Starlikeness of a product of starlike functions with non-vanishing polynomials

Theorem 2.1. If the function $f \in \mathcal{ST}(\alpha)$ , then the radius of starlikeness of order $\lambda$ for the function F defined in (2.1) is given by $$R_{\mathcal{ST}(\lambda)}(F) = \frac{2(1-\lambda)}{2-2\alpha+\beta+\sqrt{(2-2\alpha+\beta)^2+4(1-\lambda)(2\alpha+\beta-1-\lambda)}}.$$ PROOF. We start with finding the disc in which zF'(z)/F(z) lies for $z \in \overline{\mathbb{D}}_r$ , then using this disc, we determine the radius of starlikeness of order $\lambda$ for F. For the function F given by (2.1), a calculation shows that <span id="page-2-2"></span> $$\frac{zF'(z)}{F(z)} = \frac{zf'(z)}{f(z)} + \frac{\beta}{n} \frac{zQ'(z)}{Q(z)}.$$ (2.2) Since $f \in \mathcal{ST}(\alpha)$ , it is well-known that zf'(z)/f(z) has positive real part and so <span id="page-2-3"></span> $$\left| \frac{zf'(z)}{f(z)} - \frac{1 + (1 - 2\alpha)r^2}{1 - r^2} \right| \leqslant \frac{2(1 - \alpha)r}{1 - r^2}, \quad |z| \leqslant r. \tag{2.3}$$ or equivalently zf'(z)/f(z) lies in the disc $\mathbb{D}(a_f(r); c_f(r))$ for $|z| \leq r$ where $$a_f(r) = \frac{1 + (1 - 2\alpha)r^2}{1 - r^2}, \quad c_f(r) = \frac{2(1 - \alpha)r}{1 - r^2}.$$ (2.4) Let $z_k$ , $k=1,2,\ldots,n$ denote the zeros of the polynomial Q, then the polynomial Q is a constant multiple of $\prod_{k=1}^n (z-z_k)$ and so <span id="page-2-1"></span> $$\frac{zQ'(z)}{Q(z)} = \sum_{k=1}^{n} \frac{z}{z - z_k}.$$ (2.5) Since $z_k \in \mathbb{C} \setminus \mathbb{D}$ for every k, the bilinear transformation $z/(z-z_k)$ maps $\overline{\mathbb{D}}_r$ to a disc. Indeed, [6, Lemma 3.2] shows that $$\left| \frac{z}{z - z_k} + \frac{r^2}{1 - r^2} \right| \leqslant \frac{r}{1 - r^2}, \quad |z| \leqslant r$$ for every k and hence, using (2.5), we have <span id="page-2-5"></span><span id="page-2-4"></span> $$\left| \frac{zQ'(z)}{Q(z)} + \frac{nr^2}{1 - r^2} \right| \leqslant \frac{nr}{1 - r^2}, \quad |z| \leqslant r.$$ (2.6) Using (2.2), we get $$\left| \frac{zF'(z)}{F(z)} - \frac{1 - (2\alpha - 1 + \beta)r^2}{1 - r^2} \right| = \left| \frac{zf'(z)}{f(z)} - \frac{1 + (1 - 2\alpha)r^2}{1 - r^2} + \frac{\beta zQ'(z)}{n} + \frac{\beta r^2}{1 - r^2} \right| \\ \leq \left| \frac{zf'(z)}{f(z)} - \frac{1 + (1 - 2\alpha)r^2}{1 - r^2} \right| + \left| \frac{\beta zQ'(z)}{n} + \frac{\beta r^2}{1 - r^2} \right|, (2.7)$$ for $|z| \leq r$ . Since $\beta \in \mathbb{R}$ is positive, using the equations (2.3) and (2.6), the inequality (2.7) gives <span id="page-3-0"></span> $$\left| \frac{zF'(z)}{F(z)} - \frac{1 - (2\alpha - 1 + \beta)r^2}{1 - r^2} \right| \leqslant \frac{(2 - 2\alpha + \beta)r}{1 - r^2}, \quad |z| \leqslant r. \tag{2.8}$$ Define the functions $a_F$ and $c_F$ by $$a_F(r) := \frac{1 - (2\alpha - 1 + \beta)r^2}{1 - r^2}$$ and $c_F(r) := \frac{(2 - 2\alpha + \beta)r}{1 - r^2}$ . so that $zF'(z)/F(z) \in \mathbb{D}(a_F(r); c_F(r))$ . It is observed that the center $a_F$ is an increasing function of r for $2\alpha + \beta - 2 < 0$ , and is a decreasing function of r for $2\alpha + \beta - 2 \ge 0$ . From (2.8), it follows that $$\operatorname{Re} \frac{zF'(z)}{F(z)} \geqslant a_F(r) - c_F(r)$$ $$= \frac{1 - (2\alpha - 1 + \beta)r^2}{1 - r^2} - \frac{(2 - 2\alpha + \beta)r}{1 - r^2}$$ $$= \frac{1 - (1 - 2\alpha)r}{1 + r} - \frac{\beta r}{1 - r} =: \psi(r). \tag{2.9}$$ The equation $\psi(\alpha, \beta, r) = \lambda$ is simplifies to <span id="page-3-2"></span><span id="page-3-1"></span> $$(1 - 2\alpha - \beta + \lambda)r^2 - (2 - 2\alpha + \beta)r + 1 - \lambda = 0$$ (2.10) and so the smallest positive root of the equation $\psi(\alpha, \beta, r) = \lambda$ in the interval (0, 1) is given by $$\sigma_0 := \frac{2(1-\lambda)}{2 - 2\alpha + \beta + \sqrt{(2 - 2\alpha + \beta)^2 + 4(1-\lambda)(2\alpha + \beta - 1 - \lambda)}}.$$ It can be seen that $$\psi'(r) = -\frac{2(1-\alpha)(1-r)^2 + \beta(1+r)^2}{(1-r^2)^2} < 0,$$ which shows that $\psi$ is a decreasing function of $r \in (0,1)$ . Therefore, for $r < \sigma_0$ , we have $\psi(\alpha,\beta,r) > \psi(\alpha,\beta,\sigma_0) = \lambda$ and so (2.9) implies that $\text{Re}(zF'(z)/F(z)) \geqslant \psi(\alpha,\beta,r) > \lambda$ for all $r < \sigma_0$ , or in other words, the radius of starlikeness of order $\lambda$ of the function F is at least $\sigma_0$ . To show that the radius obtained is the best possible, take $f(z) = z(1-z)^{2\alpha-2} \in \mathcal{ST}(\alpha)$ and the polynomial $Q(z) = (1+z)^n$ . For these choices of the functions f and Q, we have <span id="page-3-3"></span> $$F(z) = z(1-z)^{2\alpha-2}(1+z)^{\beta},$$ which implies $$\frac{zF'(z)}{F(z)} = 1 + \frac{(2-2\alpha)z}{1-z} + \frac{\beta z}{1+z} = \frac{(1-2\alpha-\beta)z^2 + (2-2\alpha+\beta)z + 1}{1-z^2} = \lambda + \frac{(1-2\alpha-\beta+\lambda)z^2 + (2-2\alpha+\beta)z + 1-\lambda}{1-z^2}.$$ (2.11) Using the fact that $\sigma_0$ is the positive root of the polynomial in (2.10), the equation (2.11) shows that $\text{Re}(zF'(z)/F(z)) = \lambda$ for $z = -\sigma_0$ proving sharpness of $\sigma_0$ . For $\lambda = 0$ , the radius of starlikeness for the function F given by (2.1) is $$R_{\mathcal{ST}}(F) = \frac{2}{2 - 2\alpha + \beta + \sqrt{(2 - 2\alpha + \beta)^2 + 4(2\alpha + \beta - 1)}}.$$ When $\alpha$ goes to 1 in Theorem 2.1 we get that the radius of starlikeness of order $\lambda$ for the function $F(z)=z(Q(z))^{\beta/n}$ , where Q is a non-constant polynomial of degree n non-vanishing on the unit disc and $\beta>0$ , comes out to be $(1-\lambda)/(\beta+1-\lambda)$ . This result for $\beta=n$ coincides with the one obtained by Başgöze in [2, Theorem 3]. Moreover, by letting $\beta\to 0$ in Theorem 2.1, we obtain the radius of starlikeness of order $\lambda$ for the class of starlike functions of order $\alpha,0\leqslant\alpha<1$ (see [9, p. 88]).

Lemma 3.1. [16, Lemma 2.2] For 1/e < a < e, let $r_a$ be given by $$r_a = \begin{cases} a - \frac{1}{e} & \text{if } \frac{1}{e} < a \leqslant \frac{e + e^{-1}}{2} \\ e - a & \text{if } \frac{e + e^{-1}}{2} \leqslant a < e. \end{cases}$$ Then, $\{w : |w - a| < r_a\} \subset \{w : |\log w| < 1\} = \Omega_e$ , where $\Omega_e$ is the image of the unit disc $\mathbb{D}$ under the exponential function.

Theorem 3.2. If the function $f \in ST(\alpha)$ , then the radius of starlikeness associated with the exponential function for the function F defined in (2.1) is given by <span id="page-4-1"></span> $$R_{\mathcal{S}_e^*}(F) = \begin{cases} \sigma_0 & \text{if } 2\alpha + \beta - 2 \geqslant 0 \\ \sigma_0 & \text{if } 2\alpha + \beta - 2 < 0 \text{ and } X(\alpha, \beta) \leqslant 0 \\ \tilde{\sigma_0} & \text{if } 2\alpha + \beta - 2 < 0 \text{ and} X(\alpha, \beta) > 0, \end{cases}$$ where $$\sigma_0 = \frac{2(e-1)}{e(2-2\alpha+\beta) + \sqrt{(e(2-2\alpha+\beta))^2 - 4(e-1)(1-e(2\alpha-1+\beta))}},$$ $$\tilde{\sigma_0} = \frac{2(e-1)}{(2-2\alpha+\beta) + \sqrt{(2-2\alpha+\beta)^2 - 4(e-1)(2\alpha-1+\beta-e)}}$$ and $$X(\alpha,\beta) = 2(2+\beta-2\alpha)(1+e^2-2e(2\alpha+\beta-1))((-2\alpha+2+\beta) - \sqrt{(-2\alpha+2+\beta)^2-4(e-1)(2\alpha-1+\beta-e)}) + 4(2\alpha+\beta-1-e)(e^2-1)(2\alpha+\beta-2).$$ (3.1) PROOF. Our aim is to show that the $\mathbb{D}(a_F(r); c_F(r)) \subset \Omega_e$ for all $0 < r \leq R_{\mathcal{S}_e^*}(F)$ . Let $$\sigma_0 := \frac{2(e-1)}{e(2-2\alpha+\beta) + \sqrt{(e(2-2\alpha+\beta))^2 - 4(e-1)(1-e(2\alpha-1+\beta))}},$$ and $$\tilde{\sigma_0} := \frac{2(e-1)}{(2-2\alpha+\beta) + \sqrt{(2-2\alpha+\beta)^2 - 4(e-1)(2\alpha-1+\beta-e)}}.$$ It is clear that $\sigma_0$ and $\tilde{\sigma_0}$ are both positive as both $2-2\alpha+\beta$ and e-1 are positive.. For the polynomial <span id="page-5-0"></span> $$\phi(r) := (1 - e(-1 + 2\alpha + \beta))r^2 - e(2 - 2\alpha + \beta)r + (e - 1)$$ (3.2) obtained from the equivalent form $\phi(r)=0$ of the equation $c_F(r)=a_F(r)-1/e$ , it is observed that $\phi(0)=e-1>0, \ \phi(1)=-2e\beta<0$ , showing that there is a zero for $\phi$ in the interval (0,1), namely $\sigma_0$ . Also, the positive root of the equation $c_F(r)=e-a_F(r)$ or $\psi(r)=0$ with <span id="page-5-1"></span> $$\psi(r) := (-1 + 2\alpha + \beta - e)r^2 - (2 - 2\alpha + \beta)r + (e - 1)$$ (3.3) is $\tilde{\sigma_0}$ . To verify that $\psi$ has a zero in the interval (0,1), it is seen that $\psi(0)=e-1>0$ and $\psi(1)=4(\alpha-1)<0$ . Case (i): $2\alpha + \beta - 2 \geqslant 0$ . Since the center $a_F$ in (2.8) is a decreasing function of r, $a_F(r) > a_F(\sigma_0)$ , for $r \in (0, \sigma_0)$ . By definition, $\sigma_0$ is the solution the equation $c_F(r) = a_F(r) - 1/e$ , also, the radius in (2.8) satisfies $c_F(r) > 0$ , $r \in (0, 1)$ , together imply that $a_F(r) > a_F(\sigma_0) > 1/e$ , for $r \in (0, \sigma_0)$ . Further, $a_F(r) < a_F(0) = 1 < (e + e^{-1})/2 \approx 1.54308$ , for $r \in (0, \sigma_0)$ . Thus, it is established that $$\frac{1}{e} < a_F(r) \leqslant \frac{e + e^{-1}}{2e}$$ , for $r \in (0, \sigma_0)$ . Applying Lemma 3.1 we get $\mathbb{D}(a_F(\sigma_0); c_F(\sigma_0)) \subset \Omega_e$ that is, the radius of starlikeness associated with the exponential function for the function F is at least $\sigma_0$ . Case (ii): $2\alpha + \beta - 2 < 0$ and $X(\alpha, \beta) \leq 0$ . The number $$\tilde{\sigma}_1 = \sqrt{\frac{1 - 2e + e^2}{1 + 2e + e^2 - 4e\alpha - 2e\beta}} < 1$$ is the positive root of the equation $a_F(r)=(e+e^{-1})/2$ , or equivalently, $\zeta(r)=0$ with $\zeta(r):=(1+e^2-2e(2\alpha+\beta-1))r^2+2e-e^2-1$ . Here, $\zeta(0)=2e-e^2-1\approx-2.9525<0$ and $\zeta(1)=-2e(2\alpha+\beta-2)>0$ justifies the existence of a zero for the polynomial $\zeta$ in the interval (0,1). Further, $$\zeta(\tilde{\sigma_0}) = \frac{1}{4(-1 - e + 2\alpha + \beta)^2} \left( 2(2 + \beta - 2\alpha)(1 + e^2 - 2e(2\alpha + \beta - 1)) \right)$$ $$\times \left( (-2\alpha + 2 + \beta) - \sqrt{(-2\alpha + 2 + \beta)^2 - 4(e - 1)(2\alpha - 1 + \beta - e)} \right)$$ $$+ 4(2\alpha + \beta - 1 - e)(e^2 - 1)(2\alpha + \beta - 2),$$ (3.4) and from this with (3.1) we infer that $X(\alpha,\beta) \leqslant 0$ implies $\zeta(\tilde{\sigma}_0) \leqslant 0$ . Also, $\zeta(0) < 0$ , $\zeta(\tilde{\sigma}_0) \leqslant 0$ along with the fact $\zeta(\tilde{\sigma}_1) = 0$ gives $\tilde{\sigma}_0 \leqslant \tilde{\sigma}_1$ , which implies that $a_F(\tilde{\sigma}_0) \leqslant a_F(\tilde{\sigma}_1) = (e+e^{-1})/2$ . The application of Lemma 3.1 gives $\mathbb{D}(a_F(\sigma_0); c_F(\sigma_0)) \subset \Omega_e$ or in other words, the radius of starlikeness associated with the exponential function for F is at least $\sigma_0$ . Case (iii): $2\alpha + \beta - 2 < 0$ and $X(\alpha, \beta) > 0$ . Here following the same line of thought as in Case (ii), $\zeta(\tilde{\sigma_0}) > 0$ implies $a_F(\tilde{\sigma_0}) > (e + e^{-1})/2$ , and thus Lemma 3.1 gives the required radius to be at least $\tilde{\sigma_0}$ . To show that the obtained radius values are the best possible, take $f(z) = z/(1-z)^{-2\alpha+2} \in \mathcal{ST}(\alpha)$ and the polynomial $Q(z) = (1+z)^n$ , these choices give the expression for zF'(z)/F(z), as already shown in the proof of Theorem 2.1, to be $$\frac{zF'(z)}{F(z)} = \frac{(1 - 2\alpha - \beta)z^2 + (1 + (1 - 2\alpha) + \beta)z + 1}{1 - z^2}$$ $$= e - \frac{(-1 + 2\alpha + \beta - e)z^2 - (2 - 2\alpha + \beta)z + e - 1}{1 - z^2}.$$ (3.5) It is seen that (3.5) can also be written as $$\frac{zF'(z)}{F(z)} = \frac{1}{e} + \frac{(1 - e(-1 + 2\alpha + \beta))z^2 + e(2 - 2\alpha + \beta)z + e - 1}{e(1 - z^2)}.$$ (3.6) The definition of the polynomial $\phi$ in (3.2) for $r = \sigma_0$ together with (3.6) gives <span id="page-6-0"></span> $$\left|\log \frac{(-\sigma_0)F'(-\sigma_0)}{F(-\sigma_0)}\right| = \left|\log \frac{1}{e}\right| = 1,$$ proving sharpness for $\sigma_0$ . Further, the polynomial $\psi$ in (3.3) for $r = \tilde{\sigma_0}$ and (3.5) provide <span id="page-6-1"></span> $$\left|\log \frac{\tilde{\sigma}_0 F'(\tilde{\sigma}_0)}{F(\tilde{\sigma}_0)}\right| = \left|\log e\right| = 1.$$ This proves sharpness for $\tilde{\sigma_0}$ . If we let $\alpha$ goes to 1 in Theorem 3.2, then the radius of starlikeness associated with the exponential function for the function $F(z)=z(Q(z))^{\beta/n}$ where Q is a non-constant polynomial of degree n non-vanishing on the unit disc and $\beta>0$ , comes out to be $(e-1)/(e\beta+e-1)$ . Moreover, when $\beta\to 0$ in Theorem 3.2, we obtain the radius of starlikeness associated with the exponential function for the class of starlike functions of order $\alpha,0\leqslant\alpha<1$ obtained by Mendiratta et al. in [16, Theorem 3.4] and also by Khatter et al. in [10, Theorem 2.17 (2)] for $\mathcal{S}_{0,e}^*$ with $A=1-2\alpha$ and B=-1.

Lemma 4.1. [20, lemma 2.5] For 1/3 < a < 3, $$r_a = \begin{cases} \frac{3a-1}{3} & \text{if } \frac{1}{3} < a \le \frac{5}{3} \\ 3-a & \text{if } \frac{5}{3} \le a < 3. \end{cases}$$ Then $\{w: |w-a| < r_a\} \subset \Omega_c$ . Here $\Omega_c$ is the region bounded by the cadioid $\{x + \iota y: (9x^2 + 9y^2 - 18x + 5)^2 - 16(9x^2 + 9y^2 - 6x + 1) = 0\}$ . The following theorem gives the radius of starlikeness associated with a cardioid for the function F given in (2.1).

Theorem 4.2. Let the function $f \in \mathcal{ST}(\alpha)$ , the $\mathcal{S}_c^*$ radius for the function F defined in (2.1) is $$R_{\mathcal{S}_c^*}(F) = \begin{cases} \sigma_0 & \text{if } 2\alpha + \beta - 2 \geqslant 0 \\ \sigma_0 & \text{if } 2\alpha + \beta - 2 < 0 \text{ and } X(\alpha, \beta) \leqslant 0 \\ \tilde{\sigma_0} & \text{if } 2\alpha + \beta - 2 < 0 \text{ and} X(\alpha, \beta) > 0, \end{cases}$$ where <span id="page-7-1"></span> $$\sigma_0 = \frac{4}{3(2 - 2\alpha + \beta) + \sqrt{(6 - 6\alpha + 3\beta)^2 + 8(6\alpha + 3\beta - 4)}},$$ $$\tilde{\sigma_0} = \frac{4}{(2 - 2\alpha + \beta) + \sqrt{(2 - 2\alpha + \beta)^2 - 8(2\alpha + \beta - 4)}}$$ and $$X(\alpha, \beta) = 2(8 - 6\alpha - 3\beta)(6\alpha - 6 - 3\beta)(6\alpha - 6 - 3\beta) + \sqrt{(6 - 6\alpha + 3\beta)^2 + 8(6\alpha + 3\beta - 4))} + 48(6\alpha + 3\beta - 4)(2 - 2\alpha - \beta).$$ (4.1) PROOF. Consider the disc mentioned in (2.8), it needs to be shown that this disc lies in the region $\Omega_c$ for all $0 < r \leqslant R_{\mathcal{S}_c^*}(F)$ . Take the equations $c_F(r) = a_F(r) - 1/3$ and $c_F(r) = 3 - a_F(r)$ ; which are equivalent to $\phi(r) = 0$ and $\psi(r) = 0$ respectively, with the corresponding polynomials $\phi$ and $\psi$ given by <span id="page-7-2"></span> $$\phi(r) := (3(2\alpha - 1 + \beta) - 1)r^2 + 3(2 - 2\alpha + \beta)r - 2, \tag{4.2}$$ and <span id="page-7-3"></span> $$\psi(r) := (2\alpha - 1 + \beta - 3)r^2 - (2 - 2\alpha + \beta)r + 2. \tag{4.3}$$ For the polynomial $\phi$ , $\phi(0) = -2 < 0$ , $\phi(1) = 6\beta > 0$ ; thus $\phi$ has a zero in the interval (0,1), let it be denoted by $\sigma_0$ . Also, considering the polynomial $\psi$ , it is seen that $\psi(0) = 2 > 0$ and $\psi(1) = 4(\alpha - 1) < 0$ , let the zero of $\psi$ in the interval (0,1) be denoted by $\tilde{\sigma_0}$ . Then it is obtained that $$\sigma_0 := \frac{4}{3(2 - 2\alpha + \beta) + \sqrt{(6 - 6\alpha + 3\beta)^2 + 8(6\alpha + 3\beta - 4)}},$$ and $$\tilde{\sigma_0} := \frac{4}{(2 - 2\alpha + \beta) + \sqrt{(2 - 2\alpha + \beta)^2 - 8(2\alpha + \beta - 4)}}.$$ Here, it is evident from their values that both $\sigma_0$ and $\tilde{\sigma_0}$ are indeed positive. Case (i): $2\alpha + \beta - 2 \ge 0$ . First it is shown that the center $a_F$ in (2.8) satisfies <span id="page-7-0"></span> $$\frac{1}{3} < a_F(r) < \frac{5}{3}, \ r \in (0, \sigma_0). \tag{4.4}$$ The fact that the center is a decreasing function of r, implies $a_F(r) > a_F(\sigma_0)$ for $r \in (0, \sigma_0)$ . Also, $\sigma_0$ is the root of the equation $a_F(r) - 1/3 = c_F(r)$ , along with the fact that the radius satisfies $c_F(r) > 0$ , $r \in (0,1)$ gives $a_F(r) > a_F(\sigma_0) > 1/3$ for $r \in (0,\sigma_0)$ . Further, $a_F(r) < a_F(0) = 1 < 5/3$ , $r \in (0,\sigma_0)$ . This proves (4.4) and applying Lemma 4.1, infers that $\mathbb{D}(a_F(\sigma_0); c_F(\sigma_0)) \subset \Omega_c$ , that is, the radius of starlikeness associated with a cardioid for the function F is at least $\sigma_0$ . Case (ii): $2\alpha + \beta - 2 < 0$ and $X(\alpha, \beta) \le 0$ . The equation $a_F(r) = 5/3$ from Lemma 4.1, takes the form $\zeta(r) = 0$ , with $\zeta(r) := (8 - 6\alpha - 3\beta)r^2 - 2$ . Then, it is seen that $\zeta(0) = -2 < 0$ and $\zeta(1) = -3(2\alpha + \beta - 2) > 0$ . This shows that $\zeta$ has a zero in the interval (0, 1), let this be denoted by $\tilde{\sigma_1}$ , then <span id="page-8-0"></span> $$\tilde{\sigma_1} = \sqrt{\frac{2}{8 - 6\alpha - 3\beta}}.$$ Also since $$\zeta(\sigma_0) = \frac{1}{4(-4+6\alpha+3\beta)^2} (2(8-6\alpha-3\beta)(6\alpha-6-3\beta)(6\alpha-6-3\beta) + \sqrt{(6-6\alpha+3\beta)^2+8(6\alpha+3\beta-4)}) + 48(6\alpha+3\beta-4)(2-2\alpha-\beta)),$$ (4.5) from (4.1) and (4.5) it is evident that $X(\alpha, \beta) \leq 0$ is equivalent to saying $\zeta(\sigma_0) \leq 0$ . The facts that $\zeta(0) < 0$ , $\zeta(\sigma_0) \leq 0$ and $\zeta(\tilde{\sigma}_1) = 0$ together imply that $\sigma_0 \leq \tilde{\sigma}_1$ . Further, $\sigma_0 \leq \tilde{\sigma}_1$ will imply that $a_F(\sigma_0) \leq a_F(\tilde{\sigma}_1) = 5/3$ , and thus using Lemma 4.1, $\mathbb{D}(a_F(\sigma_0); c_F(\sigma_0)) \subset \Omega_c$ . Thus, proving that the required radius value is at least $\sigma_0$ . Case (iii): $2\alpha + \beta - 2 < 0$ and $X(\alpha, \beta) > 0$ . On the similar lines as in Case (ii), $X(\alpha, \beta) > 0$ implies $\zeta(\sigma_0) > 0$ , which gives that $a_F(\sigma_0) > 5/3$ , this inturn, after another application of Lemma 4.1 concludes that the radius of starlikeness associated with a cardioid for the function F is at least $\tilde{\sigma_0}$ . To verify the sharpness of the obtained radius values, take $f(z) = z/(1-z)^{-2\alpha+2} \in \mathcal{ST}(\alpha)$ and the polynomial Q as $Q(z) = (1+z)^n$ . Thus, the expression for zF'(z)/F(z) as seen in the proof for Theorem 2.1, becomes <span id="page-8-1"></span> $$\frac{zF'(z)}{F(z)} = \frac{(1 - 2\alpha - \beta)z^2 + (2 - 2\alpha + \beta)z + 1}{1 - z^2}.$$ (4.6) The polynomial $\phi$ in (4.2) gives that for $r = \sigma_0$ , <span id="page-8-2"></span> $$3((1 - 2\alpha - \beta)r^2 - (2 - 2\alpha + \beta)r + 1) = 1 - r^2.$$ (4.7) Thus, the sharpness for $\sigma_0$ is proved by using (4.6) and (4.7) which gives that $zF'(z)/F(z) = 1/3 = \varphi_c(-1)$ for $z = -\sigma_0$ . Also, the polynomial $\psi$ in (4.3) for $r = \tilde{\sigma_0}$ gives <span id="page-8-3"></span> $$(1 - 2\alpha - \beta)r^2 + (2 - 2\alpha + \beta)r + 1 = 3(1 - r^2). \tag{4.8}$$ Thus, using (4.8) in (4.6) it is seen that $\tilde{\sigma_0}F'(\tilde{\sigma_0})/F(\tilde{\sigma_0})=3=\varphi_c(1)$ . This proves the sharpness of $\tilde{\sigma_0}$ . If we let $\alpha$ goes to 1 in Theorem 4.2, then we obtain that the radius of starlikeness associated with a cardioid for the function $F(z)=z(Q(z))^{\beta/n}$ , where Q is a non-constant polynomial of degree n non-vanishing on the unit disc and $\beta>0$ , comes out to be $2/(2+3\beta)$ . Moreover, by letting $\beta\to 0$ in Theorem 4.2, we get the radius of starlikeness associated with a cardioid ([20, Theorem 4.7] with $A=1-2\alpha$ and B=-1) for the class of starlike functions of order $\alpha,0\leqslant \alpha<1$ .

Lemma 5.1. [11, lemma 2.2] For $2(\sqrt{2}-1) < a < 2$ , $$r_a = \begin{cases} a - 2(\sqrt{2} - 1) & \text{if } 2(\sqrt{2} - 1) < a \le \sqrt{2} \\ 2 - a & \text{if } \sqrt{2} \le a < 2. \end{cases}$$ Then $\{w : |w - a| < r_a\} \subset \varphi_R(\mathbb{D})$ .

Theorem 5.2. If the function $f \in \mathcal{ST}(\alpha)$ , then the $\mathcal{S}_R^*$ radius for the function F defined in (2.1) is given by <span id="page-9-1"></span> $$R_{\mathcal{S}_R^*}(F) = \begin{cases} \sigma_0 & \text{if } 2\alpha + \beta - 2 \geqslant 0 \\ \sigma_0 & \text{if } 2\alpha + \beta - 2 < 0 \text{ and } X(\alpha, \beta) \leqslant 0 \\ \tilde{\sigma_0} & \text{if } 2\alpha + \beta - 2 < 0 \text{ and} X(\alpha, \beta) > 0, \end{cases}$$ where $$\sigma_0 = \frac{2(3 - 2\sqrt{2})}{(2 - 2\alpha + \beta) + \sqrt{(-2 + 2\alpha - \beta)^2 - 4(3 - 2\sqrt{2})(2\sqrt{2} - 1 - 2\alpha - \beta)}},$$ $$\tilde{\sigma_0} = \frac{2}{(2 - 2\alpha + \beta) + \sqrt{(-2 + 2\alpha - \beta)^2 - 4(2\alpha - 3 + \beta)}}$$ and $$X(\alpha, \beta) = 2(2\alpha - 2 - \beta)(1 + \sqrt{2} - 2\alpha - \beta)((2\alpha - 2 - \beta) + \sqrt{(2\alpha - 2 - \beta)^2 - 4(3 - 2\sqrt{2})(2\sqrt{2} - 1 - 2\alpha - \beta)}) + 4(1 - 2\sqrt{2} + 2\alpha + \beta)((3 - 2\sqrt{2})(1 + \sqrt{2} - 2\alpha - \beta) + (1 - \sqrt{2})(1 - 2\sqrt{2} + 2\alpha + \beta)).$$ $$(5.1)$$ PROOF. We show that the disc mentioned in (2.8) satisfies $\mathbb{D}(a_F(r); c_F(r)) \subset \varphi_R(\mathbb{D})$ for all $0 < r \le R_{\mathcal{S}_R^*}(F)$ . Lemma 5.1 gives that the two possible values of the radius are the smallest positive roots of the equations $c_F(r) = a_F(r) - 2(\sqrt{2} - 1)$ and $c_F(r) = 2 - a_F(r)$ ; which are equivalent to $\phi(r) = 0$ and $\psi(r) = 0$ respectively, where the polynomials in r are of the form <span id="page-9-2"></span> $$\phi(r) := (2(\sqrt{2} - 1) - (2\alpha - 1 + \beta))r^2 - (2 - 2\alpha + \beta)r + 3 - 2\sqrt{2},\tag{5.2}$$ and <span id="page-9-3"></span> $$\psi(r) := (2\alpha - 3 + \beta)r^2 - (2 - 2\alpha + \beta)r + 1 \tag{5.3}$$ respectively. The fact that both the polynomials $\phi$ and $\psi$ possess zeros in the interval (0,1) can be easily verified as $\phi(0)=3-2\sqrt{2}>0$ and $\phi(1)=-2\beta<0$ , also, $\psi(0)=1>0$ , $\psi(1)=0$ $4(\alpha-1)<0$ . The respective positive zeros of $\phi$ and $\psi$ , denoted by $\sigma_0$ and $\tilde{\sigma_0}$ , are given by $$\sigma_0 := \frac{2(3 - 2\sqrt{2})}{(2 - 2\alpha + \beta) + \sqrt{(-2 + 2\alpha - \beta)^2 - 4(3 - 2\sqrt{2})(2\sqrt{2} - 1 - 2\alpha - \beta)}}$$ and $$\tilde{\sigma_0} := \frac{2}{(2 - 2\alpha + \beta) + \sqrt{(-2 + 2\alpha - \beta)^2 - 4(2\alpha - 3 + \beta)}}.$$ Case (i): $2\alpha + \beta - 2 \ge 0$ . By using the fact that the center $a_F(r)$ is decreasing function of r, it will be proved that <span id="page-10-0"></span> $$2(\sqrt{2}-1) < a_F(r) \le \sqrt{2}, \ r \in (0, \sigma_0). \tag{5.4}$$ This after the application of Lemma 5.1 will directly imply $\mathbb{D}(a_F(\sigma_0); c_F(\sigma_0)) \subset \varphi_R(\mathbb{D})$ , that is, the required radius is at least $\sigma_0$ . So, to prove (5.4), observe that $a_F(r) > a_F(\sigma_0)$ for $r \in (0, \sigma_0)$ , and $\sigma_0$ is the positive root of the equation $c_F(r) = a_F(r) - 2(\sqrt{2} - 1)$ , also since the radius $c_F(r)$ in (2.8) is positive for all $r \in (0, 1)$ , $a_F(\sigma_0) - 2(\sqrt{2} - 1) > 0$ . Lastly, $a_F(r) < a_F(0) = 1 < \sqrt{2}$ , $r \in (0, \sigma_0)$ , thus proving (5.4), and also the required result for this case. Case (ii): $2\alpha + \beta - 2 < 0$ and $X(\alpha, \beta) \le 0$ . Here, again from the Lemma 5.1, consider the equation $a_F(r) = \sqrt{2}$ , which gets simplified to $\zeta(r) = 0$ , with $\zeta(r) = (\sqrt{2} + 1 - 2\alpha - \beta)r^2 + 1 - \sqrt{2}$ . Then, for the polynomial $\zeta$ , the positive root is denoted by $\tilde{\sigma_1}$ , where <span id="page-10-1"></span> $$\tilde{\sigma_1} = \sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1 - 2\alpha - \beta}}.$$ It is observed that $\zeta(1)=-(2\alpha+\beta-2)>0$ and $\zeta(0)=1-\sqrt{2}<0$ , justifying the fact that $\tilde{\sigma_1}\in(0,1)$ . This also gives that $\sigma_0\leqslant\tilde{\sigma_1}$ if and only if $\zeta(\sigma_0)\leqslant0$ . It is seen that $$\zeta(\sigma_0) = \frac{1}{4(2\sqrt{2} - 1 - 2\alpha - \beta)^2} \left( 2(2\alpha - 2 - \beta)(1 + \sqrt{2} - 2\alpha - \beta)((2\alpha - 2 - \beta) + \sqrt{(2\alpha - 2 - \beta)^2 - 4(3 - 2\sqrt{2})(2\sqrt{2} - 1 - 2\alpha - \beta)}) + 4(1 - 2\sqrt{2} + 2\alpha + \beta)((3 - 2\sqrt{2})(1 + \sqrt{2} - 2\alpha - \beta) + (1 - \sqrt{2})(1 - 2\sqrt{2} + 2\alpha + \beta)) \right),$$ (5.5) so, combining (5.1) and (5.5), $X(\alpha, \beta) \leq 0$ is same as saying $\zeta(\sigma_0) \leq 0$ . Thus, in this case, $\sigma_0 \leq \tilde{\sigma_1}$ which implies $a_F(\sigma_0) \leq a_F(\tilde{\sigma_1}) = \sqrt{2}$ , and now, Lemma 5.1 gives that the radius of starlikeness associated with the rational function for the function F is at least $\sigma_0$ . Case (iii): $2\alpha + \beta - 2 < 0$ and $X(\alpha, \beta) > 0$ . Following similar arguments as in Case (ii), $\zeta(\sigma_0) > 0$ gives $a_F(\sigma_0) > \sqrt{2}$ , and then Lemma 5.1 implies that the required radius is at least $\tilde{\sigma_0}$ . Take $f(z) = z/(1-z)^{-2\alpha+2} \in \mathcal{ST}(\alpha)$ and the polynomial Q as $Q(z) = (1+z)^n$ . Using these choices for the function f and the polynomial Q, the expression for zF'(z)/F(z) as in the proof for Theorem 2.1, becomes, <span id="page-10-3"></span><span id="page-10-2"></span> $$\frac{zF'(z)}{F(z)} = \frac{(1 - 2\alpha - \beta)z^2 + (2 - 2\alpha + \beta)z + 1}{1 - z^2}.$$ (5.6) The polynomials $\phi$ and $\psi$ in (5.2), (5.3) for $r = \sigma_0$ and $r = \tilde{\sigma_0}$ respectively imply that $$(1 - 2\alpha - \beta)r^2 - (2 - 2\alpha + \beta)r + 1 = 2(\sqrt{2} - 1)(1 - r^2), \tag{5.7}$$ and $$(1 - 2\alpha - \beta)r^2 + (2 - 2\alpha + \beta)r + 1 = 2(1 - r^2). \tag{5.8}$$ Now, observe that using (5.7) and putting $z = -\sigma_0$ in (5.6), it is obtained that <span id="page-11-0"></span> $$\frac{(-\sigma_0)F'(-\sigma_0)}{F(-\sigma_0)} = 2(\sqrt{2} - 1) = \varphi_R(-1)$$ this proves the sharpness for the radius $\sigma_0$ . Also, considering (5.8), and replacing $z = \tilde{\sigma_0}$ in (5.6), it is seen that zF'(z)/F(z) assumes the value $2 = \varphi_R(1)$ thus proving the sharpness for $\tilde{\sigma_0}$ . When $\alpha$ goes to 1 in Theorem 5.2 we get that the radius of starlikeness associated with a rational function for the function $F(z)=z(Q(z))^{\beta/n}$ where Q is a non-constant polynomial of degree n non-vanishing on the unit disc and $\beta>0$ , comes out to be $(3-2\sqrt{2})/(3-2\sqrt{2}+\beta)$ . Further, when $\beta\to 0$ , we obtain the radius of starlikeness associated with a rational function for the class of starlike functions of order $\alpha,0\leqslant \alpha<1$ obtained by Kumar and Ravichandran in [11, Theorem 3.2] (when $A=1-2\alpha$ and B=-1).

Lemma 6.1. [21, lemma 2.2] For 1/3 < a < 5/3, $$r_a = \begin{cases} a - \frac{1}{3} & \text{if } \frac{1}{3} < a \le 1\\ \frac{5}{3} - a & \text{if } 1 \le a < \frac{5}{3}. \end{cases}$$ Then $\{w : |w-a| < r_a\} \subset \Omega_{Ne}$ , where $\Omega_{Ne}$ is the region bounded by the nephroid, that is $$\Omega_{Ne} := \left\{ \left( (u-1)^2 + v^2 - \frac{4}{9} \right)^3 - \frac{4v^2}{3} < 0 \right\}.$$

Theorem 6.2. If the function $f \in ST(\alpha)$ , then the radius of starlikeness associated with a nephroid domain for the function F defined in (2.1) is given by $$R_{\mathcal{S}_{Ne}^*}(F) = \begin{cases} \sigma_0 & \text{if } 2\alpha + \beta - 2 \geqslant 0\\ \tilde{\sigma_0} & \text{if } 2\alpha + \beta - 2 < 0, \end{cases}$$ where $$\sigma_0 = \frac{4}{3(2 - 2\alpha + \beta) + \sqrt{9(-2 + 2\alpha - \beta)^2 - 8(4 - 6\alpha - 3\beta)}},$$ and $$\tilde{\sigma_0} = \frac{4}{3(2 - 2\alpha + \beta) + \sqrt{9(-2 + 2\alpha - \beta)^2 - 8(6\alpha - 8 + 3\beta)}}.$$ PROOF. The proof aims to show that the disc in (2.8) satisfies the following condition: $$\mathbb{D}(a_F(r); c_F(r)) \subset \Omega_{Ne}, \quad 0 < r \leqslant R_{\mathcal{S}_{Ne}^*}(F).$$ Let $$\sigma_0 := \frac{4}{3(2 - 2\alpha + \beta) + \sqrt{9(-2 + 2\alpha - \beta)^2 - 8(4 - 6\alpha - 3\beta)}}.$$ The number $\sigma_0$ is the positive solution to the equation $c_F(r) = a_F(r) - 1/3$ , which transforms into $\phi(r) = 0$ with the polynomial $\phi$ in r given by <span id="page-12-0"></span> $$\phi(r) := (4 - 6\alpha - 3\beta)r^2 - 3(2 - 2\alpha + \beta)r + 2. \tag{6.1}$$ For the polynomial $\phi$ , $\phi(0)=2>0$ and $\phi(1)=-6\beta<0$ justifying the existence of a zero for $\phi$ in the interval (0,1). The number $$\tilde{\sigma_0} := \frac{4}{3(2 - 2\alpha + \beta) + \sqrt{9(-2 + 2\alpha - \beta)^2 - 8(6\alpha - 8 + 3\beta)}},$$ is the positive zero of the polynomial $\psi$ in r is given by <span id="page-12-3"></span> $$\psi(r) := (6\alpha - 8 + 3\beta)r^2 - 3(2 - 2\alpha + \beta)r + 2. \tag{6.2}$$ Here, the equation $\psi(r)=0$ is the the simplified form of the equation $c_F(r)=5/3-a_F(r)$ . The polynomial $\psi$ indeed has a zero in the interval (0,1) as $\psi(0)=2>0$ and $\psi(1)=12(\alpha-1)<0$ . It is known that the center $a_F$ in (2.8) has the property that $a_F(r)\leqslant 1$ for $2\alpha+\beta-2\geqslant 0$ ; and $a_F(r)>1$ for $2\alpha+\beta-2<0$ . Case (i): $2\alpha + \beta - 2 \geqslant 0$ . In this case, the center $a_F(r) \leqslant 1$ , thus Lemma 6.1 implies that $\mathbb{D}(a_F(\sigma_0); c_F(\sigma_0)) \subset \Omega_{Ne}$ . This proves that the $\mathcal{S}^*_{Ne}$ radius for the function F is at least $\sigma_0$ . To verify the sharpness of $\sigma_0$ , take $f(z) = z/(1-z)^{-2\alpha+2} \in \mathcal{ST}(\alpha)$ and the polynomial Q as $Q(z) = (1+z)^n$ . Using these choices for the function f and the polynomial Q, the expression for zF'(z)/F(z) as in the proof for Theorem 2.1 is, <span id="page-12-2"></span> $$\frac{zF'(z)}{F(z)} = \frac{(1 - 2\alpha - \beta)z^2 + (2 - 2\alpha + \beta)z + 1}{1 - z^2}.$$ (6.3) The polynomial $\phi$ in (6.1) provides that for $r = \sigma_0$ , <span id="page-12-1"></span> $$3((1 - 2\alpha - \beta)r^2 - (2 - 2\alpha + \beta)r + 1) = (1 - r^2). \tag{6.4}$$ Thus using (6.4), and replacing $z=-\sigma_0$ , (6.3) gives $((-\sigma_0)F'(-\sigma_0))/F(-\sigma_0)=1/3=\varphi_{Ne}(-1)$ . This proves the sharpness for $\sigma_0$ . Case (ii): $2\alpha + \beta - 2 < 0$ . Here, it is known that $a_F(r) > 1$ , so, Lemma 6.1 directly gives that the required radius is at least the positive solution of the equation $c_F(r) = 5/3 - a_F(r)$ that is $\tilde{\sigma}_0$ . To verify sharpness in this case, take $f(z)=z/(1-z)^{-2\alpha+2}\in\mathcal{ST}(\alpha)$ and the polynomial Q as $Q(z)=(1+z)^n$ . These expressions transform zF'(z)/F(z) into the form given in (6.3). The polynomial $\psi$ given in (6.2) implies that for $r=\tilde{\sigma_0}$ , <span id="page-12-4"></span> $$3((1 - 2\alpha - \beta)r^2 + (2 - 2\alpha + \beta)r + 1) = 5(1 - r^2). \tag{6.5}$$ Thus, the radius $\tilde{\sigma}_0$ is the best possible since an application of (6.5) in (6.3) gives that the expression for zF'(z)/F(z) takes the value $5/3 = \varphi_{Ne}(1)$ for $z = \tilde{\sigma}_0$ . Thus, proving the sharpness for the radius $\tilde{\sigma}_0$ . When $\alpha$ goes to 1 in Theorem 6.2 we get: The radius of starlikeness associated with a nephroid domain for the function $F(z)=z(Q(z))^{\beta/n}$ where Q is a non-constant polynomial of degree n non-vanishing on the unit disc and $\beta>0$ , comes out to be $2/(2+3\beta)$ . When $\beta\to 0$ in Theorem 6.2, we obtain the radius of starlikeness associated with a nephroid domain for the class of starlike functions of order $\alpha,0\leqslant \alpha<1$ obtained by Wani and Swaminathan [21, Theorem 3.1(ii)] when $A=1-2\alpha$ and B=-1.

Lemma 7.1. [8, lemma 2.2] Let 2/(1+e) < a < 2e/(1+e). If $$r_a = \frac{e-1}{e+1} - |a-1|,$$ then $\{w : |w-a| < r_a\} \subset \Delta_{SG}$ . In the next result, we find the radius of starlikeness associated with modified sigmoid function for the function F defined in (2.1).

Theorem 7.2. If the function $f \in \mathcal{ST}(\alpha)$ , then the $\mathcal{S}_{SG}^*$ radius for the function F given in (2.1) is given by $$R_{\mathcal{S}_{SG}^*}(F) = \begin{cases} \sigma_0 & \text{if } 2\alpha + \beta - 2 \geqslant 0\\ \tilde{\sigma_0} & \text{if } 2\alpha + \beta - 2 < 0, \end{cases}$$ where $$\sigma_0 = 2(e-1)((2-2\alpha+\beta)(e+1) + \sqrt{((e+1)(2-2\alpha+\beta))^2 - 4(e-1)(3+e-2\alpha-2\alpha e-\beta-\beta e)})^{-1},$$ and $$\tilde{\sigma_0} = 2(e-1)((2-2\alpha+\beta)(e+1) + \sqrt{((e+1)(2-2\alpha+\beta))^2 + 4(e-1)(1+3e-2\alpha-2\alpha e-\beta-\beta e)})^{-1}.$$ PROOF. We will show that $\mathbb{D}(a_F(r); c_F(r)) \subset \Delta_{SG}$ for $0 < r \leqslant R_{\mathcal{S}_{SG}^*}(F)$ . Case (i): $2\alpha + \beta - 2 \ge 0$ . It is known that in this case, the center $a_F$ satisfies the inequality $a_F(r) \le 1$ . Consequently, the equation $c_F(r) = ((e-1)/(e+1)) - |a_F(r) - 1|$ becomes $c_F(r) = ((e-1)/(e+1)) - 1 + a_F(r)$ . This equation is simplified into the form $\phi(r) = 0$ , with <span id="page-13-0"></span> $$\phi(r) := ((2 - 2\alpha - \beta)(e+1) - (e-1))r^2 - (2 - 2\alpha + \beta)(e+1)r + (e-1). \tag{7.1}$$ The polynomial $\phi$ has a zero in the interval (0,1) since, $\phi(0)=(e-1)>0$ , and $\phi(1)=-2\beta(e+1)<0$ , and this positive number is $$\sigma_0 = 2(e-1)((2-2\alpha+\beta)(e+1))$$ $$+\sqrt{((e+1)(2-2\alpha+\beta))^2-4(e-1)(3+e-2\alpha-2\alpha e-\beta-\beta e)}$$ Now, to verify the sharpness of the radius $\sigma_0$ , we take $f(z) = z/(1-z)^{-2\alpha+2} \in \mathcal{ST}(\alpha)$ and polynomial Q as $Q(z) = (1+z)^n$ . Using these choices for the function f and the polynomial Q, the expression for zF'(z)/F(z) as seen in the proof for Theorem 2.1 is obtained to be <span id="page-14-3"></span><span id="page-14-1"></span><span id="page-14-0"></span> $$\frac{zF'(z)}{F(z)} = \frac{(1 - 2\alpha - \beta)z^2 + (2 - 2\alpha + \beta)z + 1}{1 - z^2},\tag{7.2}$$ which implies that $$2 - \frac{zF'(z)}{F(z)} = \frac{(-3 + 2\alpha + \beta)z^2 - (2 - 2\alpha + \beta)z + 1}{1 - z^2}.$$ (7.3) Thus, (7.2) and (7.3) give that $$\left(\frac{zF'(z)}{F(z)}\right)\left(2 - \frac{zF'(z)}{F(z)}\right)^{-1} = \frac{(1 - 2\alpha - \beta)z^2 + (2 - 2\alpha + \beta)z + 1}{(-3 + 2\alpha + \beta)z^2 - (2 - 2\alpha + \beta)z + 1}.$$ (7.4) Further, the polynomial $\phi$ in (7.1) gets reduced to <span id="page-14-2"></span> $$((1 - 2\alpha - \beta)r^2 - (2 - 2\alpha + \beta)r + 1)e = (-3 + 2\alpha + \beta)r^2 + (2 - 2\alpha + \beta)r + 1.$$ (7.5) for $r = \sigma_0$ . Thus using (7.5) in (7.4) for $z = -\sigma_0$ , it is seen that $$\left| \log \left( \left( \frac{zF'(z)}{F(z)} \right) \left( 2 - \frac{zF'(z)}{F(z)} \right)^{-1} \right) \right| = \left| \log \frac{1}{e} \right| = 1,$$ thus proving the sharpness for $\sigma_0$ . Case (ii): $2\alpha + \beta - 2 < 0$ . Here, the center $a_F(r) > 1$ , which implies that the equation $c_F(r) = ((e-1)/(e+1)) - |a_F(r) - 1|$ converts to $c_F(r) = ((e-1)/(e+1)) + 1 - a_F(r)$ . This is equivalent to $\psi(r) = 0$ for <span id="page-14-4"></span> $$\psi(r) := ((2 - 2\alpha - \beta)(e + 1) + (e - 1))r^2 + (2 - 2\alpha + \beta)(e + 1)r - (e - 1). \tag{7.6}$$ The number $$\tilde{\sigma_0} = 2(e-1)((2-2\alpha+\beta)(e+1) + \sqrt{((e+1)(2-2\alpha+\beta))^2 + 4(e-1)(1+3e-2\alpha-2\alpha e-\beta-\beta e)})^{-1}$$ is the smallest positive zero of the polynomial $\psi$ , and the observations $\psi(0) = -(e-1) < 0$ , and $\psi(1) = 4(1-\alpha)(e+1) > 0$ justify the existence of a zero in (0,1). To verify that the radius $\tilde{\sigma}_0$ is the best possible, take the values of the function f and the polynomial Q same as in Case (i), thus the expression for (zF'(z)/F(z))/(2-(zf'(z)/F(z))) is same as given in (7.4). The polynomial $\psi$ in (7.6) gives <span id="page-14-5"></span> $$(1 - 2\alpha - \beta)r^2 + (2 - 2\alpha + \beta)r + 1 = ((-3 + 2\alpha + \beta)r^2 - (2 - 2\alpha + \beta)r + 1)e.$$ (7.7) for $r = \tilde{\sigma_0}$ . Thus, putting $z = \tilde{\sigma_0}$ in (7.4) and then using (7.7), it is obtained that $$\left| \log \left( \left( \frac{\tilde{\sigma}_0 F'(\tilde{\sigma}_0)}{F(\tilde{\sigma}_0)} \right) \left( 2 - \frac{\tilde{\sigma}_0 F'(\tilde{\sigma}_0)}{F(\tilde{\sigma}_0)} \right)^{-1} \right) \right| = \left| \log e \right| = 1.$$ This proves sharpness for $\tilde{\sigma_0}$ . When $\alpha$ goes to 1 in Theorem 7.2 we get that the radius of starlikeness associated with modified sigmoid function for the function $F(z)=z(Q(z))^{\beta/n}$ where Q is a non-constant polynomial of degree n non-vanishing on the unit disc and $\beta>0$ , is $(e-1)/(e-1+\beta(e+1))$ . When $\beta\to 0$ in Theorem 7.2, we obtain the radius of starlikeness associated with the modified sigmoid function for the class of starlike functions of order $\alpha, 0 \leqslant \alpha < 1$ . The radius of parabolic starlikeness and the radii of other related starlikeness including the one related to the lemniscate of Bernoulli can be investigated.