The Booth Lemniscate Starlikeness Radius for Janowski Starlike Functions

Corollary 3.3. Let $0 \le \alpha < 1$ . The Booth lemniscate starlikeness radius of order $\alpha$ for the class $S^*$ of starlike functions is <span id="page-8-1"></span> $$R_{\mathcal{BS}(\alpha)}(\mathcal{S}^*) = \begin{cases} \frac{1}{3 - 2\alpha}, & 0 \le \alpha \le \frac{1}{9}, \\ \frac{2\sqrt{\alpha}}{(1 + \alpha)\sqrt{1 + 16\alpha}}, & \frac{1}{9} \le \alpha \le 1. \end{cases}$$ (3.9)

Corollary 3.4. Let $0 \le \alpha < 1$ . The Booth lemniscate starlikeness radius of order $\alpha$ for the class K of convex functions is $$R_{\mathcal{BS}(\alpha)}(\mathcal{K}) = \begin{cases} \frac{1}{2-\alpha}, & 0 \le \alpha \le \frac{1}{5}, \\ \frac{2\sqrt{\alpha}}{(1+\alpha)\sqrt{1+4\alpha}}, & \frac{1}{5} \le \alpha \le 1. \end{cases}$$ PROOF. Every convex function is also starlike of order 1/2. Thus the $\mathcal{BS}(\alpha)$ -radius is at least as big as that given by Lemma 2.1 with $\beta=1/2$ . However, the extremal starlike function $k_{1/2}$ given by (3.1) is also convex, whence the result. Next let $1 < \beta < 4/3$ , and $M(\beta)$ be the class consisting of functions $f \in \mathcal{A}$ for which $\operatorname{Re}(zf'(z)/f(z)) < \beta$ . This class was introduced by Uralegaddi et al. [13] who investigated functions in the class with positive coefficients. The following result gives the $\mathcal{BS}(\alpha)$ -radius for the class $M(\beta)$ .

Theorem 3.5. Let $0 < \alpha < 1$ and $1 < \beta < 4/3$ . The Booth lemniscate starlikeness radius of order $\alpha$ for the class $M(\beta)$ is $$R_{\mathcal{BS}(\alpha)}(M(\beta)) = \begin{cases} \frac{1}{1 + 2(1 - \alpha)(\beta - 1)}, & 1 < \beta \le 1 + \frac{1 - \alpha}{8\alpha}, \\ \frac{2\sqrt{\alpha}}{(1 + \alpha)\sqrt{1 + 16\alpha(\beta - 1)^2}}, & 1 + \frac{1 - \alpha}{8\alpha} \le \beta < \frac{4}{3}. \end{cases}$$ PROOF. Every function $f \in M(\beta)$ satisfies the inequality $$\left| \frac{zf'(z)}{f(z)} - \frac{1 + (1 - 2\beta)r^2}{1 - r^2} \right| \le \frac{2(\beta - 1)r}{1 - r^2}, \quad |z| \le r < 1.$$ Define $a_f$ and $c_f$ by $$a_f(r) := \frac{1 + (1 - 2\beta)r^2}{1 - r^2}$$ and $c_f(r) := \frac{2(\beta - 1)r}{1 - r^2}$ . As $\beta > 1$ , it follows that $a_f$ is decreasing, whence $a_f(r) \le 1$ for all $0 \le r < 1$ . Recall that this function was increasing in the case of starlike functions of order $\beta$ . Since $a_f(r) \le 1$ , Lemma 2.1 shows that the disc $\mathbb{D}(a_f(r); c_f(r)) \subset G_{\alpha}(\mathbb{D})$ provided $$c_f(r) = \begin{cases} \sqrt{s(\alpha, a)}, & a_f(r) > 1 - \frac{4\alpha}{(1 - \alpha)(1 + 6\alpha + \alpha^2)}, \\ a_f(r) - 1 + \frac{1}{1 - \alpha}, & 1 - \frac{4\alpha}{(1 - \alpha)(1 + 6\alpha + \alpha^2)} \ge a_f(r), \end{cases}$$ (3.10) where $s(\alpha, a_f(r))$ is given by (3.5). Let $$\rho_0 := \frac{1}{1 + 2(1 - \alpha)(\beta - 1)} \quad \text{and} \quad \tilde{\rho}_0 := \frac{2\sqrt{\alpha}}{(1 + \alpha)\sqrt{(1 + 16\alpha(\beta - 1)^2)}}.$$ Then, $\rho_0$ satisfies the equation <span id="page-9-0"></span> $$c_f(r) = a_f(r) - 1 + \frac{1}{1 - \alpha},$$ while $\tilde{\rho_0}$ is the solution of the equation $$c_f(r)^2 = s(\alpha, a_f(r)).$$ Also, $$\rho_1 = \frac{\sqrt{4\alpha}}{\sqrt{4\alpha + (2\beta - 2)(1 - \alpha)(1 + 6\alpha + \alpha^2)}}$$ is the positive root of the equation $$a_f(r) = 1 - \frac{4\alpha}{(1-\alpha)(1+6\alpha+\alpha^2)}$$ . Evidently, $\rho_1 \leq \rho_0$ holds if and only if $$\beta \le 1 + \frac{1 - \alpha}{8\alpha}$$ . Case (i): $1 < \beta \le 1 + ((1-\alpha)/8\alpha)$ . Here $\rho_1 \le \rho_0$ , and because the center $a_f(r)$ is decreasing, then $$a_f(\rho_0) \le a_f(\rho_1) = 1 - \frac{4\alpha}{(1-\alpha)(1+6\alpha+\alpha^2)}.$$ Thus, it follows from (3.10) that $\mathbb{D}(a_f(\rho_0); c_f(\rho_0)) \subset G_\alpha(\mathbb{D})$ for every $f \in M(\beta)$ , or the $\mathcal{BS}(\alpha)$ -radius for $M(\beta)$ is at least $\rho_0$ . Case (ii): $1 + ((1 - \alpha)/8\alpha) \le \beta < 4/3$ . In this case, $\rho_1 \ge \rho_0$ , and because the center $a_f(r)$ is decreasing, then $$a_f(\rho_0) \ge a_f(\rho_1) = 1 - \frac{4\alpha}{(1-\alpha)(1+6\alpha+\alpha^2)}.$$ Thus, $\mathbb{D}(a_f(\tilde{\rho_0}); c_f(\tilde{\rho_0})) \subset G_{\alpha}(\mathbb{D})$ from (3.10). To complete the proof, we observe that the function $k_{\beta}$ given by $k_{\beta}(z) = z/(1-z)^{2-2\beta}$ shows that the radius in each case above is best possible.

Theorem 4.3. Let $0 < \alpha < 1$ , and $0 < B < A \leq 1$ . If neither condition (i) nor (ii) of Theorem 4.1 holds, then the Booth lemniscate starlikeness radius of order $\alpha$ for the class $\mathcal{S}^*[A,B]$ is $$R_{\mathcal{BS}(\alpha)}(\mathcal{S}^*[A,B]) = \begin{cases} \min\left\{1, \frac{2\sqrt{\alpha}}{(1+\alpha)\sqrt{4\alpha(A-B)^2 + B^2}}\right\}, & 4A\alpha \ge (3\alpha+1)B, \\ \min\left\{1, \frac{1}{(1-\alpha)(A-B) + B}\right\}, & 4A\alpha \le (3\alpha+1)B. \end{cases}$$ For $0<\beta\leq 1$ , the class $S^[\beta,-\beta]=:\mathcal{S}^_\beta$ consists of functions $f\in\mathcal{A}$ satisfying the inequality $$\left| \frac{zf'(z)}{f(z)} - 1 \right| < \beta \left| \frac{zf'(z)}{f(z)} + 1 \right|.$$ Parvatham [10] introduced this class in her studies on the Bernardi integral operator. The function $f(z) = z/(1-\beta z)^2$ belongs to the class $S_{\beta}^*$ . The $\mathcal{BS}(\alpha)$ -radius for this class follows readily from Theorem 4.2. <span id="page-12-3"></span>COROLLARY 4.4. For $0 \le \beta < 1$ , the Booth lemniscate starlikeness radius of order $\alpha$ for the class $S_{\beta}^*$ is $$R_{\mathcal{BS}(\alpha)}(\mathcal{S}_{\beta}^{*}) = \begin{cases} \min\left\{1, \frac{1}{\beta(3-2\alpha)}\right\}, & 0 \leq \alpha \leq \frac{1}{9}, \\ \min\left\{1, \frac{2\sqrt{\alpha}}{\beta(1+\alpha)\sqrt{1+16\alpha}}\right\}, & \frac{1}{9} \leq \alpha \leq 1. \end{cases}$$ It is worthy to note that for $\beta = 1$ , Corollary 4.4 reduces to the one given by (3.9). For $0 \le \beta < 1$ , the class $S^[1 - \beta, 0] := S^[\beta]$ consists of functions $f \in A$ satisfying the inequality $$\left| \frac{zf'(z)}{f(z)} - 1 \right| < 1 - \beta.$$ Clearly, $S^[\beta] \subset S^(\beta)$ and the function $f(z) = ze^{(1-\beta)z}$ belongs to the class $S^*[\beta]$ . This class was introduced and studied by Fournier [2], and we state its Booth lemniscate starlikeness radius.

Corollary 4.5. For $0 \le \beta < 1$ , the Booth lemniscate starlikeness radius of order $\alpha$ for the class $S^*[\beta]$ is $$R_{\mathcal{BS}(\alpha)}(\mathcal{S}^*[\beta]) = \min\left\{1; \frac{1}{(1+\alpha)(1-\beta)}\right\}.$$ In particular, $S^*[\alpha/(1+\alpha)] \subset \mathcal{BS}(\alpha)$ .