On Starlike Functions Associated with a Bean Shaped Domain

Theorem 2.1. $\mathfrak{B}(z)$ is convex for $|z| \le r$ , where $r^ \simeq 0.7074$ is the smallest positive root of $1 + 2e^{2r} + e^{4r} - (-1 + e^{2r} + 2e^{4r})r = 0$ .

Lemma 2.2. For |z|=1, the following sharp bounds hold for $\mathfrak{B}(z)$ : (i) If we define $$R(\theta) = \frac{e^{\cos \theta} \sqrt{M_{\theta} + \sqrt{M_{\theta}^2 + N_{\theta}^2}}}{\sqrt{M_{\theta}^2 + N_{\theta}^2}}$$ , then $\sqrt{\frac{2}{1 + e^2}} \le \text{Re}(\mathfrak{B}(z)) \le R(\theta_0)$ , where $R(\theta_0) \simeq 1.38846$ at $\theta_0 \simeq 1.15197$ . (ii) If we define $$I(\theta) = \frac{e^{\cos\theta}N_{\theta}}{\sqrt{M_{\theta}^2 + N_{\theta}^2}\sqrt{M_{\theta} + \sqrt{M_{\theta}^2 + N_{\theta}^2}}}$$ , then $|\text{Im}(\mathfrak{B}(z))| \leq I(\theta_1)$ , where $I(\theta_1) \simeq 0.69949$ at $\theta_1 \simeq 1.72466$ . $$0.69949 \text{ at } \theta_{1} \simeq 1.72466.$$ (iii) If we define $A(\theta) = \frac{2e^{2\cos\theta}}{\sqrt{M_{\theta}^{2} + N_{\theta}^{2}}}$ , then $\sqrt{\frac{2}{1+e^{2}}} \leq |\mathfrak{B}(z)| \leq \sqrt{A(\theta_{2})}$ , where $A(\theta_{2}) \simeq 2.0694$ at $\theta_{2} \simeq 1.31364$ . (iv) $$|\arg(\mathfrak{B}(z))| \le \frac{\beta \pi}{2}$$ , where $\beta = 0.43849139$ . This lemma provides a comprehensive set of bounds for the real and imaginary parts, modulus, and argument of $\mathfrak{B}(z)$ when |z|=1. These sharp bounds are expressed in terms of trigonometric and exponential functions and offer valuable insights into the behavior of $\mathfrak{B}(z)$ on the unit circle.  FIGURE 1. Graph indicating the sharpness of various bounds (in context of Lemma 2.2) associated with $\mathfrak{B}(\mathbb{D})$ . <span id="page-3-1"></span>Remark 2.3. It is evident that within each circle |z|=r<1, the minimum value of both Re $\mathfrak{B}(z)$ and $|\mathfrak{B}(z)|$ occurs precisely at the point z=-r. Now using results in [10], we have the following basic results for functions in $\mathcal{S}_{\mathfrak{B}}^*$ :

Theorem 2.4. Let $f \in \mathcal{S}_{\mathfrak{B}}^*$ and $f_0$ be as defined in (1.3). For any |z| = r < 1, the following fundamental results hold: - $\begin{array}{l} (i) \ \frac{f(z)}{z} \prec \frac{f_0(z)}{z} \ and \ \frac{zf'(z)}{f(z)} \prec \frac{zf_0'(z)}{f_0(z)} \ (Subordination \ Theorem). \\ (ii) \ -f_0(-r) \leq |f(z)| \leq f_0(r) \ (Growth \ Theorem). \\ (iii) \ |\arg\left(\frac{f(z)}{z}\right)| \leq \max_{|z|=r} \arg\left(\frac{f_0(z)}{z}\right) \ (Rotation \ Theorem). \\ (iv) \ Either \ f \ is \ a \ rotation \ of \ f_0 \ or \ f(\mathbb{D}) \supset \{w \in \mathbb{C} : |w| \leq -f_0(-1)\} \ (Covering \ Theorem). \\ \end{array}$ Equality holds for some non-zero z in (ii) and (iii) if and only if f is a rotation of $f_0$ . In their work [10], Ma-Minda established the distortion theorem for the class $S^*(\psi)$ , subject to the condition: <span id="page-3-0"></span> $$\psi(-r) = \min_{|z|=r} |\psi(z)| \le |\psi(z)| \le \max_{|z|=r} |\psi(z)| = \psi(r). \tag{2.2}$$ However, it is worth noting that the function $\mathfrak{B}$ does not satisfy condition (2.2) since $\max_{|z|=r} |\mathfrak{B}(z)| \neq \mathfrak{B}(r)$ . Recently, Gangania and Kumar [1] modified the distortion theorem by relaxing the condition (2.2), leading to a more general result, as follows:

Theorem 2.5. [1] Let $\psi$ be a Ma-Minda function. Suppose $\min_{|z|=r} |\psi(z)| = |\psi(z_1)|$ and $\max_{|z|=r} |\psi(z)| = |\psi(z_2)|$ , where $z_1 = re^{i\theta_1}$ and $z_2 = re^{i\theta_2}$ for some $\theta_1, \theta_2 \in [0, \pi]$ . For $f \in \mathcal{S}^*(\psi)$ and $|z_0| = r < 1$ , the following inequality holds: $$|\psi(z_1)| \left(\frac{-f_0(-r)}{r}\right) \le |f'(z_0)| \le \left(\frac{f_0(r)}{r}\right) |\psi(z_2)|.$$ To establish the distortion theorem for functions in $\mathcal{S}_{\mathfrak{B}}^*$ , we consider the expression for $|\mathfrak{B}(re^{i\theta})|^2$ , denoted as $g_r(\theta)$ : $$|\mathfrak{B}(re^{i\theta})|^2 = \frac{2e^{2r\cos\theta}}{\sqrt{1 + e^{4r\cos\theta} + 2e^{2r\cos\theta}\cos(2r\sin\theta)}} =: g_r(\theta).$$ Next, we calculate the derivative of $g_r(\theta)$ , $$g_r'(\theta) = -\frac{4e^{2r\cos\theta}r(\sin\theta + e^{2r\cos\theta}\cos(2r\sin\theta)\sin\theta - e^{2r\cos\theta}\sin(2r\sin\theta)\cos\theta)}{(1 + e^{4r\cos\theta} + 2e^{2r\cos\theta}\cos(2r\sin\theta))^{3/2}}.$$ By simple observation, we can see that, the critical points of $g_r(\theta)$ are the zeroes of $$h_r(\theta) := \sin \theta + e^{2r\cos \theta} \cos(2r\sin \theta) \sin \theta - e^{2r\cos \theta} \sin(2r\sin \theta) \cos \theta.$$ Clearly, $\theta=0$ and $\pi$ are zeroes of $h_r(\theta)$ for all r. Remark 2.3 confirms that $|\mathfrak{B}(z)|$ attains its minimum at $\theta=\pi$ for |z|=r. However, the maximum is not necessarily attained at $\theta=0$ , as $h_r(\theta)$ possesses roots other than 0 and $\pi$ for $r>r_0\simeq 0.639$ . Consequently, we can now state the distortion theorem for functions in $\mathcal{S}_{\mathfrak{B}}^*$ as follows:

Theorem 2.6. (Distortion Theorem) Let $\max_{|z|=r} |\mathfrak{B}(z)| = |\mathfrak{B}(z_0)|$ , where $z_0 = re^{i\theta_0}$ for some $\theta_0 \in [0, \pi]$ . For $f \in \mathcal{S}_{\mathfrak{B}}^*$ and $f_0$ as defined in (1.3), the following inequalities hold for |z| = r < 1: $$f'_0(-r) \le |f'(z)| \le \frac{|\mathfrak{B}(re^{i\theta_0})|f_0(r)|}{r} \quad \text{for} \quad r > r_0 \simeq 0.639,$$ $f'_0(-r) \le |f'(z)| \le f'_0(r) \quad \text{for} \quad r \le r_0.$ The following lemma aims to find the maximal disk centered at the sliding point (a,0) on the real line, that can fit within $\mathfrak{B}(\mathbb{D})$ .

Lemma 2.7. Let $\sqrt{2/(1+e^2)} < \alpha < e\sqrt{2/(1+e^2)}$ . Then we have the inclusion: <span id="page-4-3"></span> $$\{w: |w-\alpha| < r_{\alpha}\} \subset \mathfrak{B}(\mathbb{D}),$$ (2.3) where $r_{\alpha}$ is defined as follows: <span id="page-4-2"></span> $$r_{\alpha} = \begin{cases} \alpha - \sqrt{\frac{2}{(1+e^2)}}, & \sqrt{\frac{2}{(1+e^2)}} < \alpha \le \frac{1+e}{\sqrt{2(1+e^2)}} \\ e\sqrt{\frac{2}{(1+e^2)}} - \alpha, & \frac{1+e}{\sqrt{2(1+e^2)}} \le \alpha < e\sqrt{\frac{2}{(1+e^2)}}. \end{cases}$$ (2.4)

Theorem 2.10. Let p(z) := (1 + Az)/(1 + Bz), where $-1 < B < A \le 1$ , then $p(z) < \mathfrak{B}(z)$ if and only if <span id="page-5-0"></span> $$A \le \begin{cases} 1 - \sqrt{\frac{2}{1+e^2}} (1-B) & \text{if } \frac{1-AB}{1-B^2} \le \frac{1+e}{\sqrt{2(1+e^2)}} \\ e\sqrt{\frac{2}{1+e^2}} (1+B) - 1 & \text{if } \frac{1-AB}{1-B^2} \ge \frac{1+e}{\sqrt{2(1+e^2)}}. \end{cases}$$ (2.6)

Theorem 2.12. Let $f \in \mathcal{S}_{\mathfrak{B}}$ , then $f \in \mathcal{S}_e$ , which is $\mathcal{S}_e^* := \{ f \in \mathcal{A}; zf'(z)/f(z) \prec e^z \}$ .

Theorem 2.13. The class $\mathcal{S}_{\mathfrak{B}}^*$ satisfies the following inclusion relations: - (i) $\mathcal{S}_{\mathfrak{B}}^ \subset \mathcal{S}^(\alpha) \subset \mathcal{S}$ , whenever $\alpha = \sqrt{2/(1+e^2)}$ . (ii) $\mathcal{S}_{\mathfrak{B}}^ \subset \mathcal{R}\mathcal{S}^(1/\beta) \subset \mathcal{M}(\beta)$ , whenever $\beta \geq R(\theta_0)$ (cf. Lemma 2.2). (iii) $\mathcal{S}_{\mathfrak{B}}^ \subset \mathcal{S}\mathcal{S}^(\beta) \subset \mathcal{S}$ for $\beta_0 \leq \beta < 1$ , where $\beta_0 = 0.438491$ (cf. Lemma 2.2). - (iv) $\mathcal{S}_{\mathfrak{B}}^{\mathfrak{F}} \subset \mathcal{SP}(\rho)$ , whenever $\rho \geq 0.13186$ . (v) $$k - \mathcal{ST} \subset \mathcal{S}_{\mathfrak{B}}^*$$ , whenever $k \geq 2e/(2e - \sqrt{2(1+e^2)})$ .

Theorem 2.14. Let $f \in S^(q_c)$ , $c \in (0,1]$ , then $f \in S^_{\mathfrak{B}}$ if and only if $c < (e^2 - 1)/(e^2 + 1)$ .

Lemma 3.1. [14, Theorem 3.4h,p.132] Let g(z) be univalent in $\mathbb{D}$ , $\Phi$ and $\Theta$ be analytic in a domain $\Omega$ containing $g(\mathbb{D})$ such that $\Phi(w) \neq 0$ , when $w \in g(\mathbb{D})$ . Now letting $Q(z) = zg'(z) \cdot \Phi(g(z))$ , $h(z) = \Theta(g(z)) + Q(z)$ and either h is convex or G is starlike. In addition, if $$\operatorname{Re} \frac{zh'(z)}{Q(z)} = \operatorname{Re} \left( \frac{\Theta'(g(z))}{\Phi(g(z))} + \frac{zQ'(z)}{Q(z)} \right) > 0$$ and let p be analytic in $\mathbb{D}$ , with p(0) = q(0), $p(\mathbb{D}) \subset \Omega$ and $$\Theta(p(z)) + zp'(z) \cdot \Phi(p(z)) \prec \Theta(g(z)) + zg'(z) \cdot \Phi(g(z)) =: h(z)$$ then $p \prec q$ , and g is the best dominant. Now we have the following lemma which is needed to prove our results:

Lemma 3.2. Let $R_0 = e\sqrt{2/(1+e^2)} \simeq 1.32725$ . Then $$\left|\log\left(\frac{w^2}{2-w^2}\right)\right| \ge 2$$ if and only if $|w| \ge R_0$ .

Theorem 3.3. Let $\alpha \in [0,1], \ \gamma \in (0,1], \ k \in \{-1,0\}$ and $\beta \in \mathbb{C} \setminus \{0\}$ with $\operatorname{Re} \beta > 0$ . Suppose $p \in \mathcal{H}$ satisfies $$(1 - \alpha)p(z) + \alpha p^{2}(z) + \beta \frac{zp'(z)}{p^{k}(z)} \prec \sqrt{1 + \tanh z},$$ then $$p(z) \prec \left(\frac{1+Az}{1+Bz}\right)^{\gamma} \qquad -1 < B < A \le 1,$$ whenever <span id="page-9-0"></span> $$|\beta| \ge \frac{1}{\gamma(A-B)} \left( R_0 + \alpha \left( \frac{1+A}{1+B} \right)^{2\gamma} + (1-\alpha) \left( \frac{1+A}{1+B} \right)^{\gamma} \right) \frac{(1+A)^{\gamma(k-1)+1}}{(1+B)^{\gamma(k-1)-1}}, \tag{3.1}$$ where $R_0 = e\sqrt{2/(1+e^2)} \simeq 1.32725$ . Proof. Let $q(z) = ((1 + Az)/(1 + Bz))^{\gamma}$ . Taking $\Theta(w) = (1 - \alpha)w + \alpha w^2$ and $\Phi(w) = \beta/w^k$ in Lemma 3.1, the function $Q: \mathbb{D} \to \mathbb{C}$ becomes $$Q(z) = \frac{\beta z q'(z)}{q^k(z)} = \frac{\beta \gamma (A - B) z (1 + Bz)^{\gamma(k-1)-1}}{(1 + Az)^{\gamma(k-1)+1}}.$$ Since, $$\frac{zQ'(z)}{Q(z)} = 1 + (\gamma(k-1) - 1)\frac{Bz}{1 + Bz} - (\gamma(k-1) + 1)\frac{Az}{1 + Az}$$ and after a computation we obtain. $$\operatorname{Re} \frac{zQ'(z)}{Q(z)} \ge \frac{1 - AB + \gamma(A - B)(1 - k)}{(1 - A)(1 - B)} \ge 0,$$ therefore Q(z) is starlike for k = -1, 0, 1. Further, for the function h defined by $$h(z) := \Theta(q(z)) + Q(z) = (1 - \alpha)q(z) + \alpha q^{2}(z) + Q(z),$$ we have $$\frac{zh'(z)}{Q(z)} = \frac{(1-\alpha)q^k(z)}{\beta} + \frac{2\alpha}{\beta}q^{k+1}(z) + \frac{zQ'(z)}{Q(z)}.$$ Since Re $q^k(z) > 0$ for $k = \{-1, 0, 1\}$ , thus Re zh'(z)/Q(z) > 0 for $k = \{-1, 0\}$ . Hence, by Lemma 3.1, we have $p(z) \prec q(z)$ , whenever $$(1 - \alpha)p(z) + \alpha p^2(z) + \beta \frac{zp'(z)}{p^k(z)} \prec (1 - \alpha)q(z) + \alpha q^2(z) + \beta \frac{zq'(z)}{q^k(z)}.$$ Thus, it is enough to show that now $$\sqrt{1 + \tanh z} \prec (1 - \alpha)q(z) + \alpha q^2(z) + \beta \frac{zq'(z)}{q^k(z)} = h(z).$$ Let $w = \mathfrak{B}(z) = \sqrt{1 + \tanh(z)}$ . Then $\mathfrak{B}^{-1}(w) = 1/2 \log(w^2/(2 - w^2))$ . Therefore the subordination $\mathfrak{B}(z) \prec h(z)$ is equivalent to $z \prec \mathfrak{B}^{-1}(h(z))$ . Thus we only need to show $|\mathfrak{B}^{-1}(h(e^{i\theta}))| \geq 1$ where $0 \leq \theta \leq 2\pi$ which is true if and only if $$\left| \log \left( \frac{\left( (1 - \alpha) \left( \frac{1 + Ae^{i\theta}}{1 + Be^{i\theta}} \right)^{\gamma} + \alpha \left( \frac{1 + Ae^{i\theta}}{1 + Be^{i\theta}} \right)^{2\gamma} + \frac{\beta \gamma (A - B)e^{i\theta} (1 + Be^{i\theta})^{\gamma(k-1)-1}}{(1 + Ae^{i\theta})^{\gamma(k-1)+1}} \right)^{2}}{2 - \left( (1 - \alpha) \left( \frac{1 + Ae^{i\theta}}{1 + Be^{i\theta}} \right)^{\gamma} + \alpha \left( \frac{1 + Ae^{i\theta}}{1 + Be^{i\theta}} \right)^{2\gamma} + \frac{\beta \gamma (A - B)e^{i\theta} (1 + Be^{i\theta})^{\gamma(k-1)+1}}{(1 + Ae^{i\theta})^{\gamma(k-1)+1}} \right)^{2}} \right) \right| \ge 2.$$ By Lemma 3.2, it follows that the above inequality holds whenever $$\left| (1-\alpha) \left( \frac{1+Ae^{i\theta}}{1+Be^{i\theta}} \right)^{\gamma} + \alpha \left( \frac{1+Ae^{i\theta}}{1+Be^{i\theta}} \right)^{2\gamma} + \frac{\beta \gamma (A-B)e^{i\theta} (1+Be^{i\theta})^{\gamma(k-1)-1}}{(1+Ae^{i\theta})^{\gamma(k-1)+1}} \right| \ge R_0,$$ which is true whenever conditions in (3.1) holds. Note that the theorem stated above, Theorem 3.3, is applicable for k=1 whenever $\gamma \in (0,1/2]$ . Specifically, we derive the following extended result from Theorem 3.3 when A=1, B=0, and $\gamma=1/2$ .

Theorem 3.4. Let $\alpha \in [0,1]$ , $k \in \{-1,0,1\}$ and $\beta \in \mathbb{C} \setminus \{0\}$ with $\operatorname{Re} \beta > 0$ . Suppose $p \in \mathcal{H}$ satisfies $$(1 - \alpha)p(z) + \alpha p^{2}(z) + \beta \frac{zp'(z)}{p^{k}(z)} \prec \sqrt{1 + \tanh z},$$ then $p(z) \prec \sqrt{1+z}$ , whenever $$|\beta| \ge (R_0 + 2\alpha + (1 - \alpha)\sqrt{2})2^{(k+3)/2},$$ where $R_0 = e\sqrt{2/(1+e^2)} \simeq 1.32725$ . Since the proof is much akin to the Theorem 3.3, it is omitted here.

Theorem 3.5. Let $\delta \in \{0,1\}$ , $k \in \{-1,0,1\}$ , $\gamma \in (0,1]$ and $\beta \in \mathbb{C} \setminus \{0\}$ with $\operatorname{Re} \beta > 0$ . Suppose $p \in \mathcal{H}$ satisfies $$(p(z))^{\delta} + \beta \frac{zp'(z)}{(p(z))^k} \prec \sqrt{1 + \tanh(z)},$$ then $$p(z) \prec \left(\frac{1+Az}{1+Bz}\right)^{\gamma} \qquad -1 < B < A \le 1,$$ whenever <span id="page-10-0"></span> $$|\beta| \ge \frac{1}{\gamma(A-B)} \left( R_0 + \left( \frac{1+A}{1+B} \right)^{\delta \gamma} \right) \frac{(1+A)^{\gamma(k-1)+1}}{(1+B)^{\gamma(k-1)-1}},$$ (3.2) where $R_0 = e\sqrt{2/(1+e^2)} \simeq 1.32725$ .

Theorem 4.1. The sharp $$S_{\mathfrak{B}}^*$$ -radius for the class $S^[A, B]$ is given by (i) $R_{S_{\mathfrak{B}}}(S^*[A, B]) = \min \left\{1; \frac{\sqrt{2}e - \sqrt{1+e^2}}{A\sqrt{1+e^2} - \sqrt{2}eB}\right\} =: R_1$ , whenever $-1 \le B < 0 < A \le 1$ . (ii) $$R_{\mathcal{S}_{\mathfrak{B}}}(\mathcal{S}^[A, B]) = \begin{cases} R_1 & R_1 \leq R_0 \\ R_2 & R_1 > R_0, \end{cases}$$ whenever $0 \leq B < A \leq 1$ , where $$R_0 := \sqrt{\frac{\sqrt{2(1+e^2)} - e - 1}{B(\sqrt{2(1+e^2)}A - Be - B)}} \quad and \quad R_2 := \min\left\{1, \frac{\sqrt{1+e^2} - \sqrt{2}}{A\sqrt{1+e^2} - \sqrt{2}B}\right\}.$$

Theorem 4.4. The sharp $S_{\mathfrak{B}}$ -radius for the classes $S_e$ , $S_{SG}$ , $S_{\mathcal{J}}$ and $S_{\rho}^*$ are given by - (i) $R_{\mathcal{S}_{\infty}}(\mathcal{S}_e^) = r_e = 1 + \log(\sqrt{2/(1+e^2)}) \simeq 0.28311.$ - (ii) $R_{S_{\mathfrak{R}}}(S_{SG}^) = r_{SG} = -\log(e^{-1}\sqrt{2(1+e^2)} 1) \simeq 0.679492.$ - (iii) $R_{\mathcal{S}_{\mathfrak{B}}^{}}(\mathcal{S}_{\mathcal{Q}}^{}) = r_{\mathcal{Q}} = (e^{2} 1)/(2e\sqrt{2(e^{2} + 1)}) \simeq 0.2869.$ - (iv) $R_{\mathcal{S}_{\mathfrak{B}}^{}}(\mathcal{S}_{\varrho}^{}) = r_{\varrho} \simeq 0.253877$ , where $r_{\varrho}$ is the smallest positive root of the equation $$\sqrt{1+e^2}(1+re^r) = \sqrt{2}e.$$

Theorem 4.5. The sharp $\mathcal{S}_{\mathfrak{B}}$ -radius for the classes $\mathcal{S}_L^(\alpha)$ and $\mathcal{S}_e^(\alpha)$ , are given by (i) $R_{\mathcal{S}_{\mathfrak{B}}}(\mathcal{S}_L^(\alpha)) = r_L(\alpha) := 1 ((\sqrt{2/(1+e^2)} \alpha)/(1-\alpha))^2$ , where $\alpha \in [0, 1-e/\sqrt{2(1+e^2)})$ . (ii) $R_{\mathcal{S}_{\mathfrak{B}}}(\mathcal{S}_e^*(\alpha)) = r_e(\alpha) := \log\left((e\sqrt{2/(1+e^2)} \alpha)/(1-\alpha)\right)$ , where $\alpha \in [0, 1)$ .

Theorem 4.6. The sharp $\mathcal{S}_{\mathfrak{B}}$ -radius for the classes $\mathcal{BS}^(\alpha)$ , $\mathcal{S}_{cs}^*(\alpha)$ and $\mathcal{ST}_L(s)$ are given by - (i) $R_{\mathcal{S}_{\mathfrak{B}}}(\mathcal{BS}^(\alpha)) = r_{BS}(\alpha) = 1 \sqrt{1 + 4\alpha(1 e\sqrt{2/(1 + e^2)})^2/(2\alpha(1 e\sqrt{2/(1 + e^2)}))}$ , where $\alpha \in [0, 1)$ . - (ii) $R_{\mathcal{S}_{\infty}}(\mathcal{S}_{cs}^(\alpha)) = r_{cs}(\alpha)$ , where $r_{cs}(\alpha)$ is the smallest positive root of $$\frac{(1+e^2)(1+\alpha r - \alpha r^2)}{(1-r)(1+\alpha r)} - e\sqrt{2(1+e^2)} = 0,$$ where $\alpha \in [0,1)$ . (iii) $R_{\mathcal{S}_{\mathfrak{B}}}(\mathcal{ST}_L(s)) = \min\{1, r^\}$ , where $r^*$ is the smallest positive root of $$(1+e^2)(1+sr)^2 - e\sqrt{2(1+e^2)} = 0,$$ where $|s| \le 1 \setminus \{0\}$ .

Theorem 4.7. If $f \in \mathcal{S}_{\mathfrak{B}}^*$ . Then (i) $f \in \mathcal{C}(\alpha)$ in $|z| < r_{\alpha}$ , where $r_{\alpha}$ is the smallest positive root of the equation <span id="page-15-1"></span> $$1 - \tanh r = \left(\alpha + \frac{r(1 + \tan r)}{2(1 - r^2)}\right)^2, \quad \alpha \in (0, 1]. \tag{4.2}$$ (ii) $f \in \mathcal{S}^*(\alpha)$ in $|z| < (\log((2 - \alpha^2)/\alpha^2))/2$ , where $\alpha \in (\sqrt{2/(1 + e^2)}, 1)$ .

Definition 1.1. Let $\mathcal{S}_{\mathfrak{B}}^*$ be the class of normalized starlike functions, defined as follows: $$\mathcal{S}_{\mathfrak{B}}^* := \left\{ f \in \mathcal{S} : \frac{zf'(z)}{f(z)} \prec \mathfrak{B}(z) := \sqrt{1 + \tanh z} \right\}.$$ Note that $\mathfrak{B}(z) := \sqrt{1 + \tanh z} = \sqrt{2/(1 + e^{-2z})}$ and $\mathfrak{B}(z)$ conformally maps $\mathbb{D}$ onto the region $$\Omega_{\mathfrak{B}} := \left\{ w \in \mathbb{C} : \left| \log \left( \frac{w^2}{2 - w^2} \right) \right| < 2 \right\}.$$ The functions in the classes $\mathcal{S}_{\mathfrak{B}}$ can be represented through an integral formula as follows: A function $f \in \mathcal{S}_{\mathfrak{B}}$ if and only if there exists an analytic function $p(z) \prec \mathfrak{B}(z)$ such that <span id="page-1-0"></span> $$f(z) = z \exp \int_0^z \frac{p(t) - 1}{t} dt.$$ (1.2) By taking $p(z) = \mathfrak{B}(z)$ in (1.2), we obtain an extremal function, defined as: <span id="page-1-1"></span> $$f_0(z) = z + \frac{z^2}{2} + \frac{z^3}{16} - \frac{13z^4}{288} - \frac{11z^5}{1152} + \dots \in \mathcal{S}_{\mathfrak{B}}^*.$$ (1.3) The class $\mathcal{S}_{\mathfrak{B}}$ and its integral representation (1.2) offer valuable insights into the properties and behavior of starlike functions associated with the intriguing bean-shaped domain. The extremal function $f_0(z)$ represents a notable example within this class, paving the way for further exploration and analysis. The present investigation explores the properties of $\mathfrak{B}(z)$ and the class $\mathcal{S}_{\mathfrak{B}}$ . It establishes inclusion relations between $\mathcal{S}_{\mathfrak{B}}^*$ and well-known classes, accompanied by diagrammatic representations. The study also derives sharp conditions under which $\psi(p) \prec \sqrt{1 + \tanh(z)}$ implies $p(z) \prec ((1 + Az)/(1 + Bz))^{\gamma}$ , where $\psi(p)$ is defined as: $$(1-\alpha)p(z) + \alpha p^2(z) + \beta \frac{zp'(z)}{p^k(z)}$$ and $(p(z))^{\delta} + \beta \frac{zp'(z)}{(p(z))^k}$ . Additionally, find sharp $\mathcal{S}_{\mathfrak{B}}^*$ -radii estimates for certain geometrically defined function classes available in the literature.