On Geometric properties and Coefficient bounds for starlike functions associated with balloon-shaped domain

Theorem 2.4. The radius of convexity for the function B(z) is given by $$r_c = 1 - 1/e \approx 0.63212.$$

Lemma 2.5. (Function Bounds) Let $B_R(\theta)$ and $B_I(\theta)$ denote the real and imaginary part of the function $B(e^{i\theta})$ respectively and given as $$B_R(\theta) = \frac{1 - \log(2\cos\theta/2)}{(1 - \log(2\cos\theta/2))^2 + (\theta/2)^2}, \quad and \quad B_I(\theta) = \frac{\theta/2}{(1 - \log(2\cos\theta/2))^2 + (\theta/2)^2}. \quad (2.4)$$ Then we have - (i) $0 < \Re(B(z)) < 1/(1 \log 2) \approx 3.25889$ at $\theta = 0$ and $\pi$ , we get maximum and minimum of $B_R(\theta)$ respectively. - (ii) $|\Im(B(z))| \le B_I(\theta_0) \approx 1.41379$ at $\theta_0 = 0.524085$ . (iii) $|\arg(B(z))| \le \left|\arctan\left(\frac{\theta_0/2}{1-\log(2\cos\theta_0/2)}\right)\right| \approx 0.88329$ at $\theta_0 = 1.37501$ . (iv) $|B(z)| \le \left|\frac{1}{1-\log(1+r)}\right|$ , where |z| = r < 1. $$(iv) |B(z)| \le \left| \frac{1}{1 - \log(1+r)} \right|, where |z| = r < 1$$ These bounds are best possible. We now study the geometric extent of the image domain $B(\mathbb{D})$ via disk containment. In particular, we determine the largest disk contained in $B(\mathbb{D})$ and the smallest disk containing $B(\mathbb{D})$ .

Theorem 2.6. For $B(z) = 1/(1 - \log(1+z))$ , then we have (1) $\mathbb{D}_L := \{w : |w - a| < r_a\} \subset B(\mathbb{D}), where$ $$r_a = \begin{cases} \sqrt{d(\theta_a)}, & 0 < a < \frac{2 - \log 2}{(1 - \log 2)(3 - \log 2)}, \\ \frac{1}{1 - \log 2} - a, & \frac{2 - \log 2}{(1 - \log 2)(3 - \log 2)} \le a < \frac{1}{1 - \log 2}. \end{cases}$$ and $\theta_a \in (0,\pi)$ , is the root of the given equation <span id="page-5-0"></span> $$\sin B(a(A^2 - B^2) - A) + B\cos B(2aA - 1) = 0, (2.5)$$ with $A := 1/(1 - \log(2\cos(\theta/2)))$ , $B := \theta/2$ . (2) $B(\mathbb{D}) \subset \{w : |w-a| < R_a\} =: \mathbb{D}_S, \text{ where }$ $$R_a = \begin{cases} \frac{1}{1 - \log 2} - a, & 0 < a < \frac{1}{2(1 - \log 2)}, \\ a, & \frac{1}{2(1 - \log 2)} \le a < \frac{1}{1 - \log 2}. \end{cases}$$ Proof. Let $B(e^{i\theta}) = B_R(\theta) + iB_I(\theta)$ , where $B_R(\theta)$ and $B_I(\theta)$ are defined in (2.4), represents the boundary of $B(\mathbb{D})$ and is symmetric about real axis. Take $d(\theta)$ denote the square of the distance of (a,0) from the points on the curve $B(e^{i\theta})$ is given by $$d(\theta) := a^2 + \frac{1 - 2aA}{A^2 + B^2}.$$ where A and B are given in (2.5). (1) We'll start with $\{w: |w-a| < r_a\} \subset B(\mathbb{D})$ . For $0 < a < \frac{2-\log 2}{(1-\log 2)(3-\log 2)}$ , $$d'(\theta) = \frac{\sin B(a(A^2 - B^2) - A) + B\cos B(2aA - 1)}{\cos B(A^2 + B^2)^2} = 0,$$ A mathematical computation shows that $d'(\theta)$ changes sign exactly at $\theta_a$ , where $\theta_a$ is the root of the equation $$\sin B(a(A^2 - B^2) - A) + B\cos B(2aA - 1) = 0.$$ Since $d'(\theta) < 0$ for $\theta \in (0, \theta_a)$ and $d'(\theta) > 0$ for $\theta \in (\theta_a, \pi)$ , we get $d(\theta)$ is decreasing in $(0, \theta_a)$ and increasing in $(\theta_a, \pi)$ . For the largest disk inside $B(\mathbb{D})$ , we need $r_a = \min_{\theta \in [0, \pi]} \sqrt{d(\theta)}$ . Therefore $$\min_{\theta \in [0,\pi]} d(\theta) = d(\theta_a) \implies r_a = \sqrt{d(\theta_a)}.$$ For $\frac{2-\log 2}{(1-\log 2)(3-\log 2)} \le a < \frac{1}{1-\log 2}$ , $d'(\theta) \ge 0$ for all $\theta \in (0,\pi)$ , hence $d(\theta)$ is monotonic increasing. Therefore, we get $$\min_{\theta \in [0,\pi]} d(\theta) = d(0) = \left(\frac{1}{1 - \log 2} - a\right)^2 \implies r_a = \frac{1}{1 - \log 2} - a.$$ (2) For $B(\mathbb{D}) \subset \{w : |w-a| < R_a\}$ , we need to find the smallest disk that contain $B(\mathbb{D})$ , we need $R_a = \sup_{\theta \in [0,\pi]} \sqrt{d(\theta)}$ . For $0 < a < \frac{1}{2(1-\log 2)}$ , the supremum of $d(\theta)$ over $[0,\pi]$ is $$\sup_{\theta \in [0,\pi]} d(\theta) = d(0) = \left(\frac{1}{1 - \log 2} - a\right)^2 \implies R_a = \frac{1}{1 - \log 2} - a.$$ For $\frac{1}{2(1-\log 2)} \le a < \frac{1}{1-\log 2}$ , $d(\theta) \le a^2$ for all $\theta \in (0,\pi)$ . Hence $R_a = a$ . This completes the proof. Remark 2.7. The largest disk $\mathbb{D}_L := \{w : |w - a_1| < r_{a_1}\}$ contained in $B(\mathbb{D})$ is obtained when $a_1 \approx 1.90$ and $r_{a_1} = 1/(1 - \log 2) - 1.90 \approx 1.3589$ , such that $\mathbb{D}_L \subset B(\mathbb{D})$ . And the smallest disk $\mathbb{D}_S := \{w : |w - a_2| < R_{a_2}\}$ containing $B(\mathbb{D})$ , is obtained when $a_2 \approx 1.6295$ and $R_{a_2} \approx 1.6295$ , such that $B(\mathbb{D}) \subset \mathbb{D}_S$ as shown in Figure 1.  <span id="page-6-0"></span>FIGURE 1. Visualization of the image $B(\mathbb{D})$ under the mapping $B(z) = 1/(1 - \log(1+z))$ , together with the largest inscribed disk and the smallest disk containing $B(\mathbb{D})$ . The parameters $a_1$ and $a_2$ denote the centers of the respective disks on the real axis.

Theorem 2.8. Let $f \in \mathcal{S}_B^*$ , then <span id="page-6-1"></span> $$|a_2^2 - a_3| \le \frac{1}{2}.$$ This inequality is sharp.

Theorem 2.9. Let $f \in \mathcal{S}_B^*$ , then <span id="page-7-2"></span> $$|a_3^2 - a_5| \le \frac{1}{4}.$$ This inequality is sharp.

Theorem 2.10. Let $f \in \mathcal{S}_B^*$ , then <span id="page-8-0"></span> $$|H_{3,1}(f)| \le \frac{1}{9}.$$ This inequality is sharp.

Theorem 2.11. Let $f \in \mathcal{S}_B^*$ , then $$|a_4 - a_2^3| \le \frac{17}{36}.$$ This inequality is sharp.

Theorem 2.12. Let $f \in \mathcal{S}_B^*$ , then <span id="page-14-0"></span> $$|a_5 - a_2^4| \le \frac{187}{288}.$$ This inequality is sharp.

Theorem 2.13. Let $f \in \mathcal{S}_B^*$ . Then <span id="page-15-0"></span> $$|T_{3,1}(f)| \le 1.$$ The above estimate is sharp.

Theorem 2.14. Let $f \in \mathcal{S}_B^*$ , then $$-\frac{1}{16} \le H_{3,1}^T(f) \le 1.$$ This inequality is sharp.