The Booth Lemniscate Starlikeness Radius for Janowski Starlike Functions

Lemma 2.1. Let $0 \le \alpha < 1$ and $(1-2\alpha)/(2-2\alpha) < a < (3-2\alpha)/(2-2\alpha)$ . Then the following inclusions hold: $$\mathbb{D}(a; r_a) \subseteq G_{\alpha}(\mathbb{D}) \subseteq \mathbb{D}(a; R_a),$$ where $r_a$ and $R_a$ are given by $$r_{a} = \begin{cases} a - 1 + \frac{1}{1 - \alpha}, & \frac{1 - 2\alpha}{2 - 2\alpha} < a \le 1 - \frac{4\alpha}{(1 - \alpha)(1 + 6\alpha + \alpha^{2})}, \\ \sqrt{s(\alpha, a)}, & 1 - \frac{4\alpha}{(1 - \alpha)(1 + 6\alpha + \alpha^{2})} < a < 1 + \frac{4\alpha}{(1 - \alpha)(1 + 6\alpha + \alpha^{2})}, \\ 1 - a + \frac{1}{1 - \alpha}, & 1 + \frac{4\alpha}{(1 - \alpha)(1 + 6\alpha + \alpha^{2})} \le a < \frac{3 - 2\alpha}{2 - 2\alpha}, \end{cases}$$ and $$R_a = \begin{cases} 1 - a + \frac{1}{1 - \alpha}, & \frac{1 - 2\alpha}{2 - 2\alpha} < a \le 1, \\ a - 1 + \frac{1}{1 - \alpha}, & 1 \le a < \frac{3 - 2\alpha}{2 - 2\alpha}, \end{cases}$$ with $$s(\alpha, a) = \frac{\sqrt{\alpha[\alpha - (1 - a)^2(1 - \alpha^2)^2]} + \alpha(1 + 2(1 + \alpha)^2(1 - a)^2)}{2\alpha(1 + \alpha)^2}, \quad \alpha \neq 0.$$ PROOF. As noted earlier, the result for a=1 was proved in [1, Lemma 3.4]. If $\alpha=0$ , then $G_0(z)=1+z$ . In this case, readily $r_a=a$ and $R_a=2-a$ for $1/2< a\leq 1$ . Also, for $1\leq a<3/2$ , it is readily seen that $r_a=2-a$ and $R_a=a$ . Thus, assume next that $\alpha \neq 0$ and $a \neq 1$ . The boundary $\partial G_{\alpha}(\mathbb{D})$ of the image of the unit disc $\mathbb{D}$ in parametric form is given by $$G_{\alpha}(e^{it}) = 1 + \frac{e^{it}}{1 - \alpha e^{2it}} = 1 + \frac{(1 - \alpha)\cos t + i(1 + \alpha)\sin t}{1 + \alpha^2 - 2\alpha\cos(2t)}.$$ The result is proved by showing the minimum and maximum distance from the point (a, 0) to the point on the boundary $\partial G_{\alpha}(\mathbb{D})$ are respectively $r_a$ and $R_a$ . Thus consider the function $$H(\cos t) = \left(1 - a + \frac{(1 - \alpha)\cos t}{1 + \alpha^2 - 2\alpha\cos(2t)}\right)^2 + \left(\frac{(1 + \alpha)\sin t}{1 + \alpha^2 - 2\alpha\cos(2t)}\right)^2$$ $$= (1 - a)^2 + \frac{1 + 2(1 - a)(1 - \alpha)\cos(t)}{1 + \alpha^2 - 2\alpha\cos(2t)},$$ (2.1) that is, $$H(x) = (1-a)^2 + \frac{1+2(1-a)(1-\alpha)x}{(1+\alpha)^2 - 4\alpha x^2}, \quad x = \cos t \in [-1, 1].$$ (2.2) A computation using (2.2) shows that $$H'(x) = \frac{2(1-a)(1-\alpha)}{(\alpha+1)^2 - 4\alpha x^2} + \frac{8\alpha x(2(1-a)(1-\alpha)x+1)}{((\alpha+1)^2 - 4\alpha x^2)^2}$$ $$= \frac{8\alpha(1-\alpha)(1-a)(x-x_1)(x-x_2)}{((\alpha+1)^2 - 4\alpha x^2)^2},$$ (2.3) where $x_1$ and $x_2$ are the two zeros of H'(x). These are the zeros of the polynomial $$4\alpha(1-\alpha)(1-a)x^2 + 4\alpha x + (1-\alpha)(1+\alpha)^2(1-a) = 0$$ (2.4) given by $$x_1 = -\frac{\alpha + \sqrt{\alpha(\alpha - (1-a)^2(1-\alpha^2)^2)}}{2\alpha(1-a)(1-\alpha)},$$ and $$x_2 = -\frac{\alpha - \sqrt{\alpha(\alpha - (1-a)^2(1-\alpha^2)^2)}}{2\alpha(1-a)(1-\alpha)}.$$ The zeros $x_1$ and $x_2$ satisfy $x_1x_2 = (1 + \alpha)^2/(4\alpha) \ge 1$ . Further, $x_1$ , $x_2$ are real if $\alpha$ and $\alpha$ satisfy $\alpha \ge (1 - \alpha)^2(1 - \alpha^2)^2$ , or equivalently, whenever $$|a-1| \le \frac{\sqrt{\alpha}}{1-\alpha^2}.$$ The following notations are introduced to give greater clarity to the proof. Let $$\alpha_0 := 1 - \frac{\sqrt{\alpha}}{1 - \alpha^2}, \quad \alpha_1 := 1 - \frac{4\alpha}{(1 - \alpha)(1 + 6\alpha + \alpha^2)},$$ $$\tilde{\alpha}_0 := 1 + \frac{\sqrt{\alpha}}{1 - \alpha^2}, \quad \tilde{\alpha}_1 := 1 + \frac{4\alpha}{(1 - \alpha)(1 + 6\alpha + \alpha^2)}.$$ A little computations shows that $x_1 < -1$ for $\alpha_0 < a < 1$ , while $x_1 > 1$ for $1 < a < \tilde{\alpha}_0$ . Similarly, $x_2 < 0$ for a < 1; indeed, $x_2 < -1$ for $\alpha_0 \le a < \alpha_1$ , and $-1 \le x_2 \le 0$ for $\alpha_1 \le a < 1$ . Also, $x_2 > 0$ for a > 1; indeed, $0 \le x_2 \le 1$ for $1 < a \le \tilde{\alpha}_1$ and $x_2 > 1$ for $\tilde{\alpha}_1 < a \le \tilde{\alpha}_0$ . These observations together with (2.3) will be helpful in the following cases. Case (i). If $\alpha_0 \le a \le \alpha_1$ , it follows that the function H is increasing and therefore $$r_a = \sqrt{H(-1)} = a - 1 + \frac{1}{1 - \alpha}$$ , and $R_a = \sqrt{H(1)} = 1 - a + \frac{1}{1 - \alpha}$ . Case (ii). If $\alpha_1 < a < 1$ , then H'(x) < 0 for $x < x_2$ , while H'(x) > 0 for $x > x_2$ . Thus, $x_2$ is the minimum point. Since a < 1, the maximum of H occurs at x = 1. Therefore, $$r_a = \sqrt{H(x_2)}$$ and $R_a = \sqrt{H(1)}$ Note that $$\sqrt{H(x_2)} = \sqrt{\frac{\sqrt{\alpha[\alpha - (1-a)^2(1-\alpha^2)^2]} + \alpha(1+2(1+\alpha)^2(1-a)^2)}{2\alpha(1+\alpha)^2}}.$$ Case (iii). If $1 < a < \tilde{\alpha}_1$ , then H'(x) < 0 for $x < x_2$ , while H'(x) > 0 for $x > x_2$ . Thus, $x_2$ is the minimum point. Since a > 1, the function H attains its maximum at x = -1 so that $$r_a = \sqrt{H(x_2)}$$ and $R_a = \sqrt{H(-1)}$ . Case (iv). If $\tilde{\alpha}_1 \leq a \leq \tilde{\alpha}_0$ , then the function H is decreasing, whence $$r_a = \sqrt{H(1)}$$ and $R_a = \sqrt{H(-1)}$ . It remains next to consider the range $$|a-1| > \sqrt{\alpha}/(1-\alpha^2).$$ In this case, H' is non-vanishing in [-1, 1]. Since $$H'(0) = \frac{8\alpha(1-\alpha)(1-a)x_1x_2}{(\alpha+1)^4},$$ and (2.4) yields $x_1x_2 = (1+\alpha)^2/(4\alpha) > 0$ , it follows that H'(0) < 0 for a > 1, while H'(0) > 0 for a < 1. Since H' is non-vanishing, we deduce for $x \in [-1, 1]$ that H'(x) < 0 whenever a > 1, and H'(x) > 0 for a < 1. Therefore, for a > 1, $$r_a = \min \sqrt{H(x)} = \sqrt{H(1)}$$ and $R_a = \max \sqrt{H(x)} = \sqrt{H(-1)}$ . Similarly, for a < 1, $$r_a = \sqrt{H(-1)}$$ and $R_a = \sqrt{H(1)}$ . The results now follow because $$\sqrt{H(-1)} = a - 1 + \frac{1}{1 - \alpha}$$ , and $\sqrt{H(1)} = 1 - a + \frac{1}{1 - \alpha}$ . It is worth noting that the assumption in Lemma 2.1 on the center a ensures that the disc $\mathbb{D}(a; r_a)$ contains the point w = 1.

Theorem 3.1. Let $0 < \alpha < 1$ and $0 \le \beta < 1$ . If $f \in S^*(\beta)$ , then f is Booth lemniscate starlike of order $\alpha$ in the disk of radius $$R_{\mathcal{BS}(\alpha)}(\mathcal{S}^*(\beta)) = \begin{cases} \frac{2\sqrt{\alpha}}{(1+\alpha)\sqrt{1+16\alpha(1-\beta)^2}}, & 0 \le \beta < \max\left\{0; \frac{9\alpha-1}{8\alpha}\right\}, \\ \frac{1}{1+2(1-\alpha)(1-\beta)}, & \max\left\{0; \frac{9\alpha-1}{8\alpha}\right\} \le \beta < 1. \end{cases}$$ (3.2) PROOF. It is known (see [11]) that functions $f \in \mathcal{S}^*(\beta)$ satisfy $$\left| \frac{zf'(z)}{f(z)} - \frac{1 + (1 - 2\beta)r^2}{1 - r^2} \right| \le \frac{2(1 - \beta)r}{1 - r^2}, \quad |z| \le r < 1.$$ Thus $zf'(z)/f(z) \in \mathbb{D}(a_f(r); c_f(r))$ where $$a_f(r) := \frac{1 + (1 - 2\beta)r^2}{1 - r^2}$$ and $c_f(r) := \frac{2(1 - \beta)r}{1 - r^2}$ . (3.3) We wish to find $\rho$ so that $\mathbb{D}(a_f(\rho); c_f(\rho)) \subseteq G_{\alpha}(\mathbb{D})$ for every $f \in \mathcal{S}^*(\beta)$ . Since $a_f'(r) > 0$ , we note that $a_f(r) \geq 1$ . Now Lemma 2.1 shows that the disc $\mathbb{D}(a_f(r); c_f(r)) \subset G_{\alpha}(\mathbb{D})$ provided $$c_f(r) = \begin{cases} \sqrt{s(\alpha, a)}, & a_f(r) < 1 + \frac{4\alpha}{(1 - \alpha)(1 + 6\alpha + \alpha^2)}, \\ 1 - a_f(r) + \frac{1}{1 - \alpha}, & 1 + \frac{4\alpha}{(1 - \alpha)(1 + 6\alpha + \alpha^2)} \le a_f(r), \end{cases}$$ (3.4) where $$s(\alpha, a_f(r)) = \frac{\sqrt{\alpha[\alpha - (1 - a_f(r))^2(1 - \alpha^2)^2]} + \alpha(1 + 2(1 + \alpha)^2(1 - a_f(r))^2)}{2\alpha(1 + \alpha)^2}.$$ (3.5) Let us write $$\rho_0 := \frac{2\sqrt{\alpha}}{(1+\alpha)\sqrt{(1+16\alpha(1-\beta)^2)}} \quad \text{and} \quad \tilde{\rho}_0 := \frac{1}{1+2(1-\alpha)(1-\beta)}.$$ The number $\rho_0 < 1$ is the positive root of the equation $c_f(r) = \sqrt{s(\alpha, a_f(r))}$ given by (3.5). Indeed, this equation has the form $$(2\alpha(1+\alpha)^2(c_f(r)^2-(1-a_f(r))^2))^2=\alpha^2-\alpha(1-\alpha^2)^2(1-a_f(r))^2,$$ which upon solving and replacing $a_f$ and $c_f$ by the expressions given by (3.3), yields the solution $\rho_0$ . Also, the number $\tilde{\rho}_0 < 1$ is the root of the equation $$c_f(r) = 1 - a_f(r) + \frac{1}{1 - \alpha}.$$ Further, $$\rho_1 := \frac{\sqrt{2\alpha}}{\sqrt{2\alpha + (1-\alpha)(1-\beta)(1+6\alpha+\alpha^2)}} < 1$$ is the positive root of the equation $$\frac{(1-\beta)r^2}{1-r^2} = \frac{2\alpha}{(1-\alpha)(1+6\alpha+\alpha^2)},$$ obtained from rewriting the equation $$a_f(r) = 1 + \frac{4\alpha}{(1-\alpha)(1+6\alpha+\alpha^2)}.$$ Let us also write $$\beta_0 := 1 - \frac{1 - \alpha}{8\alpha} = \frac{9\alpha - 1}{8\alpha}, \qquad \tilde{\beta}_0 := 1 - \frac{1 - \alpha}{2(1 + \alpha)^2} = \frac{1 + 5\alpha + 2\alpha^2}{2(1 + \alpha)^2}.$$ Then, $\rho_0 = \tilde{\rho_0}$ if $\beta = \beta_0$ . Since $0 < \alpha < 1$ , a calculation shows that $\beta_0 < \tilde{\beta}_0$ . For $\alpha < 1/9$ , or equivalently for $\beta_0 < 0$ , we shall show that $R_{\mathcal{BS}(\alpha)} = \tilde{\rho}_0$ for all $0 \le \beta < 1$ . When $\alpha \ge 1/9$ , or equivalently $\beta_0 \ge 0$ , we shall show that $R_{\mathcal{BS}(\alpha)} = \rho_0$ for $0 \le \beta < \beta_0$ , while $R_{\mathcal{BS}(\alpha)} = \tilde{\rho}_0$ for $\beta_0 \le \beta < 1$ . Thus there are two cases to consider: $0 \le \beta < \beta_0$ , and $\beta_0 \le \beta < 1$ . Case (i). Let $0 \le \beta < \beta_0$ . A little calculations shows that the inequality $\rho_0 < \rho_1$ is equivalent to $$(2(1+\alpha)^2\beta - (1+5\alpha+2\alpha^2))(1-9\alpha+8\alpha\beta) > 0.$$ (3.6) Inequality (3.6) holds if and only if $$\beta < \min \left\{ \beta_0, \tilde{\beta}_0 \right\} = \beta_0, \quad \text{or} \quad \beta > \max \left\{ \beta_0, \tilde{\beta}_0 \right\} = \tilde{\beta}_0.$$ In this case, $\rho_0 \leq \rho_1$ and $$a_f(\rho_0) \le a_f(\rho_1) = 1 + \frac{4\alpha}{(1-\alpha)(1+6\alpha+\alpha^2)}.$$ From Lemma 2.1, it follows that $\mathbb{D}(a_f(\rho_0); c_f(\rho_0)) \subset G_{\alpha}(\mathbb{D})$ , whence the $\mathcal{BS}(\alpha)$ -radius for $\mathcal{S}^*(\beta)$ is at least $\rho_0$ . To validate sharpness of $\rho_0$ , we consider the generalized Koebe function $k_\beta$ given by (3.1). We shall show the existence of a point on $|z| = \rho_0$ that is mapped to a point on $\partial G_\alpha(\mathbb{D})$ . In other words, we prove that there is some t such that the image of the point $z = \rho_0 e^{it}$ under the map $zk'_\beta(z)/k_\beta(z)$ belongs to $G_\alpha(e^{it})$ . For this purpose, let us write $G_{\alpha}(e^{it}) = u(t) + iv(t)$ . Then u and v satisfy the equation $$((u-1)^2 + v^2)^2 = \left(\frac{u-1}{1-\alpha}\right)^2 + \left(\frac{v}{1+\alpha}\right)^2.$$ (3.7) The representation of the function $zk'_{\beta}(z)/k_{\beta}(z)$ at $z=\rho_0 e^{it}$ in Cartesian coordinates is $$\frac{zk_{\beta}'(z)}{k_{\beta}(z)} = \frac{1 + (1 - 2\beta)z}{1 - z} = \frac{1 - \overline{z} + (1 - 2\beta)z - (1 - 2\beta)|z|^2}{|1 - z|^2} = \frac{1 - 2\beta\rho_0\cos t - (1 - 2\beta)(\rho_0)^2}{1 + (\rho_0)^2 - 2\rho_0\cos t} + i\frac{2(1 - \beta)\rho_0\sin t}{1 + (\rho_0)^2 - 2\rho_0\cos t}.$$ By taking $$u(t) = \frac{1 - 2\beta \rho_0 \cos t - (1 - 2\beta)(\rho_0)^2}{1 + (\rho_0)^2 - 2\rho_0 \cos t} \quad \text{and} \quad v(t) = \frac{2(1 - \beta)\rho_0 \sin t}{1 + (\rho_0)^2 - 2\rho_0 \cos t},$$ it is readily seen that $$((u-1)^2 + v^2)^2 = \frac{16(\rho_0)^4 (1-\beta)^4}{(1+(\rho_0)^2 - 2\rho_0 \cos t)^2},$$ and $$\left(\frac{u-1}{1-\alpha}\right)^2 + \left(\frac{v}{1+\alpha}\right)^2 = \frac{4(\rho_0)^2(1-\beta)^2\left[(\rho_0-\cos t)^2(1+\alpha)^2 + (\sin t)^2(1+\alpha)^2\right]}{(1+(\rho_0)^2 - 2\rho_0\cos t)^2(1-\alpha)^2(1+\alpha)^2}.$$ From (3.7), we seek to find a t satisfying the equation $$4(\rho_0)^2(1-\beta)^2 = \frac{(\rho_0 - \cos t)^2}{(1-\alpha)^2} + \frac{(\sin t)^2}{(1+\alpha)^2}.$$ Replacing the value $\rho_0$ and writing $x = \cos t$ yields $$\frac{16\alpha(1-\beta)^2}{(1+\alpha)^2(1+16\alpha(1-\beta)^2)} - \frac{1-x^2}{(1+\alpha)^2} - \frac{1}{(1-\alpha)^2} \left(x - \frac{2\sqrt{\alpha}}{(1+\alpha)\sqrt{1+16\alpha(1-\beta)^2}}\right)^2 = 0,$$ or equivalently, the equation $$4\alpha \left(1 + 16\alpha (1-\beta)^2\right) x^2 - 4(1+\alpha)\sqrt{\alpha (1 + 16\alpha (1-\beta)^2)} x + (1+\alpha)^2 = 0.$$ (3.8) Clearly, the number $$x_0 = \frac{1 + \alpha}{2\sqrt{\alpha(1 + 16\alpha(1 - \beta)^2)}}$$ is the positive double real root of (3.8). Since $0 < \beta \le \beta_0$ , a computation shows that $x_0 < 1$ . With $t = \arccos x_0$ and $z = \rho_0 e^{it}$ , the point $zk'_{\beta}(z)/k_{\beta}(z)$ lies on $\partial G_{\alpha}(\mathbb{D})$ . This proves sharpness for $\rho_0$ . Case (ii). Let $\beta_0 \leq \beta < 1$ . Here $\tilde{\rho}_0 \geq \rho_1$ and $$a_f(\tilde{\rho}_0) \ge a_f(\rho_1) = 1 + \frac{4\alpha}{(1-\alpha)(1+6\alpha+\alpha^2)}.$$ From Lemma 2.1, it follows that $\mathbb{D}(a_f(\tilde{\rho}_0); c_f(\tilde{\rho}_0)) \subset G_{\alpha}(\mathbb{D})$ showing that the $\mathcal{BS}(\alpha)$ -radius for the class of starlike functions of order $\beta$ is at least $\tilde{\rho}_0$ . To show sharpness of $\tilde{\rho}_0$ , consider again the generalized Koebe function $k_{\beta}$ given by (3.1). We shall find a point on $|z| = \tilde{\rho}_0$ such that it is mapped to a point on $\partial G_{\alpha}(\mathbb{D})$ . Evidently, $$\frac{zk'_{\beta}(z)}{k_{\beta}(z)} = 1 + 2(1 - \beta)\frac{z}{1 - z}.$$ Since $$\frac{\tilde{\rho}_0}{1-\tilde{\rho}_0} = \frac{1}{2(1-\alpha)(1-\beta)},$$ evaluating at $z = \tilde{\rho}_0$ gives $$\frac{zk_{\beta}'(z)}{k_{\beta}(z)} = 1 + \frac{1}{1-\alpha} = G_{\alpha}(1) \in \partial G_{\alpha}(\mathbb{D}).$$ This proves sharpness of $\tilde{\rho}_0$ . The condition (3.2) suggests that the $\mathcal{BS}(\alpha)$ -radius in the case $\alpha = 0$ is $1/(3-2\beta)$ . That this is indeed the case follows easily from Lemma 2.1.

Theorem 3.2. Let $0 \le \beta < 1$ . The Booth lemniscate starlikeness radius (of order 0) for the class of starlike functions of order $\beta$ is $1/(3-2\beta)$ . Theorem 3.1 and Theorem 3.2 also readily yield the following results for starlike and convex 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 $$R_{\mathcal{BS}(\alpha)}(\mathcal{S}^*) = \begin{cases} \frac{1}{3 - 2\alpha}, & 0 \le \alpha \le \frac{1}{9}, \\ \frac{2\sqrt{\alpha}}{(1 + \alpha)\sqrt{1 + 16\alpha}}, & \frac{1}{9} \le \alpha \le 1. \end{cases}$$ (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 $$c_f(r) = a_f(r) - 1 + \frac{1}{1 - \alpha},$$ 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 4.1. Let $-1 < B < A \le 1$ . The inclusion $S^*[A, B] \subset \mathcal{BS}(\alpha)$ holds if either (i) $$(1-\alpha)(1+6\alpha+\alpha^2)|B|(A-B) \le 4\alpha(1-B^2)$$ and $(1+\alpha)^2(4\alpha(A-B)^2+B^2) \le 4\alpha$ , (ii) $$(1-\alpha)(1+6\alpha+\alpha^2)|B|(A-B) \ge 4\alpha(1-B^2)$$ and $(1-\alpha)(A-B)+|B| \le 1$ . PROOF. Every function $f \in \mathcal{S}^*[A, B]$ satisfies (see [11]) $$\left| \frac{zf'(z)}{f(z)} - \frac{1 - ABr^2}{1 - B^2r^2} \right| \le \frac{(A - B)r}{1 - B^2r^2}, \quad |z| \le r < 1. \tag{4.1}$$ This shows that $zf'(z)/f(z) \in \mathbb{D}(a_f; c_f)$ where $$a_f = \frac{1 - AB}{1 - B^2}$$ and $c_f = \frac{A - B}{1 - B^2}$ We first prove the result for B < 0. Here note that $a_f > 1$ . Case (i). Assume that $(1-\alpha)(1+6\alpha+\alpha^2)B(A-B) \le 4\alpha(1-B^2)$ and $(1+\alpha)^2(4\alpha(A-B)^2+B^2) \le 4\alpha$ . The first inequality reduces to $$a_f \le 1 + \frac{4\alpha}{(1-\alpha)(1+6\alpha+\alpha^2)},$$ and so the result will follow if $$c_f^2 \le \frac{\sqrt{\alpha[\alpha - (1 - a_f)^2(1 - \alpha^2)^2]} + \alpha(1 + 2(1 + \alpha)^2(1 - a_f)^2)}{2\alpha(1 + \alpha)^2}.$$ The latter inequality is the statement of the second inequality $(1+\alpha)^2(4\alpha(A-B)^2+B^2) \leq 4\alpha$ . Case (ii). Assume that $(1-\alpha)(1+6\alpha+\alpha^2)B(A-B) \ge 4\alpha(1-B^2)$ and $(1-\alpha)(A-B)-B \le 1$ . Then $$a_f \ge 1 + \frac{4\alpha}{(1-\alpha)(1+6\alpha+\alpha^2)},$$ whence the result will follow if $$c_f \le 1 - a_f + \frac{1}{1 - \alpha},$$ or equivalently, when $(1 - \alpha)(A - B) - B \le 1$ . When $B \geq 0$ , the center $a_f \leq 1$ , and the proof proceeds similarly as before, and is thus omitted. We next turn our attention when the conditions in Theorem 4.1 fail to hold. In this case, we seek the $\mathcal{BS}(\alpha)$ -radius for the class $\mathcal{S}^*[A,B]$ . The following result is also an extension of Theorem 3.1.

Theorem 4.2. Let $0 < \alpha < 1$ , $-1 < B \le 0$ and $B < A \le 1$ . 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 $$R_{\mathcal{BS}(\alpha)}(\mathcal{S}^*[A,B]) = \begin{cases} \min\left\{1, \frac{2\sqrt{\alpha}}{(1+\alpha)\sqrt{4\alpha(A-B)^2 + B^2}}\right\}, & 4A\alpha \ge (5\alpha - 1)B, \\ \min\left\{1, \frac{1}{(1-\alpha)(A-B) - B}\right\}, & 4A\alpha \le (5\alpha - 1)B. \end{cases}$$ PROOF. The inequality (4.1) gives $zf'(z)/f(z) \in \mathbb{D}(a_f(r); c_f(r))$ , where $$a_f(r) := \frac{1 - ABr^2}{1 - B^2r^2}$$ and $c_f(r) := \frac{(A - B)r}{1 - B^2r^2}$ The result follows easily for B=0, and so assume that B<0. Since $$a'_f(r) = -\frac{2B(B-A)r}{(1-B^2r^2)^2},$$ and -1 < B < 0, it follows that $a_f$ is increasing with $a_f(r) \ge 1$ for $0 \le r < 1$ . Only a brief outline of the proof will be given here because the proof is similar to Theorem 3.1. The numbers $$\tilde{\rho_0} = \frac{2\sqrt{\alpha}}{(1+\alpha)\sqrt{4\alpha(A-B)^2 + B^2}}$$ and $\rho_0 = \frac{1}{(1-\alpha)(A-B) - B}$ satisfy respectively the equations $$c_f(r)^2 = s(\alpha, a_f(r))$$ with $s(\alpha, a_f(r))$ given by (3.5), and $$c_f(r) = 1 - a_f(r) + \frac{1}{1 - \alpha}.$$ Also, the number $$\rho_1 = \frac{\sqrt{4\alpha}}{\sqrt{4\alpha B^2 + (1 - \alpha)(1 + 6\alpha + 6\alpha^2)(B^2 - AB)}}$$ is the solution to the equation $$a_f(r) = 1 + \frac{4\alpha}{(1-\alpha)(1+6\alpha+\alpha^2)}.$$ Here, the condition $\rho_1 \leq \rho_0$ holds if and only if $$A \le \left(1 - \frac{1 - \alpha}{4\alpha}\right) B. \tag{4.2}$$ Thus $a_f(\rho_1) \leq a_f(\rho_0)$ if and only if (4.2) holds. The result follows by an application of Lemma 2.1 and is sharp for the function $f \in \mathcal{S}^*[A,B]$ given by $f(z) = z/(1+Bz)^{(B-A)/B}$ for $B \neq 0$ , while $f(z) = ze^{Az}$ for B = 0. The result in the case B > 0 is similar, which we state without

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.

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, $S^*[\alpha/(1+\alpha)] \subset \mathcal{BS}(\alpha)$ .