Radii problems for Ma-Minda Starlikeness

Lemma 1.1. ([29]) Let $\wp(z) = 1 + ze^z$ . Then we have $\{w : |w - a| < R_a\} \subset \wp(\mathbb{D})$ , where $$R_a = \begin{cases} (a-1) + 1/e, 1 - 1/e < a \le 1 + (e - e^{-1})/2; \\ e - (a-1), & 1 + (e - e^{-1})/2 \le a < 1 + e. \end{cases}$$ In the present investigation, we find the $S^(\psi)$ -radii for the normalized Special functions given by (1.2), (1.4), (1.8) and (1.10) using the Assumption 1.1. Further, the generalization of a classical problem of maximization of Goluzin for the class $S^(\psi)$ is established. Various non-trivial radius problems using the concept of convolution (i.e, term by term multiplication between coefficients of two power series) are studied for the case of starlike domains $\psi(\mathbb{D})$ (for example $\wp(\mathbb{D})$ ) which also show the importance of radius of convexity. Moreover, we find the sufficient conditions for some normalized functions f in $\mathcal{A}$ to be in $S^*(\psi)$ in terms of it's coefficients.

Theorem 3.1. Suppose $\Phi$ is a non-constant entire function and $0 < |z_0| < 1$ and assume that the class $S^*(\psi)$ is closed. Then maximum of either <span id="page-10-2"></span> $$\Re\Phi\left(\log\frac{f(z_0)}{z_0}\right) \quad or \quad \left|\Phi\left(\log\frac{f(z_0)}{z_0}\right)\right|$$ (3.1) for functions in the class $S^*(\psi)$ is attained only when the function is of the form $$f(z) = z \exp \int_0^{\zeta z} \frac{\psi(t) - 1}{t} dt, \tag{3.2}$$ where $|\zeta| = 1$ . Proof Since the class $S^*(\psi)$ is compact, therefore the problem under consideration has a solution. Moreover, in view of a result of Goluzin [24], in (3.1) it suffices to consider the continuous functional $$\Re \Phi \left( \log \frac{f(z_0)}{z_0} \right).$$ Let $f \in \mathcal{S}^*(\psi)$ . Then using a result from [31], $f(z)/z \prec f_0(z)/z =: F(z)$ , where $f_0(z) = z \exp \int_0^z \frac{\psi(t)-1}{t} dt$ or equivalently $\log(f(z)/z) \prec \log F(z)$ . Thus, $$g(z) = \varPhi\left(\log\frac{f(z)}{z}\right) \prec \varPhi(\log F(z)) = G(z).$$ Note that G is also non-constant as is $\Phi$ . Thus for each $r \in (0,1)$ by subordination principle, we obtain $g(\overline{\mathbb{D}}_r) \subset G(\overline{\mathbb{D}}_r) = \Omega$ . Since $G(xz) \prec G(z)$ for $|x| \leq 1$ is obvious, therefore for $|z_0| = r$ , we have $\{g(z_0) : g \prec G \text{ in } \mathbb{D}\} = \Omega$ . Now by considering a support line to the compact set $\Omega$ , we conclude that $$\max_{f \in \mathcal{S}^*(\psi)} \Re \Phi \left( \log \frac{f(z_0)}{z_0} \right) = \Re w_1, \quad w_1 \in \partial \Omega.$$ Since G is also an open map, therefore there exists a point $z_1$ where $|z_1| = r$ and $G(z_1) = w_1$ such that among finitely many $w_1$ , for one suitable $w_1$ , we have $$\Phi\left(\log\frac{f(z_0)}{z_0}\right) = w_1,$$ where f is the solution for the extremal problem. Now by the well known Lindelöf Principle, we have <span id="page-11-0"></span> $$\Phi\left(\log\frac{f(z)}{z}\right) = \Phi(\log F(xz)),$$ (3.3) that is, if f is the desired solution, then (3.3) holds for some x, |x| = 1. Since $\Phi$ is non-constant analytic function, so we may write $$\Phi(w) = c_0 + c_n w^n + c_{n+1} w^{n+1} + \cdots ; c_n \neq 0.$$ If we set $\log(f(z)/z) = \alpha_1 z + \alpha_2 z^2 + \cdots$ and $\log(F(z)) = \beta_1 z + \beta_2 z^2 + \cdots$ , then from (3.3), comparing the coefficients, we get $c_n \alpha_1^n = c_n \beta_1^n$ . Or equivalently, $\alpha_1^n = \beta_1^n$ , which in particular implies that $|\alpha_1| = |\beta_1|$ . Since $\log(f(z)/z) \prec \log F(xz)$ , $|\alpha_1| = |\beta_1|$ is possible only if $\log(f(z)/z) = \log F(xyz)$ for some |y| = 1. Therefore, we conclude that $$f(z) = z \exp \int_0^{uz} \frac{\psi(t) - 1}{t} dt,$$ where |u|=1 if f is a solution to the extremal problem. Remark 3.1 Note that the analogous result for the class $C(\psi)$ also holds. Now as an application of the Theorem 3.1, we obtain the result due to MacGregor [32]: Corollary 3.1 [32] Suppose $\Phi$ is a non-constant entire function and $0 < |z_0| < 1$ . Then the maximum of the expression (3.1) for functions in the class $S^*(\alpha)$ is attained only when the function is of the form $$f(z) = \frac{z}{(1 - \zeta z)^{2-2\alpha}}, \ |\zeta| = 1.$$ Proof If $f \in S^*(\alpha)$ , then $f(z)/z \prec 1/(1-z)^{2-2\alpha}$ and the result follows. Corollary 3.2 Suppose $\Phi$ is a non-constant entire function and $0 < |z_0| < 1$ . Then the maximum of the expression (3.1) for functions in the class $S_{\wp}^*$ is attained only when the function is of the form <span id="page-12-0"></span> $$f(z) = z \exp(e^{\zeta z} - 1), \ |\zeta| = 1.$$ Proof If $f \in \mathcal{S}_{\wp}^*$ , then $f(z)/z \prec \exp(e^z - 1)$ and the result follows.

Theorem 4.1. [31] Let $\psi(\mathbb{D})$ be convex, $g \in \mathcal{C}$ and $f \in \mathcal{S}^(\psi)$ . Then $f g \in \mathcal{S}^*(\psi)$ . For instance, for $\psi(z)=1+ze^z$ , $z+\sqrt{1+z^2}$ , $e^{e^z-1}$ and $1+4z/3+2z^2/3$ Theorem 4.1 is not valid. Therefore, we need to modify Theorem 4.1 to accommodate such cases to further derive various radius problems. But first we need to recall an important result due to Ruscheweyh and Sheil-Small: Lemma 4.1 ([35], p. 126) Suppose that either $g \in C$ , $h \in S$ or else $g, h \in S^_{1/2}$ . Then for any analytic function G in $\mathbb{D}$ , we have <span id="page-12-2"></span><span id="page-12-1"></span> $$\frac{g hG(z)}{q h(z)} \in \overline{co}G(\mathbb{D}),$$ where $\overline{co}G(\mathbb{D})$ is the closed convex hull of $G(\mathbb{D})$ . Keenly observing the proof of Lemma 4.1, we see that the unit disk $\mathbb{D}$ can be replaced by the sub-disk $\mathbb{D}_r := \{z : |z| < r\}$ , where $0 < r \le 1$ and consequently, we obtain the following modified result. Since the proof is similar, so it is omitted here.

Lemma 4.2. Suppose either $g \in C$ , $h \in S$ or else $g, h \in S^_{1/2}$ . Then for any analytic function G in $\mathbb{D}_r$ , we have $(g hG(z))/(g h(z)) \in \overline{co}G(\mathbb{D}_r)$ , where $r \in [0,1]$ . This immediately gives the following fundamental result Theorem 4.2 (Imrovement of Theorem 4.1) Let $r_0$ be the radius of convexity of $\psi$ . If $g \in \mathcal{C}$ and $f \in \mathcal{S}^(\psi)$ . Then $f g \in \mathcal{S}^*(\psi)$ in |z| < r, where $r = \min\{r_0, 1\}.$ Corollary 4.1 Let $f \in \mathcal{S}_{\wp}$ and $g \in \mathcal{C}$ . Then $f g \in \mathcal{S}_{\wp}^*$ in $\mathbb{D}_{r_0}$ , where $r_0 =$ $(3-\sqrt{5})/2$ is the radius of convexity of $\wp$ . Now consider the operators $\mathcal{F}_i: \mathcal{A} \to \mathcal{A}$ defined by $$\mathcal{F}_1(f)(z) = f * g_1(z) = zf'(z)$$ $$\mathcal{F}_2(f)(z) = f * g_2(z) = \frac{1}{2}(f(z) + zf'(z))$$ $$\mathcal{F}_3(f)(z) = f * g_3(z) = \frac{k+1}{z^k} \int_0^z t^{k-1} f(t) dt, \quad \Re k > 0,$$ where $g_3(z) = \sum_{n=1}^{\infty} (k+1)/(k+n)z^n$ , $g_2(z) = (z-z^2/2)/(1-z^2)^2$ and $g_1(z) = z/(1-z)^2$ . Note that the function $g_1$ is convex in $|z| < 2 - \sqrt{3}$ , $g_2(z) = (z-z^2/2)/(1-z^2)^2$ is convex in |z| < 1/2 while $g_3 \in \mathcal{C}$ . The above defined operators were introduced by Alexander, Livingston and Bernardi, respectively. Now we obtain the following result, where $S_{SG}^ := S^(\frac{2}{e^{-z}+1}), S_C^ := S^(1+4z/3+2z^2/3),$ $S_b^ = S^(e^{e^z-1}) \text{ and } S_s^ := S^(1+\sin z):$ Corollary 4.2 Let $\mathcal{F}_i$ , i = 1 to 3 be the operators as defined above. - (i) Let $f \in \mathcal{S}_{\omega}$ . Then $\mathcal{F}_i(f) \in \mathcal{S}_{\omega}$ in $\mathbb{D}_{r_i}$ , where $r_1 = 2 \sqrt{3}$ , $r_2 = (3 \sqrt{5})/2$ and $r_3 = (3 - \sqrt{5})/2$ - (ii) Let $f \in \mathcal{S}_C$ . Then $\mathcal{F}_i(f) \in \mathcal{S}_C$ in $\mathbb{D}_{r_i}$ , where $r_1 = 2 \sqrt{3}$ , $r_2 = 1/2$ and - (iii) Let $f \in \mathcal{S}_s$ . Then $\mathcal{F}_i(f) \in \mathcal{S}_s$ in $\mathbb{D}_{r_i}$ , where $r_1 = 2 \sqrt{3}$ , $r_2 = 0.345$ and - (iv) Let $f \in \mathcal{S}_{SG}$ . Then $\mathcal{F}_i(f) \in \mathcal{S}_{SG}$ in $\mathbb{D}_{r_i}$ , where $r_1 = 2 \sqrt{3}$ , $r_2 = 1/2$ and The radii are sharp. In 2010, Ali et al. [2] dealt with the problem of finding $S^(\psi)$ -radii of the convolution f \ g, between two starlike functions. In fact, they showed that if $f,g \in S$ and $h_{\rho}(z) = f g(\rho z)/\rho$ , then $h_{\rho} \in \mathcal{SL}^*$ for $0 \leq \rho \leq$ $(\sqrt{5}-2)/(\sqrt{2}-1)\approx 0.09778$ . They used the property of the function $\psi$ being convex. Now using Theorem 4.2, we can obtain the result even for the case when $\psi(\mathbb{D})$ is starlike. Here, we have shown the usability of the radius of convexity of $\psi$ .

Theorem 4.3. Let $f, g \in S$ and $h_{\rho}(z) := f g(\rho z)/\rho$ . Then - (i) $h_{\rho} \in \mathcal{S}_{\rho}^{*}$ for $0 \leq \rho \leq (2e \sqrt{4e^2 2e + 1})/(2e 1) \approx 0.0957$ , - (ii) $h_{\rho} \in \mathcal{S}_{C}^{}$ for $0 \le \rho \le (3 \sqrt{7})/2 \approx 0.177124$ , (iii) $h_{\rho} \in \mathcal{S}_{s}^{}$ for $0 \le \rho \le (\sqrt{\sin 1^{2} + 2\sin 1 + 4} 2)/(2 + \sin 1) \approx 0.185835$ , (iv) $h_{\rho} \in \mathcal{S}_{b}^{}$ for $0 \leq \rho \leq (2e - \sqrt{3e^{2} + e^{2/e}})/(e + e^{1/e}) \approx 0.122919$ , (v) $h_{\rho} \in \mathcal{S}_{SG}^{}$ for $0 \leq \rho \leq (\sqrt{7e^{2} + 6e + 3} - 2(1 + e))/(3e + 1) \approx 0.108309$ . The constants are best possible. Proof We only prove first part and rest part's proof also follow in a similar fashion. (i): Let $H(z) = z + \sum_{n=2}^{\infty} n^2 z^n = (z(1+z))/(1-z)^3$ . It is easy to see that <span id="page-14-0"></span> $$\left| \frac{zH'(z)}{H(z)} - \frac{1+r^2}{1-r^2} \right| \le \frac{4r}{1-r^2}, \quad |z| = r < 1. \tag{4.1}$$ Now by Lemma 1.1, the disk (4.1) lies inside the cardioid $\wp(\mathbb{D})$ , provided $$\frac{4r}{1-r^2} \le \frac{1+r^2}{1-r^2} - 1 + \frac{1}{e}$$ which in turn gives $r \leq r_0 := (2e - \sqrt{4e^2 - 2e + 1})/(2e - 1)$ . Define the function $h: \mathbb{D} \to \mathbb{C}$ by h(z) := f(z) \ g(z). Then h(z) = F(z) \ G(z) \ H(z), where F and G are, respectively defined as zF'(z) = f(z) and zG'(z) = g(z). Since $f, g \in \mathcal{S}$ , it follows that $F G \in \mathcal{C}$ . Also, $H(r_0z)/r_0 \in \mathcal{S}^_{\wp}$ . Hence, using Theorem 4.2, we have $$F(z) G(z) H(\rho_0 z)/\rho_0 \in \mathcal{S}_{\wp}^*$$ where $\rho_0 = \min\{r_0, r_c\} = r_0$ and $r_c = (3 - \sqrt{5})/2$ is the radius of convexity of $\wp$ . For $z = -\rho_0$ , $zH'(z)/H(z) = (1 + 4z + z^2)/(1 - z^2) = 1 - 1/e$ , which implies that $\rho_0$ is sharp. Remark 4.1 It is worthy to mention that in Theorem 4.3, we need $r_c$ , radius of convexity. However, the sharp radius of convexity for the class $\mathcal{S}^*(\psi)$ is an open problem.

Theorem 4.4. Let $f, g \in \mathcal{S}$ and $h_{\rho}(z) := f g(\rho z)/\rho$ . Then $h_{\rho}(z) \in \mathcal{S}^*\left(\frac{1+Az}{1+Bz}\right)$ for $$0 \le \rho \le \frac{2(B^2 - 1) + \sqrt{4(1 - B^2)^2 + (A - B)^2}}{A - B} =: \rho_0,$$ where $-1 < B < A \le 1$ . Proof Since for the function $p(z) \prec (1 + Az)/(1 + Bz)$ , we have <span id="page-14-1"></span> $$\left| p(z) - \frac{1 - AB}{1 - B} \right| \le \frac{A - B}{1 - B^2}.$$ (4.2) Therefore, for the disk (4.1) to lie inside the disk (4.2), we must have $$\frac{1 - AB}{1 - B^2} - \frac{A - B}{1 - B^2} \le \frac{1 + r^2}{1 - r^2} \le \frac{1 - AB}{1 - B^2} + \frac{A - B}{1 - B^2} \text{ and } \frac{4r}{1 - r^2} \le \frac{A - B}{1 - B^2},$$ which upon simplification hold for $r \le r_0 = \sqrt{(A-B)/(2+A+B)}$ and $r \le \rho_0$ respectively, where $\rho_0$ is the smallest positive root of the following equation $$(A - B)r^{2} + 4(1 - B^{2})r - (A - B) = 0.$$ Since $\min\{r_0, \rho_0\} = \rho_0$ and the class $\mathcal{S}^*\left(\frac{1+Az}{1+Bz}\right)$ is closed under convolution with convex functions, now the result follows in a similar way as in the part (i) of Theorem 4.3.

Theorem 5.1. Let $f(z) = z/(1 + \sum_{k=1}^{\infty} a_k z^k)$ . If the coefficients of f satisfy $$|1-a| + \sum_{k=1}^{\infty} (R_a + |1-a-k|)|a_k| \le R_a,$$ where a and $R_a$ are as defined in the Assumption 1.1. Then $f \in \mathcal{S}^*(\psi)$ . Proof For $f(z) = z/(1 + \sum_{k=1}^{\infty} a_k z^k)$ , we have $$\left| \frac{zf'(z)}{f(z)} - a \right| = \left| 1 - a - \frac{\sum_{k=1}^{\infty} k a_k z^k}{1 + \sum_{k=1}^{\infty} a_k z^k} \right|.$$ Thus by Assumption 1.1, $f \in \mathcal{S}^*(\psi)$ , if $\left|1 - a - \frac{\sum_{k=1}^{\infty} k a_k z^k}{1 + \sum_{k=1}^{\infty} a_k z^k}\right| \leq R_a$ . The above inequality holds whenever $$|1-a| + \sum_{k=1}^{\infty} |1-a-k| |a_k| r^k \le R_a (1 - \sum_{k=1}^{\infty} |a_k| r^k)$$ or equivalently, $|1-a|+\sum_{k=1}^{\infty}(|1-a-k|+R_a)|a_k|r^k\leq R_a$ . Letting r tends to $1^-$ , completes the

Theorem 5.2. Let $f(z) = z/(1+z^k)^n$ , where $n, k \in \mathbb{Z}^+$ are fixed. Then $f \in \mathcal{S}^*(\psi)$ in $$|z| < \left(\frac{R_a - |1 - a|}{R_a + |1 - a - kn|}\right)^{1/k},$$ where a and $R_a$ are as defined in the Assumption 1.1. Proof For $f(z) = z/(1+z^k)^n$ , we have $$\left| \frac{zf'(z)}{f(z)} - a \right| = \left| 1 - a \frac{knz^k}{1 + z^k} \right|.$$ Thus by Assumption 1.1, $f \in \mathcal{S}^*(\psi)$ , if $\left|1-a-\frac{knz^k}{1+z^k}\right| < R_a$ . This inequality holds whenever $|1-a|+|1-a-kn||z|^k < R_a(1-|z|^k)$ which upon simplification yields that $$|z|^k < \frac{R_a - |1 - a|}{R_a + |1 - a - kn|}.$$ Hence the result follows.

Theorem 5.3. Let p(z) be a polynomial such that p(0) = 1 and $\deg p(z) = m$ . Let $R = \min\{|z| : p(z) = 0, z \neq 0\}$ . Then the function $$f(z) = z(p(z))^{\beta/m} \in \mathcal{S}^*(\psi)$$ in $$|z| < \frac{R(R_a - |1 - a|)}{|\beta| + R_a - |1 - a|},$$ where a and $R_a$ are as defined in the Assumption 1.1. Proof Assume that $z_k$ , (k = 1, 2, ..., m) are zeros of the polynomial p(z). For the function $f(z) = z(p(z))^{\beta/m}$ , we have $$\frac{zf'(z)}{f(z)} = 1 + \frac{\beta}{m} \sum_{k=1}^{\infty} \frac{z}{z - z_k}$$ or equivalently, $$\frac{zf'(z)}{f(z)} - a = 1 - a + \frac{\beta}{m} \sum_{k=1}^{\infty} \left( \frac{z}{z - z_k} + \frac{r^2}{R^2 - r^2} - \frac{r^2}{R^2 - r^2} \right).$$ Thus by Assumption 1.1, $f \in \mathcal{S}^*(\psi)$ whenever $$(|1 - a| - R_a)(R^2 - r^2) + |\beta|(Rr + r^2) < 0,$$ which is satisfied if $|z| = r < R(R_a - |1 - a|)/(|\beta| + R_a - |1 - a|)$ . Kuroki and Owa [30] introduced and studied the class $S(\alpha, \beta)$ of functions f, which satisfy the condition $zf'(z)/f(z) \prec p_{\alpha,\beta}(z)$ , where $$p_{\alpha,\beta}(z) := 1 + \frac{\beta - \alpha}{\pi} i \log \frac{1 - e^{2\pi i \frac{1 - \alpha}{\beta - \alpha}} z}{1 - z},$$ $\alpha < 1, \, \beta > 1$ and $p_{\alpha,\beta}$ maps $\mathbb D$ onto the convex domain $\{w \in \mathbb C : \alpha < \Re w < \beta\}$ . Note that if $\alpha \ngeq 0$ then this class also contains non-univalent functions, and univalent starlike if $1 > \alpha \ge 0$ . Remark 5.1 Note that we can extend Theorem 5.4 $(\psi(z) \neq (1+z)/(1-z))$ and Theorem 5.5 for $\mathcal{S}^*(\psi)$ -radius if we replace the radius 1/e by $r_1$ , where $r_1$ is given by the Assumption 1.1. <span id="page-17-0"></span>We now conclude this section some results explicitly for the class $\mathcal{S}_{\wp}^*$ .

Theorem 5.4. Let $f \in \mathcal{S}(\alpha, \beta)$ . Then $f \in \mathcal{S}_{\wp}^*$ in $\mathbb{D}_{r_0}$ , where $r_0$ is the least positive root of the equation <span id="page-17-4"></span> $$\frac{\beta - \alpha}{\pi} \left( \log \frac{1 + \sqrt{2(1 + \cos(2\pi \frac{1 - \alpha}{\beta - \alpha}))r + r^2}}{1 - r^2} + 2 \arctan \frac{r}{1 - r} \right) - \frac{1}{e} = 0. (5.1)$$ Proof Consider the analytic function $p_{\alpha,\beta}(z) := 1 + \frac{\beta - \alpha}{\pi} i \log q(z)$ , where $$q(z) = \frac{1 - cz}{1 - z}$$ and $c = \exp\left(2\pi i \frac{1 - \alpha}{\beta - \alpha}\right)$ . Note that q(z) is a bilinear transformation, maps $\mathbb{D}_r$ onto the disk: $$\left| q(z) - \frac{1 + cr^2}{1 - r^2} \right| \le \frac{|1 + c|r}{1 - r^2},$$ which implies $$|q(z)| \le \frac{1 + |1 + c|r + r^2}{1 - r^2},$$ and therefore. <span id="page-17-1"></span> $$\log|q(z)| \le \log\left(\frac{1 + |1 + c|r + r^2}{1 - r^2}\right). \tag{5.2}$$ For any $\delta \in \mathbb{C}$ with $|\delta| = 1$ , we have $1 + \delta z < 1 + z$ . So to maximize $|\arg(1 + \delta z)|$ , it suffices to consider $|\arg(1 + z)|$ . Now for |z| = r, we have <span id="page-17-2"></span> $$|\arg(1+z)| \le \arctan\frac{r}{1-r}.\tag{5.3}$$ Hence to apply Lemma 1.1, we need to maximize $|p_{\alpha,\beta}(z)-1|$ , that is, <span id="page-17-3"></span> $$|p_{\alpha,\beta} - 1| = \frac{\beta - \alpha}{\pi} \left| \log |q(z)| + i \arg \frac{1 - cz}{1 - z} \right|. \tag{5.4}$$ Using (5.2) and (5.3) in (5.4), we see that $$|p_{\alpha,\beta} - 1| \le \frac{\beta - \alpha}{\pi} \left( \log \frac{1 + |1 + c|r + r^2}{1 - r^2} + 2 \arctan \frac{r}{1 - r} \right) \le \frac{1}{e}$$ holds in $|z| < r_0$ whenever $r_0$ is the smallest positive root of (5.1). Note that if we choose $\alpha = 1 + \frac{\delta - \pi}{2 \sin \delta}$ and $\beta = 1 + \frac{\delta}{2 \sin \delta}$ , where $\pi/2 \le \delta < \pi$ , then $S(\alpha, \beta)$ reduces to the class $V(\delta)$ introduced by Kargar et al. [26]. Corollary 5.1 Let $f \in \mathcal{V}(\delta)$ . Then $f \in \mathcal{S}_{\wp}^*$ in $\mathbb{D}_{r_{\delta}}$ , where $r_{\delta}$ is the least positive root of the equation $$\frac{1}{2\sin\delta}\left(\log\frac{1+\sqrt{2(1+\cos(2(\pi-\delta)))}r+r^2}{1-r^2}+2\arctan\frac{r}{1-r}\right)-\frac{1}{e}=0.$$ Now we consider the following class introduced in [13]: $$S_{\lambda} := \left\{ f \in \mathcal{A} : \frac{f(z)}{z} \in P_{\lambda} \right\}, \tag{5.5}$$ where $P_{\lambda} := \{ p \in \mathcal{A}_0 : \Re(e^{i\lambda}p(z)) > 0, -\pi/2 \le \lambda \le \pi/2 \}$ denotes the class of tilted Carathéodory functions [38]. Note that $P_0$ reduces to $\mathcal{P}$ , the class of Carathéodory functions. For the function $p \in P_{\lambda}$ , upper bound on the quantity zp'(z)/p(z) is given by the following lemma that will be used for our next result:

Lemma 5.1. [38] If $p \in P_{\lambda}$ , then $|zp'(z)/p(z)| \leq M(\lambda, r)$ , where <span id="page-18-1"></span> $$M(\lambda,r) = \begin{cases} \frac{2r\cos\lambda}{r^2 - 2r|\sin\lambda| + 1} & for \ r < |\tan\frac{\lambda}{2}|;\\ \frac{2r}{1 - r^2} & for \ r \ge |\tan\frac{\lambda}{2}|. \end{cases}$$ The equality holds for some point $z=re^{i\theta},\ r\in(0,1)$ if and only if $p(z)=p_{\lambda}(yz),\ where\ p_{\lambda}(z)=\frac{1+e^{-2i\lambda}z}{1-z}$ and $y=e^{i(\theta_0-\theta)}$ with <span id="page-18-0"></span> $$\theta_0 = \begin{cases} \frac{\pi}{2} + \lambda & \text{for } r < -\tan\frac{\lambda}{2}; \\ -\frac{\pi}{2} + \lambda & \text{for } r < \tan\frac{\lambda}{2}; \\ \arcsin\left(\frac{1+r^2}{r^2-1}\right) + \lambda \text{ for } r \ge |\tan\frac{\lambda}{2}|. \end{cases}$$ Next, we determine the largest radius r such that the function $F(z) := f(z)g(z)/z \in \mathcal{S}^*_{\wp}$ in |z| < r, whenever $f, g \in \mathcal{S}_{\lambda}$ .

Theorem 5.5. Let $c_{\lambda} = \cos \lambda$ , $s_{\lambda} = \sin \lambda$ and $t_{\lambda} = |\tan(\lambda/2)|$ . If $f, g \in \mathcal{S}_{\lambda}$ , then $F \in \mathcal{S}_{\wp}^*$ in $\mathbb{D}_{r_0}$ , where $$r_0 := \begin{cases} 2ec_{\lambda} + |s_{\lambda}| + \sqrt{((4e^2 - 1)c_{\lambda} + 4e|s_{\lambda}|)c_{\lambda}}, & \text{if } r < t_{\lambda}; \\ \sqrt{4e^2 + 1} - 2e, & \text{if } r > t_{\lambda}. \end{cases}$$ Proof Since $f, g \in \mathcal{S}_{\lambda}$ , it follows that the functions p(z) = f(z)/z and q(z) = g(z)/z belong to the class $P_{\lambda}$ such that F(z) = zp(z)q(z). Thus $$\frac{zF'(z)}{F(z)} - 1 = \frac{zp'(z)}{p(z)} + \frac{zq'(z)}{q(z)}.$$ Now from Lemma 5.1, we obtain $$\left| \frac{zF'(z)}{F(z)} - 1 \right| \le 2M(\lambda, r).$$ Therefore, using Lemma 1.1, we conclude that if $2M(\lambda, r) \leq 1/e$ , then $F \in \mathcal{S}_{\wp}^*$ . Since $2M(\lambda, r) \leq 1/e$ holds whenever $\frac{2rc_{\lambda}}{r^2 - 2|s_{\lambda}|r + 1} \leq \frac{1}{2e}$ if $r < t_{\lambda}$ , and $\frac{2r}{1 - r^2} \leq \frac{1}{2e}$ if $r \geq t_{\lambda}$ ; or equivalently $$r^2 - 2(|s_{\lambda}| + 2ec_{\lambda})r + 1 \ge 0$$ , if $r < t_{\lambda}$ and $$r^2 + 4er - 1 \le 0$$ , if $r \ge t_\lambda$ , respectively. Hence the result follows with $r_0$ as given in the hypothesis. Further, for the functions $$f(z) = g(z) = \frac{z(1 + e^{-2i\lambda}yz)}{1 - yz},$$ sharpness hold in view of Lemma 5.1.

Definition 1. For the subfamilies $\mathcal{G}_1$ and $\mathcal{G}_2$ of $\mathcal{A}$ , we say that $r_0$ is the $\mathcal{G}_1$ radius of the class $\mathcal{G}_2$ , if $r_0 \in (0,1)$ is largest number such that $r^{-1}f(rz) \in \mathcal{G}_1$ , $0 < r \le r_0$ for all $f \in \mathcal{G}_2$ . Enormous interest in the radius problems regarding special functions started from the work of Brown [11,12], Wilf [39], and Kreyszig and Todd [27]. Recently, the radii of starlikeness and convexity of some normalized special functions were studied widely for certain Ma-Minda sub-classes as they can be represented as Hadamard factorization under certain conditions. See the work on Bessel functions [1,37], Struve functions [1,4], Wright functions [6], Lommel functions [1,4] and Legendre polynomials of odd degree [9]. We also refer to see [10,15]. For more on recent radius problems, see [18,19,20,21,22,23]. Bessel Function: The Bessel function $\mathcal{J}_{\beta}$ of first kind of order $\beta \in \mathbb{C}$ is a particular solution of the homogeneous Bessel differential equation $$z^2w''(z) + zw'(z) + (z^2 - \beta^2)w(z) = 0$$ and have the following series expansion: $$\mathcal{J}_{\beta}(z) := \sum_{n \ge 1} \frac{(-1)^n}{n! \Gamma(n+\beta+1)} \left(\frac{z}{2}\right)^{2n+\beta},$$ where $z \in \mathbb{C}$ and $\beta \notin \mathbb{Z}^-$ . Let us consider the following three normalized functions expressed in terms of $\mathcal{J}_{\beta}(z)$ <span id="page-2-1"></span> $$\begin{cases} f_{\beta}(z) = (2^{\beta} \Gamma(\beta + 1) \mathcal{J}_{\beta}(z))^{1/\beta} = z - \frac{1}{4\beta(\beta + 1)} z^{3} + \cdots, & \beta \neq 0 \\ g_{\beta}(z) = 2^{\beta} \Gamma(\beta + 1) z^{1-\beta} \mathcal{J}_{\beta}(z) = z - \frac{1}{4(\beta + 1)} z^{3} + \cdots, & \\ h_{\beta}(z) = 2^{\beta} \Gamma(\beta + 1) z^{1-\beta/2} \mathcal{J}_{\beta}(\sqrt{z}) = z - \frac{1}{4(\beta + 1)} z^{2} + \cdots. \end{cases}$$ (1.2) Since the zeros of $\mathcal{J}_{\beta}$ are real if $\beta > 0$ , therefore using the Weierstrass decomposition, we have for $\beta > 0$ : $$\mathcal{J}_{\beta}(z) := \frac{z^{\beta}}{2^{\beta} \Gamma(\beta+1)} \prod_{n>1} \left( 1 - \frac{z^2}{j_{\beta,n}^2} \right),$$ where $j_{\beta,n}$ is the n-th positive zero of $\mathcal{J}_{\beta}$ and satisfies $j_{\beta,n} < j_{\beta,n+1}$ for $n \in \mathbb{N}$ . Thus we have <span id="page-2-3"></span> $$\frac{z\mathcal{J}_{\beta}'(z)}{\mathcal{J}_{\beta}(z)} = \beta - \sum_{n>1} \frac{2z^2}{j_{\beta,n}^2 - z^2}.$$ (1.3) <u>Struve function</u>: The Struve function $\mathbf{H}_{\beta}$ of first kind is a particular solution of the second-order inhomogeneous Bessel differential equation $$z^{2}w''(z) + zw'(z) + (z^{2} - \beta^{2})w(z) = \frac{4(\frac{z}{2})^{\beta+1}}{\sqrt{\pi}\Gamma(\beta + \frac{1}{2})}$$ and have the following form: $$\mathbf{H}_{\beta}(z) := \frac{(\frac{z}{2})^{\beta+1}}{\sqrt{\frac{\pi}{4}} \Gamma(\beta + \frac{1}{2})} {}_{1}F_{2}\left(1; \frac{3}{2}, \beta + \frac{3}{2}; -\frac{z^{2}}{4}\right),$$ where $-\beta-\frac{3}{2}\notin\mathbb{N}$ and ${}_1F_2$ is a hypergeometric function. Since it is not normalized, so we consider the following normalized functions involving $\mathbf{H}_\beta$ : <span id="page-2-2"></span> $$\begin{cases} U_{\beta}(z) = \left(\sqrt{\pi}2^{\beta}(\beta + \frac{3}{2})\mathbf{H}_{\beta}(z)\right)^{\frac{1}{\beta+1}}, \\ V_{\beta}(z) = \sqrt{\pi}2^{\beta}z^{-\beta}\Gamma(\beta + \frac{3}{2})\mathbf{H}_{\beta}(z), \\ W_{\beta}(z) = \sqrt{\pi}2^{\beta}z^{\frac{1-\beta}{2}}\Gamma(\beta + \frac{3}{2})\mathbf{H}_{\beta}(\sqrt{z}). \end{cases} (1.4)$$ Moreover, for $|\beta| \leq \frac{1}{2}$ , it has the Hadamard factorization given by <span id="page-2-0"></span> $$\mathbf{H}_{\beta}(z) = \frac{z^{\beta+1}}{\sqrt{\pi} 2^{\beta} \Gamma(\beta + \frac{3}{2})} \prod_{n>1} \left( 1 - \frac{z^2}{z_{\beta,n}^2} \right), \tag{1.5}$$ where $z_{\beta,n}$ is the n-th positive root of $\mathbf{H}_{\beta}$ such that $z_{\beta,n+1} > z_{\beta,n}$ and $z_{\beta,1} > 1$ and also from (1.5), we obtain <span id="page-2-4"></span> $$\frac{z\mathbf{H}'_{\beta}(z)}{\mathbf{H}_{\beta}(z)} = (\beta + 1) - \sum_{n>1} \frac{2z^2}{z_{\beta,n}^2 - z^2}.$$ (1.6) Lommel function: The Lommel function $\mathcal{L}_{u,v}$ of first kind is a particular solution of the second-order inhomogeneous Bessel differential equation $$z^{2}w''(z) + zw'(z) + (z^{2} - v^{2})w(z) = z^{u+1}.$$ where $u \pm v \notin \mathbb{Z}^-$ and is given by $$\mathcal{L}_{u,v} = \frac{z^{u+1}}{(u-v+1)(u+v+1)} {}_{1}F_{2}\left(1; \frac{u-v+3}{2}, \frac{u+v+3}{2}; -\frac{z^{2}}{4}\right),$$ where $\frac{1}{2}(-u \pm v - 3) \notin \mathbb{N}$ and ${}_1F_2$ is a hypergeometric function. Since it is not normalized, so we consider the following normalized functions involving $\mathcal{L}_{u,v}$ : <span id="page-3-3"></span> $$\begin{cases} f_{u,v}(z) = ((u-v+1)(u+v+1)\mathcal{L}_{u,v}(z))^{\frac{1}{u+1}}, \\ g_{u,v}(z) = (u-v+1)(u+v+1)z^{-u}\mathcal{L}_{u,v}(z), \\ h_{u,v}(z) = (u-v+1)(u+v+1)z^{\frac{1-u}{2}}\mathcal{L}_{u,v}(\sqrt{z}). \end{cases}$$ (1.7) Authors in [1,4] obtained the radius of starlikeness for the following normalized functions expressed in terms of $\mathcal{L}_{u,v}$ : <span id="page-3-0"></span> $$f_{u-\frac{1}{2},\frac{1}{2}}(z), \quad g_{u-\frac{1}{2},\frac{1}{2}}(z) \quad \text{and} \quad h_{u-\frac{1}{2},\frac{1}{2}}(z), \tag{1.8}$$ where $0 \neq u \in (-1,1)$ Legendre polynomial: The Legendre polynomials $P_n$ are the solutions of the Legendre differential equation: $$((1-z^2)P'_n(z))' + n(n+1)P_n(z) = 0,$$ where $n \in \mathbb{Z}^+$ and using Rodrigues' formula, $P_n$ can be represented in the form: $$P_n(z) = \frac{1}{2^n n!} \frac{d^n (z^2 - 1)^n}{dz^n}$$ and it also satisfies the geometric condition $P_n(-z) = (-1)^n P_n(z)$ . Moreover, the odd degree Legendre polynomials $P_{2n-1}(z)$ have only real roots which satisfy <span id="page-3-4"></span><span id="page-3-2"></span> $$0 = z_0 < z_1 < \dots < z_{n-1} \quad \text{or} \quad -z_1 > \dots > -z_{n-1}. \tag{1.9}$$ Thus, the normalized form is as follows: <span id="page-3-1"></span> $$\mathcal{P}_{2n-1}(z) := \frac{P_{2n-1}(z)}{P'_{2n-1}(0)} = z + \sum_{k=2}^{2n-1} a_k z^k = a_{2n-1} z \prod_{k=1}^{n-1} (z^2 - z_k^2).$$ (1.10) At this conjunction, motivated from the work [1,4,6,9,10,21,37] it is natural to consider the radius problem : Problem 1.1 Find the $S^*(\psi)$ -radii for the normalized functions given in (1.2), (1.4), (1.8) and (1.10). That is, $S^*(\psi)$ -radius and $C(\psi)$ -radius of $g \in A$ is defined as follows: $$r_0(g) = \sup\{r \in (0, r_0) : \frac{zg'(z)}{g(z)} \in \psi(\mathbb{D}), z \in \mathbb{D}_{r_0}\}$$ and $$r_0(g) = \sup\{r \in (0, r_0) : 1 + \frac{zg''(z)}{g'(z)} \in \psi(\mathbb{D}), z \in \mathbb{D}_{r_0}\}.$$ Till date, for a specific given function $\psi$ the above problem was considered, see [10]. Certain special functions's radius of starlikeness of order $\alpha \in [0,1)$ is given in [1,4,6,9,37]. To solve this in general, we need to consider the following assumption: <span id="page-4-0"></span>Assumption 1.1 Consider the Ma-Minda function $\psi$ as defined in (1.1). Let $a \in \psi(\mathbb{D}) \cap \mathbb{R}$ , $r_a$ is the radius depending on a and assume the maximal disk $|w-a| < r_a$ such that <span id="page-4-2"></span> $$\{w : |w - a| < r_a\} \subseteq \psi(\mathbb{D}).$$ The following is an example of the Assumption 1.1: