le R_{\mathcal{S}_{SG}}(\mathcal{G}_2)$ , we have $n(r) \ge 2/1 + e$ . That is, $$\frac{r(r+5)}{1-r^2} \le \frac{e-1}{e+1}.$$ For the class $\mathcal{G}_2$ , the centre of the disc is 1, therefore the disc obtained in (2.12) is contained in the region bounded by the modified sigmoid, by Lemma 3.12. For the function $f_2$ defined in (2.8), at $z = R_{\mathcal{S}_{SG}^*}(\mathcal{G}_2) = \rho$ , $$\left|\log \frac{zf_2'(z)/f_2(z)}{2 - (zf_2'(z)/f_2(z))}\right| = \left|\log \frac{(1+5\rho)/(1-\rho^2)}{2 - ((1+5\rho)/(1-\rho^2))}\right| = 1.$$ (iii) The function defined by $s(r) = 1 - (4r(1-r^2)^{-1}), 0 \le r < 1$ is a decreasing function. Let $\rho = R_{\mathcal{S}_{SG}}(\mathcal{G}_3)$ is the root of the equation s(r) = 2/(1+e). For $0 < r \le R_{\mathcal{S}_{SG}}(\mathcal{G}_3)$ , we have $s(r) \ge 2/(1+e)$ . That is, $$\frac{4r}{1-r^2} \le \frac{e-1}{e+1}.$$ For the class $\mathcal{G}_3$ , the centre of the disc is 1, therefore the disc obtained in (2.15) is contained in the region bounded by the modified sigmoid, by Lemma 3.12. For the function $g_1$ defined in (2.3), at $z = R_{\mathcal{S}_{SG}^*}(\mathcal{G}_3) = \rho$ , $$\left|\log \frac{zg_1'(z)/g_1(z)}{2 - (zg_1'(z)/g_1(z))}\right| = \left|\log \frac{(1 + 4\rho - \rho^2)/(1 - \rho^2)}{2 - ((1 + 4\rho - \rho^2)/(1 - \rho^2))}\right| = 1.$$ Though we have no proof, we believe that the sharp radii for the class $\mathcal{G}_2$ are the following: - (1) $R_{\mathcal{S}_n}(\mathcal{G}_2) = 5 2\sqrt{6} \approx 0.1010.$ - (2) $R_{\mathcal{S}_{e}^{*}}(\mathcal{G}_{2}) = 2(e^{2} + e + 2)/(2 + e) \left(5e + \sqrt{-8 + 4e + 29e^{2}}\right) \approx 0.1276.$ - (3) $R_{S_{}^{}}(\mathcal{G}_{2}) = (15 \sqrt{217})/2 \approx 0.1345.$ - (4) $R_{\mathcal{S}_{\mathbb{Q}}^*}(\mathcal{G}_2) = (6\sqrt{2} 8)/(\sqrt{2} 1)\left(5 + \sqrt{41 12\sqrt{2}}\right) \approx 0.1183.$ - (5) $R_{\mathcal{S}_R^*}(\mathcal{G}_2) = (10\sqrt{2} 14)/(\sqrt{2} 1)\left(5 + \sqrt{81 40\sqrt{2}}\right) \approx 0.0345.$ Our estimate for these radii are respectively 0.0972, 0.1213, 0.1279, 0.1131, 0.0340 and are very much close to the above mentioned values.