Radius Constants of Sigmoid Starlike Functions

Lemma 1.1. [3] Let $$2/(1+e) < a < 2e/(1+e)$$ . If $$r_a = \frac{e-1}{e+1} - |a-1|,$$ $^2020\ Mathematics\ Subject\ Classification.\ 30{\rm C}45,\ 30{\rm C}80.$ Key words and phrases. Radius problem, Sigmoid function, Starlike functions. ^\*Corresponding Author The first author is supported by The Council of Scientific and Industrial Research(CSIR). Ref.No.:08/133(0018)/2017-EMR-I.. then <span id="page-1-2"></span> $$\{w \in \mathbb{C} : |w - a| < r_a\} \subset \Delta_{SG}. \tag{1.1}$$

Lemma 2.1. [15, Lemma 2.1] (i) If $p \in \mathcal{P}_n[A, B]$ , then for |z| = r, $$\left| p(z) - \frac{1 - ABr^{2n}}{1 - B^2r^{2n}} \right| \le \frac{(A - B)r^n}{1 - B^2r^{2n}}.$$ (ii) In particular, if $p \in \mathcal{P}_n(\alpha) := \mathcal{P}[1 - 2\alpha, -1]$ , then $$\left| p(z) - \frac{1 + (1 - 2\alpha)r^{2n}}{1 - r^{2n}} \right| \le \frac{2(1 - \alpha)r^n}{1 - r^{2n}}.$$

Lemma 2.2. [17, Lemma 2] If $p \in \mathcal{P}_n(\alpha)$ , then for |z| = r, $$\left| \frac{zp'(z)}{p(z)} \right| \le \frac{2nr^n(1-\alpha)}{(1-r^n)(1+(1-2\alpha)r^n)}.$$

Theorem 2.3. The sharp $\mathcal{S}_{SG,n}$ -radius of the class $\mathcal{S}_n^[A,B]$ is given by (i) $$R_{\mathcal{S}_{SG,n}}(\mathcal{S}_n^[A,B]) = \min\left\{1, \left(\frac{e-1}{A(1+e)-2B}\right)^{\frac{1}{n}}\right\}, \text{ when } 0 \le B < A \le 1.$$ (ii) $$R_{\mathcal{S}_{SG,n}}(\mathcal{S}_n^[A,B]) = \min\left\{1, \left(\frac{e-1}{A(1+e)-2Be}\right)^{\frac{1}{n}}\right\}, \text{ when } -1 \le B < A \le 1 \text{ with } B \le 0.$$ In particular, for the class $S$ , we have $R_{S_{SG}}(S^*) = (e-1)/(3e+1)$ .

Theorem 2.6. The radius estimates of Sigmoid starlikeness, for the classes $\mathcal{BS}^(\alpha)$ , $\mathcal{S}_L^(\alpha)$ and $\mathcal{S}_{\alpha,e}^*$ are given by - (i) $R_{\mathcal{S}_{SG}}(\mathcal{BS}^(\alpha)) = r_{\mathcal{BS}}(\alpha) := 2(e-1)/((1+e) + \sqrt{(1+e)^2 + 4\alpha(e-1)^2}), \text{ where } \alpha \in [0,1).$ (ii) $R_{\mathcal{S}_{SG}}(\mathcal{S}_L^(\alpha)) = r_L(\alpha) := ((e-1)(3+e-2\alpha(1+e)))/((1-\alpha)^2(1+e)^2), \text{ where } \alpha \in [0,1).$ - $[0, (3+e)/2(1+e)). \text{ In particular, } R_{\mathcal{S}_{SG}}(\mathcal{S}_L^) = ((e-1)(3+e))/(1+e)^2.$ $(iii) R_{\mathcal{S}_{SG}}(\mathcal{S}_{\alpha,e}^) = r_e(\alpha) := \log(2e \alpha(1+e))/(1+e)(1-\alpha), \text{ where } \alpha \in [0, (e(1+e)-2e)/((1+e)(1+e))]$ - e)(e-1). In particular, $R_{\mathcal{S}_{SG}}(\mathcal{S}_e^) = \log(2e/(1+e))$ . All estimates are sharp. (i) Let $f \in \mathcal{BS}^*(\alpha)$ . Then $zf'(z)/f(z) \prec 1 + z/(1 - \alpha z^2)$ and thus Proof. $$\left|\frac{zf'(z)}{f(z)} - 1\right| \le \left|\frac{z}{1 - \alpha z^2}\right| \le \left|\frac{r}{1 - \alpha r^2}\right| \quad \text{on } |z| = r.$$ Using Lemma 1.1, it can be said that the above disk lies in $\Delta_{SG}$ if $r/(1-\alpha r^2) \leq (e-1)^2$ 1)/(e+1). This further implies $r \leq r_{\mathcal{BS}}(\alpha)$ . Sharpness holds for the function $$f_{\mathcal{BS}}(z) = \begin{cases} z \left( \frac{1 + \sqrt{\alpha z}}{1 - \sqrt{\alpha z}} \right)^{1/(2\sqrt{\alpha})}, & \alpha \in (0, 1) \\ z e^z, & \alpha = 0. \end{cases}$$ It can be verified with the following graph that $zf'_{\mathcal{BS}}(z)/f_{\mathcal{BS}}(z)$ touches the boundary of $\Delta_{SG}$ at the points $\pm 2(e-1)/((1+e)+\sqrt{(1+e)^2+4\alpha(e-1)^2})$ . Note that the domain $\Omega_{BS}$ denotes the image of $\mathbb{D}$ mapped by the function $1 + z/(1 - \alpha z^2)$  Sharpness for $\alpha = 0.5$ Sharpness for $\alpha = 0.9$ Figure 1 (ii) Let $$f \in \mathcal{S}_L^*(\alpha)$$ , then $zf'(z)/f(z) \prec \alpha + (1-\alpha)\sqrt{1+z}$ and therefore on $|z| = r$ $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \leq |(1-\alpha)(1-\sqrt{1+z})| \leq (1-\alpha)(1-\sqrt{1-r}).$$ By Lemma 1.1, it is clear that for the above disk to lie in $\Delta_{SG}$ , we need $(1-\alpha)(1-\sqrt{1-r}) \le (e-1)/(e+1)$ , which upon simplification yields $r \le ((e-1)(3+e-2\alpha(1+e)))/((1-e-2\alpha(1+e)))$ $(\alpha)^2(1+e)^2$ ). Note that for the function $$f_L(z) = z + (1 - \alpha)z^2 + \frac{1}{16}(1 - \alpha)(1 - 2\alpha)z^3 + \cdots,$$ the result is sharp. The sharpness of this result can be verified by the following graph, where $\Omega_L$ denotes the image of $\mathbb{D}$ mapped by $\alpha + (1-\alpha)\sqrt{1+z}$ .  Figure 2 (iii) Let $$f \in \mathcal{S}_{\alpha,e}^*$$ , then $zf'(z)/f(z) \prec \alpha + (1-\alpha)e^z$ . So on $|z| = r$ $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \leq (1-\alpha)|e^z - 1| \leq (1-\alpha)(e^r - 1).$$ By Lemma 1.1, $f \in \mathcal{S}_{SG}^*$ if $(1-\alpha)(e^r-1) \leq (e-1)/(e+1)$ , which is equivalent to $r \leq r_e(\alpha)$ . The result is sharp for the function $$f_e(z) = z + (1 - \alpha)z^2 + \frac{1}{4}(1 - \alpha)(3 - 2\alpha)z^3 + \cdots$$ and is validated by the following graph. The image of $\mathbb D$ mapped by $\alpha + (1-\alpha)e^z$ is denoted by $\Omega_e$ .  Figure 3 Before we proceed further, let us recall the following classes: In [14], Mendiratta et al. considered the class of starlike functions associated with right lemniscate of Bernoulli, denoted by $\mathcal{S}_{RL}^ =$ $\mathcal{S}^(\phi)$ , where $\phi$ is given by $$\phi(z) = \sqrt{2} - (\sqrt{2} - 1)\sqrt{\frac{1 - z}{1 + 2(\sqrt{2} - 1)z}}.$$ The class of cardioid starlike functions, denoted by $S_C^ := S^(1 + 4z/3 + 2z^2/3)$ , defined by Sharma et al. [18]. For $k = \sqrt{2} + 1$ , Kumar and Ravichandran [10] introduced $\mathcal{S}_R^*$ by taking $\phi$ as 1 + z(k+z)/k(k-z).

Theorem 2.7. The sharp $\mathcal{S}_{SG}$ -radius for the classes $\mathcal{S}_{RL}$ , $\mathcal{S}_{C}$ and $\mathcal{S}_{R}$ is given by: - (i) $R_{\mathcal{S}_{SG}}(\mathcal{S}_{RL}^) =: r_{RL} = \left(4\sqrt{2} 7e 5\right) \frac{(e-1)}{(32\sqrt{2} 7e^2 + 6e(4\sqrt{2} 5) 47)} \approx 0.738309.$ (ii) $R_{\mathcal{S}_{SG}}(\mathcal{S}_C^) =: r_C = -1 + \sqrt{\frac{(-1 + 5e)}{(2 + 2e)}} \approx 0.301221.$ - (iii) $R_{\mathcal{S}_{SG}}(\mathcal{S}_R^) =: r_R = (\sqrt{(2\sqrt{2}+3)(2e^2-1)} (\sqrt{2}+1)e)/(1+e) \approx 0.645131.$ (i) Let $f \in \mathcal{S}_{RL}^*$ . Then Proof. $$\frac{zf'(z)}{f(z)} \prec \sqrt{2} - (\sqrt{2} - 1)\sqrt{\frac{1 - z}{1 + 2(\sqrt{2} - 1)z}}.$$ Thus on |z|=r, we have $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le 1 - \left( \sqrt{2} - (\sqrt{2} - 1)\sqrt{\frac{1+r}{1 - 2(\sqrt{2} - 1)r}} \right).$$ By Lemma 1.1, f is in $\mathcal{S}_{SG}^*$ if $$1 - \left(\sqrt{2} - (\sqrt{2} - 1)\sqrt{\frac{1+r}{1 - 2(\sqrt{2} - 1)r}}\right) \le \frac{e - 1}{e + 1},$$ which is equivalent to $r \leq r_{RL}$ . This result is sharp for the following function $$f_{RL}(z) = z \left( \frac{\sqrt{1-z} + \sqrt{1+2(\sqrt{2}-1)z}}{2} \right)^{2\sqrt{2}-2} \exp\left(\sqrt{2(\sqrt{2}-1)} \tan^{-1} \Psi(z)\right),$$ where $$\Psi(z) = \frac{\sqrt{2(\sqrt{2}-1)}\left(\sqrt{2(\sqrt{2}-1)z+1} - \sqrt{1-z}\right)}{2(\sqrt{2}-1)\sqrt{1-z} + \sqrt{2(\sqrt{2}-1)z+1}}.$$ The sharpness of this bound can be verified from Figure 4(i). (ii) Let $f \in \mathcal{S}_C^*$ , then we have $zf'(z)/f(z) \prec 1 + 4z/3 + 2z^2/3$ . Therefore on |z| = r, we get $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{2(r^2 + 2r)}{3},$$ which if not exceeds (e-1)/(e+1), implies that f lies in $\mathcal{S}_{SG}^*$ , by Lemma 1.1. Solving this, we get $r \leq r_C$ . In order to verify the sharpness of this result, we consider the following function. $$f_C(z) = z \exp\left(\frac{4z}{3} + \frac{z^2}{3}\right).$$ Clearly $f_C \in \mathcal{S}_C^*$ and moreover $zf_C'(z)/f_C(z)$ touches the boundary of $\Delta_{SG}$ at the point $z_0 = -1 + \sqrt{(-1 + 5e/(2 + 2e))}$ , as shown in Figure 4(ii). (iii) Let $f \in \mathcal{S}_R^*$ , then $$\frac{zf'(z)}{f(z)} \prec 1 + \frac{z(k+z)}{k(k-z)},$$ where $k = \sqrt{2} + 1$ . Thus on |z| = r, $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{r(k+r)}{k(k-r)}.$$ In view of Lemma 1.1, $f \in \mathcal{S}_{SG}^*$ if $r(k+r)/k(k-r) \leq (e-1)/(e+1)$ . Solving this inequality, we obtain $r \leq r_R$ . The equality of the radius estimate holds for the function $$f_R(z) = \frac{k^2 z}{(k-z)^2} e^{-z/k}, \quad k = \sqrt{2} + 1.$$ Figure 4(iii) verifies the sharpness of the result. Note that $\Omega_{RL}$ , $\Omega_c$ and $\Omega_R$ denote the image of $\mathbb D$ mapped by zf'(z)/f(z) for $f_{RL}$ , $f_C$ and $f_R$ repectively. Let us recall the following classes in order to obtain our next result. By taking $\phi(z) = z + \sqrt{1+z^2}$ , Sharma et al. [19] introduced $\mathcal{S}^_{\emptyset}$ . Similarly, Kumar and Gangania [9] introduced $\mathcal{S}^_{\wp}$ by taking $\phi$ as $1+ze^z$ , the cardioid function.

Theorem 2.8. The sharp $\mathcal{S}_{SG}$ -radii for the classes $\mathcal{S}_{\mathcal{A}}$ and $\mathcal{S}_{\wp}^*$ is given by (i) $$R_{\mathcal{S}_{SG}}(\mathcal{S}_{\mathcal{C}}^) = \frac{-1-2e+3e^2}{4e+4e^2} \approx 0.389089.$$ <span id="page-6-0"></span> Figure 4 (ii) $R_{\mathcal{S}_{SG}}(\mathcal{S}_{\wp}^) = r_{\wp} \approx 0.331672$ , where $r_{\wp}$ is the smallest positive root of the equation $(e + 1)re^r = e - 1$ . Proof. (i) Let $f \in \mathcal{S}_{\mathbb{Q}}^*$ , then $zf'(z)/f(z) \prec z + \sqrt{1+z^2}$ . Therefore on |z| = r, $\left| \frac{zf'(z)}{f(z)} - 1 \right| = |z + \sqrt{1+z^2} - 1| \le r + \sqrt{1+r^2} - 1.$ Now by using Lemma 1.1, the above disk lies inside the domain $\Delta_{SG}$ if $r + \sqrt{1 + r^2} - 1 \le (e-1)/(e+1)$ . Solving this equation, we obtain the desired bound of r. The result is sharp for the function $$f_{\mathbb{C}}(z) = z \exp\left(\int_0^z \frac{t + \sqrt{1 + t^2} - 1}{t} dt\right) = z + z^2 + \frac{3z^3}{4} + \frac{5z^4}{12} + \frac{z^5}{6} + \cdots$$ (ii) Let $f \in \mathcal{S}_{\wp}^*$ , then it is clear that $zf'(z)/f(z) \prec 1 + ze^z$ . So we have $$\left| \frac{zf'(z)}{f(z)} - 1 \right| = |ze^z| \le re^r \quad \text{on } |z| = r.$$ By using Lemma 1.1, we can say that $f \in \mathcal{S}_{SG}^*$ if $re^r \leq (e-1)/(e+1)$ , which is equivalent to $r \leq r_{\wp}$ . The sharpness of the result can be verified by the function $f_{\wp}(z) = ze^{e^z-1}$ . The following graph depicts the sharpness of both the estimates. Note that the image of $\mathbb{D}$ mapped by $z + \sqrt{1+z^2}$ and $1 + ze^z$ are respectively denoted by $\Omega_{\emptyset}$ and $\Omega_{\wp}$ .  Figure 5 Now we consider the following classes for our next result. The class $\mathcal{S}_{Ne}$ , defined by Wani and Swaminathan [21] by taking $\phi$ as $1+z-z^3/3$ and the class $\mathcal{S}_S^ = \mathcal{S}^*(1+\sin z)$ , introduced by Cho et al. [2].

Theorem 2.9. The $S_{SG}$ -radius for the classes $S_{Ne}$ and $S_S^*$ is given by: (i) $R_{\mathcal{S}_{SG}}(\mathcal{S}_{Ne}^) = r_{Ne} \approx 0.43473$ , which is the smallest positive root of the equation $(e+1)(3r+r^3) = 3(e-1)$ . (ii) $$R_{\mathcal{S}_{SG}}(\mathcal{S}_S^) = \log\left(\frac{\sqrt{2(1+e^2)}+e-1}{1+e}\right) \approx 0.447074.$$

Theorem 2.10. The $S_{SG,n}^*$ radius of the class $G_1$ is given by $$R_{\mathcal{S}_{SG,n}^*}(\mathcal{G}_1) = r_{\mathcal{G}_1} := \frac{(e-1)^{\frac{1}{n}}}{(2n(1+e) + \sqrt{4n^2(1+e)^2 + (e-1)^2})^{\frac{1}{n}}}.$$ The estimate is sharp.

Theorem 2.11. The $\mathcal{S}_{SG,n}^*$ radius of the class $\mathcal{G}_2$ is given by $$R_{\mathcal{S}_{SG,n}^*}(\mathcal{G}_2) = r_{\mathcal{G}_2} := \frac{(2(e-1))^{\frac{1}{n}}}{(3n(1+e) + \sqrt{(3n(e+1))^2 + 4(e-1)(n(e+1) + (e-1))})^{\frac{1}{n}}}.$$ The estimate is sharp.

Theorem 2.12. The $S_{SG,n}^*$ radius of the class $G_3$ is given by $$R_{\mathcal{S}_{SG,n}^*}(\mathcal{G}_3) = r_{\mathcal{G}_3} := \frac{(2(e-1))^{\frac{1}{n}}}{(3n(1+e) + \sqrt{(3n(e+1))^2 + 4(e-1)(n(e+1) + (e-1))})^{\frac{1}{n}}}.$$ The estimate is sharp.

Theorem 2.13. The $S_{SG,n}^*$ radius of the class $G_4$ is given by $$R_{\mathcal{S}_{SG,n}^*}(\mathcal{G}_4) = r_{\mathcal{G}_4} := \frac{(2(e-1))^{\frac{1}{n}}}{((n+1)(1+e) + \sqrt{(n+1)^2(1+e)^2 + 4(e-1)((e+1)n-2)})^{\frac{1}{n}}}.$$ The estimate is sharp.

Theorem 2.14. The sharp $\mathcal{S}_{SG,n}$ radius for the class $\mathcal{CS}_n^(\alpha)$ is given by $$R_{\mathcal{S}_{SG,n}}(\mathcal{CS}_n^(\alpha)) = r_{cs} := (e-1)/((1+e)(1+n-\alpha) + \sqrt{(e+1)^2(1+n-\alpha)^2 + (e-1)((1-2\alpha)(e+1)+2e)})$$

Theorem 2.15. The sharp $\mathcal{S}_{SG,n}^*$ radius for $\mathcal{W}_n$ is $$R_{\mathcal{S}_{SG,n}^*}(\mathcal{W}_n) := r_w = \left(\frac{e-1}{\sqrt{n^2(e+1)^2 + (e-1)^2} + n(e+1)}\right)^{1/n}.$$

Theorem 2.16. The $S_{SG,n}^*$ radius for the class $\mathcal{M}(\beta)$ is given by $$S_{SG}^*(\mathcal{M}_n(\beta)) = \left(\frac{e-1}{(e-1)+(e+1)(\beta)-1}\right)^{1/n}.$$ The result is sharp for the function $f(z) = z(1-z^n)^{(2(\beta-1)/n)}$ . The proof of the above theorem is omitted here as it is much similar to the proof of Theorem 2.14. Next we consider the class $C(\alpha)$ ( $0 \le \alpha < 1$ ), of convex functions of order $\alpha$ . Note that this class is a generalization of the class C, which can be obtained by setting $\alpha = 0$ .

Theorem 2.17. Let $f \in \mathcal{S}_{SG}^*$ , then f is convex of order $\alpha$ in $|z| < r_{\mathcal{C}}(\alpha)$ , where $r_{\mathcal{C}}(\alpha) \in (0,1)$ is the smallest positive root of $$e^r(r+\alpha) - 2 + \alpha = 0.$$