Coefficient problems of Starlike Functions Related to a Balloon-Shaped Domain

Lemma 2.1. [7]: If $p \in \mathcal{P}$ is of the form (2.1), then $$p_1 = 2\zeta_1, (2.2)$$ $$p_2 = 2\zeta_1^2 + 2(1 - \zeta_1^2)\zeta_2, (2.3)$$ $$p_3 = 2\zeta_1^3 + 4(1 - \zeta_1^2)\zeta_1\zeta_2 - 2(1 - \zeta_1^2)\zeta_1\zeta_2^2 + 2(1 - \zeta_1^2)(1 - |\zeta_2|^2)\zeta_3, \tag{2.4}$$ for some $\zeta_1, \ \zeta_2, \ \zeta_3 \in \overline{\mathbb{D}}$ . For $\zeta_1 \in \mathbb{T} := \{z \in \mathbb{C} ; |z| = 1\}$ , there is a unique function $p \in \mathcal{P}$ with $p_1$ as in (2.2), namely, $$p(z) = \frac{1 + \zeta_1 z}{1 - \zeta_1 z}, \quad z \in \mathbb{D}$$ (2.5) For $\zeta_1 \in \mathbb{D}$ and $\zeta_2 \in \mathbb{T}$ , there is a unique function $p \in \mathcal{P}$ with $p_1$ and $p_2$ as in (2.2) and (2.3), namely, $$p(z) = \frac{1 + (\overline{\zeta_1} \zeta_2 + \zeta_1)z + \zeta_2 z^2}{1 + (\overline{\zeta_1} \zeta_2 - \zeta_1)z - \zeta_2 z^2}, \quad z \in \mathbb{D}.$$ (2.6)

Lemma 2.2. [8]: If A, B, $C \in \mathbb{R}$ , let us consider $$Y(A, B, C) := \max\{|A + Bz + Cz^2| + 1 - |z|^2, z \in \overline{\mathbb{D}}\}$$ Case 1: If $AC \geq 0$ , then $$Y(A, B, C) = \begin{cases} |A| + |B| + |C|, & |B| \ge 2(1 - |C|), \\ 1 + |A| + \frac{B^2}{4(1 - |C|)}, & |B| < 2(1 - |C|). \end{cases}$$ Case 2: If AC < 0, then $$Y(A,B,C) = \begin{cases} 1 - |A| + \frac{B^2}{4(1-|C|)}, & -4AC(C^{-2}-1) \le B^2 \land |B| < 2(1-|C|), \\ 1 + |A| + \frac{B^2}{4(1+|C|)}, & B^2 < \min\{4(1+|C|)^2, -4AC(C^{-2}-1)\}, \\ R(A,B,C), & Otherwise, \end{cases}$$ where $$R(A,B,C) = \begin{cases} |A| + |B| - |C|, & |C|(|B| + 4|A|) \le |AB|, \\ -|A| + |B| + |C|, & |AB| \le |C|(|B| - 4|A|), \\ (|C| + |A|)\sqrt{1 - \frac{B^2}{4AC}}, & Otherwise. \end{cases}$$

Lemma 2.3. [5, 36] Let $p \in \mathcal{P}$ . Then, the following inequalities hold true $$|p_t| \leq 2, \qquad t \geq 1, |p_{t+2k} - \rho p_t p_k^2| \leq 2(1+2\rho), \quad 0 \leq \rho \leq 1, |p_2 - \frac{p_1^2}{2}| \leq 2 - \frac{|p_1|^2}{2},$$ and $$|c_{n+k} - \mu c_n c_k| \le 2 \max\{1, |2\mu - 1|\} = \begin{cases} 2, & \text{if } 0 \le \mu \le 1, \\ 2|2\mu - 1|, & \text{Otherwise.} \end{cases}$$

Lemma 2.4. [44] Let $w \in \mathcal{H}$ , are said to be schwarz function such that w(0) = 0 and |w(z)| < 1 for all $z \in \mathbb{D}$ , and have the following series: $$w(z) = \sum_{n=1}^{\infty} b_n z^n \tag{2.7}$$ Then, the following inequalities hold true $$|b_1| \le 1,$$ $|b_2| \le 1 - |b_1|^2,$ $|b_3| \le 1 - |b_1|^2 - \frac{|b_2|^2}{1 + |b_1|}.$

Theorem 3.1. Let $f \in \mathcal{S}_B^*$ . Then, the following inequalities hold true $$|a_2| \le 1$$ , $|a_3| \le \frac{3}{4}$ , $|a_4| \le \frac{19}{36}$ , $|a_5| \le \frac{101}{288}$ These inequalities are sharp.

Theorem 3.2. Let $f \in \mathcal{S}_B^*$ . Then for any complex number $\mu \in \mathbb{C}$ , the following inequality holds: $$|a_3 - \mu a_2^2| \le \frac{1}{2} \max\{1, |\mu - \frac{3}{4}|\}.$$ This inequality is sharp.

Theorem 3.3. Let $f \in \mathcal{S}_B^*$ . Then $$|\mathcal{H}_{2,2}(f)| \le \frac{1}{4}.$$ (3.6) This inequality is sharp.

Theorem 3.4. Let $f \in \mathcal{S}_B^*$ . Then, the following inequalities for the logarithmic coefficients are true: $$|\gamma_1| \le \frac{1}{2}, \quad |\gamma_2| \le \frac{1}{4}, \quad |\gamma_3| \le \frac{1}{8}.$$ These bounds are sharp.

Theorem 3.5. Let $f \in \mathcal{S}_B^*$ . Then $$|\mathcal{H}_{2,1}(F_f/2)| \le \frac{1}{16}.$$ (3.12) This inequality is sharp. Proof. Let $f \in \mathcal{S}_B$ . Since the class $\mathcal{S}_B$ is invariant under rotation and $\mathcal{H}_{2,1}(F_f/2)$ is given by (3.11), it follows that $|\mathcal{H}_{2,1}(F_f/2)|$ is also rotationally invariant. Therefore, without loss of generality, we may assume that $a_2 \geq 0$ . Consequently, by (3.2), we have $p_1 \geq 0$ , which, in view of (2.2), implies that $\zeta_1 \in [0,1]$ . Hence, upon substituting (3.2), (3.3), and (3.4) into (3.10), we obtain $$\mathcal{H}_{2,1}(F_f/2) = \left(\frac{1}{2}a_2\right) \left(\frac{1}{2}\left(a_4 - a_2a_3 + \frac{1}{3}a_2^3\right)\right) - \left(\frac{1}{2}\left(a_3 - \frac{1}{2}a_2^2\right)\right)^2$$ $$= \frac{1}{4}\left(a_2a_4 - a_3^2 + \frac{1}{12}a_2^4\right)$$ $$= \frac{1}{9216}\left(7p_1^4 - 24p_1^2p_2 - 144p_2^2 + 192p_1p_3\right)$$ (3.13) Now, applying Lemma 2.1 to (3.13), we get $$\mathcal{H}_{2,1}(F_f/2) = \frac{1}{576} \left( 12\zeta_2^2(\zeta_1^4 + 2\zeta_1^2 - 3) + 12\zeta_1^2\zeta_2(1 - \zeta_1^2) + 7\zeta_1^4 + 48\zeta_1(1 - \zeta_1^2)(1 - |\zeta_2|^2)\zeta_3 \right)$$ (3.14) (1) Since $|\zeta_3| \leq 1$ , from (3.14), we have the following cases for $\zeta_1 = 0$ and $\zeta_1 = 1$ : $$|\mathcal{H}_{2,1}(F_f/2)| = \begin{cases} \frac{|\zeta_2|^2}{16} \le \frac{1}{16}, & \zeta_1 = 0, \\ \frac{7}{576}, & \zeta_1 = 1. \end{cases}$$ (2) When $\zeta_1 \in (0,1)$ , since $|\zeta_3| \leq 1$ , applying the triangle inequality to (3.14) gives $$\begin{aligned} |\mathcal{H}_{2,1}(F_f/2)| &\leq \frac{1}{576} \Big| 12\zeta_2^2(\zeta_1^4 + 2\zeta_1^2 - 3) + 12\zeta_1^2\zeta_2(1 - \zeta_1^2) + \\ &\qquad \qquad 7\zeta_1^4 + 48\zeta_1(1 - \zeta_1^2)(1 - |\zeta_2|^2)\zeta_3 \Big| \\ &= \frac{1}{7}\zeta_1(1 - \zeta_1^2)\Psi(A, B, C) \end{aligned}$$ (3.15) where $$\Psi(A, B, C) := |A + B\zeta_2 + C\zeta_2^2| + 1 - |\zeta_2|^2,$$ and $$A = \frac{7\zeta_1^3}{48(1-\zeta_1^2)}, \quad B = \frac{\zeta_1}{4}, \quad C = -\frac{3+\zeta_1^2}{4\zeta_1}.$$ Since AC < 0, by applying Case 2 of Lemma 2.2, we proceed as follows. We define $$T_2(\zeta_1) := -4AC\left(\frac{1}{C^2} - 1\right) - B^2 = -\frac{\zeta_1^2(18 - \zeta_1^2)}{12(3 + \zeta_1^2)} \le 0,$$ which gives $$-4AC\left(\frac{1}{C^2} - 1\right) \le B^2.$$ A. For each $\zeta_1 \in (0,1)$ $$T_1(\zeta_1) := |B| - 2(1 - |C|) = \frac{3}{2\zeta} + \frac{3\zeta_1}{4} - 2 > 0,$$ implying |B| > 2(1 - |C|). Furthermore, $$T_2(\zeta_1) := -4AC\left(\frac{1}{C^2} - 1\right) - B^2 = -\frac{\zeta_1^2(18 - \zeta_1^2)}{12(3 + \zeta_1^2)} \le 0,$$ which gives $-4AC\left(\frac{1}{C^2}-1\right) \leq B^2$ . Thus, $T_1(\zeta_1) \cap T_2(\zeta_1) = \emptyset$ , and this case does not occur for any $\zeta_1 \in (0,1)$ , as stated in Lemma 2.2. B. For $\zeta_1 \in (0,1)$ , we have $$T_3(\zeta_1) := 4(1+|C|)^2 = \frac{(3+4\zeta_1+\zeta_1^2)^2}{4\zeta_1^2} > 0,$$ $$T_4(\zeta_1) := -4AC\left(\frac{1}{C^2} - 1\right) = -\frac{7\zeta_1^2(9 - \zeta_1^2)}{48(3 + \zeta_1^2)} < 0.$$ Therefore, $\min\{T_3(\zeta_1), T_4(\zeta_1)\} = T_4(\zeta_1)$ . Since $-4AC\left(\frac{1}{C^2} - 1\right) \leq B^2$ , this case is also not valid for any $\zeta_1 \in (0, 1)$ . C. Considering $$T_5(\zeta_1) := |AB| - |C|(|B| + 4|A|) = -\frac{12 + 20\zeta_1^2 + 3\zeta_1^4}{64(1 - \zeta_1^2)} < 0,$$ we get |AB| < |C|(|B| + 4|A|), implying this case is impossible for $\zeta_1 \in (0, 1)$ . D. Finally, define $$T_6(\zeta_1) := |AB| - |C|(|B| - 4|A|) = -\frac{36 - 108\zeta_1^2 - 47\zeta_1^4}{192(1 - \zeta_1^2)} \le 0.$$ which holds for $$0 < \zeta_1 \le \zeta' = \left(\sqrt{\frac{6}{47}(8\sqrt{2} - 9)}\right)$$ . Hence, by Lemma 2.2, $$\Psi(A, B, C) \le |A| + |B| + |C|.$$ Using this in (3.15), we get $$|\mathcal{H}_{2,1}(F_f/2)| \leq \frac{1}{7}\zeta_1(1-\zeta_1^2)(|A|+|B|+|C|)$$ $$= \frac{1}{576}(36-12\zeta_1^2-31\zeta_1^4)$$ $$\leq \frac{1}{16} \approx 0.0625.$$ For $\zeta' < \zeta_1 < 1$ , applying Lemma 2.2 again yields $$|\mathcal{H}_{2,1}(F_f/2)| \leq \frac{1}{12}\zeta_1(1-\zeta_1^2)\left(|C|+|A|\right)\sqrt{1-\frac{B^2}{4AC}}$$ $$= \frac{1}{168}\sqrt{\frac{6+\zeta_1^2}{21+7\zeta_1^2}}\left(5\zeta_1^4+24\zeta_1^2-36\right) := \phi_2(\zeta_1)$$ For $\zeta_1 \in (\zeta', 1)$ , we find that $$\phi_2(\zeta_1) \le 0.0516512 \le \frac{1}{16}$$ at $\zeta_1 = \sqrt{\frac{6}{47}(8\sqrt{2} - 9)}$ . Therefore, it follows that the inequality (3.12) holds. In lemma 2.1, on replacing $p_1 = p_3 = 0$ and $p_2 = 2$ . The corresponding extremal function $f \in \mathcal{S}_B^*$ described as $$\frac{zf'(z)}{f(z)} = \frac{1}{1 - \log(2p(z)/(p(z) + 1))}$$ (3.16) Where p(z) is given in (2.6) with $\zeta_1 = 0$ and $\zeta_2 = 1$ , we get $p(z) = (1+z^2)/(1-z^2)$ . On solving (3.16), we get the function (1.4). We now proceed to establish the bounds for the Second-order Hankel determinants, where the entries are the logarithmic coefficients of the inverse of $f \in \mathcal{S}_B^*$ . Using the equation (1.7) in (1.11), the logarithmic coefficients of the inverse functions are derived as follows: $$\mathcal{H}_{2,1}(F_{f^{-1}}/2) = \begin{vmatrix} \Gamma_1 & \Gamma_2 \\ \Gamma_2 & \Gamma_3 \end{vmatrix} = \Gamma_1 \Gamma_3 - \Gamma_2^2 = \frac{1}{48} \left( 13a_2^4 - 12a_2^2 a_3 - 12a_3^2 + 12a_2 a_4 \right). \tag{3.17}$$ Similarly, we can verify that $|\mathcal{H}_{2,1}(F_{f^{-1}}/2)|$ is also invariant under rotation. Indeed, for the rotated function $f_{\theta}(z) := e^{-i\theta} f(e^{i\theta}z)$ , where $f \in \mathcal{S}$ and $\theta \in \mathbb{R}$ , we obtain $$\mathcal{H}_{2,1}(F_{f_{\theta}^{-1}}/2) = \frac{e^{4i\theta}}{48} \left( 13a_2^4 - 12a_2^2a_3 - 12a_3^2 + 12a_2a_4 \right) = e^{4i\theta}\mathcal{H}_{2,1}(F_{f^{-1}}/2). \tag{3.18}$$ In the following, we obtain the sharp bounds of the second-order Hankel determinant related to the logarithmic coefficients of the inverse function for functions f belonging to the class $\mathcal{S}_B^*$ :

Theorem 3.6. Let $f \in \mathcal{S}_B^*$ . Then $$|\mathcal{H}_{2,1}(F_{f^{-1}}/2)| \le \frac{43}{576}.$$ (3.19) This inequality is sharp.

Theorem 4.1. Let $f \in \mathcal{S}_B^*$ . Then $$|\mathcal{T}_{2,1}(f)| < 2.$$ This inequality is sharp.

Theorem 4.2. Let $f \in \mathcal{S}_B^*$ . Then $$|\mathcal{T}_{2,2}(f)| \le \frac{25}{16}.$$ This inequality is sharp.

Theorem 4.3. Let $f \in \mathcal{S}_B^*$ . Then $$|\mathcal{T}_{2,3}(f)| \le \frac{545}{648}.$$ This inequality is sharp.

Theorem 4.4. Let $f \in \mathcal{S}_B^*$ . Then $$|\mathcal{T}_{2,1}(F_f/2)| \le \frac{17}{64}.$$ This inequality is sharp.

Theorem 4.5. Let $f \in \mathcal{S}_B^*$ , then $$|\mathcal{T}_{2,1}(F_{f^{-1}}/2)| \le \frac{25}{64}.$$ This inequality is sharp.