On a Subclass of Starlike Functions Associated with a Strip Domain

Lemma 2.1. For 0 < r < 1, the function $\tau(z)$ satisfies $$\min_{|z|=r} \operatorname{Re} \tau(z) = \tau(-r) \quad and \quad \max_{|z|=r} \operatorname{Re} \tau(z) = \tau(r).$$

Theorem 2.2. The function $\tau(z) = 1 + \arctan z$ is a convex univalent function.

Theorem 2.9. (A). Let $f \in \mathcal{S}_{\tau}^*$ and $\bar{\tau}(z)$ given by equation (2.4). Then the following results hold: - (1) Subordination results: $\frac{zf'(z)}{f(z)} \prec \frac{z\bar{\tau}'(z)}{\bar{\tau}(z)}$ and $\frac{f(z)}{z} \prec \frac{\bar{\tau}(z)}{z}$ . - (2) Growth theorem: For $|z| = r < 1, -\bar{\tau}(-r) \le |f(z)| \le \bar{\tau}(r)$ . - (3) Covering theorem: Either f is a rotation of $\bar{\tau}$ or $\{w : |w| < -\bar{\tau}(-1)\} \subset f(\mathbb{D})$ , where $-\bar{\tau}(-1) = \lim_{r \to 1} -\bar{\tau}(-r)$ . - (4) Rotation theorem: For |z| = r < 1, $\left| \arg \frac{f(z)}{z} \right| \le \max_{|z|=r} \arg \frac{\bar{\tau}(z)}{z}$ . Equality holds in (2) and (4) for some non-zero z if and only if f(z) is a rotation of $\bar{\tau}(z)$ .

Lemma 2.10. (Disk Containment) Let us consider the disk $D(a,r) := \{w \in \mathbb{C} : |w-a| = r\}$ with real a and $1 \in D(a,r)$ , then we observe that $D(a,r) \in \Omega_{\tau}$ if and only if <span id="page-3-4"></span> $$1 - \frac{\pi}{4} \le a - r < a + r \le 1 + \frac{\pi}{4}. \tag{2.5}$$ Moreover, the disk $D(1,\pi/4)$ is the largest disk contained in the region $\Omega_{\tau}$ , further $D(a,r) \subset D(1,\pi/4)$ if and only if $$|a-1| \le \frac{\pi}{4} - r.$$ It can also be viewed as, <span id="page-3-5"></span> $$1 - \frac{\pi}{4} \le a - r \quad when \quad a \le \frac{\pi}{4} \tag{2.6}$$ and <span id="page-3-6"></span> $$a+r \le 1+\frac{\pi}{4} \quad when \quad a > \frac{\pi}{4}. \tag{2.7}$$  FIGURE 1. The largest disk that can be inscribed inside the domain $\Omega_{\tau}$ is $D(1, \pi/4)$ . The proof of the Lemma 2.10 is skipped here as it directly follows from Theorem 2.2, Theorem 2.5 and Remark 2.3. Corollary 2.11. The disk $\{w : |w-1| \le \arctan(r)\}\$ is contained in $\tau(|z| \le r)$ and is maximal. Next, we present a conclusion in the form of the following theorem.

Theorem 2.12. Let $-1 < B < A \le 1$ . Then, $p(z) := (1 + Az)/(1 + Bz) \in \mathcal{S}_{\tau}^*$ if and only if <span id="page-4-0"></span> $$1 - \frac{\pi}{4} \le \frac{1 - A}{1 - B} \le \frac{1 + A}{1 + B} \le 1 + \frac{\pi}{4},\tag{2.8}$$ or equivalently, <span id="page-4-1"></span> $$A \le \begin{cases} \frac{\pi}{4} + (1 - \frac{\pi}{4})B & \text{if} \quad \frac{1 - AB}{1 - B^2} \le \frac{\pi}{4} \\ \frac{\pi}{4} + (1 + \frac{\pi}{4})B & \text{if} \quad \frac{1 - AB}{1 - B^2} > \frac{\pi}{4}. \end{cases}$$ (2.9)

Theorem 3.1. The class $S_{\tau}^*$ satisfies the following inclusion relations: - (1) $\mathcal{S}_{\tau}^ \subset \mathcal{S}^(\alpha)$ , for $0 \le \alpha \le 1 \pi/4$ . - (2) $S_{\tau}^ \subset \mathcal{RS}^(1/\alpha) \subset \mathcal{M}(\alpha)$ , for $\alpha \geq 1 + \pi/4$ . - (3) If k > 1 and $0 \le \alpha < 1$ , then $\mathcal{ST}(k, \alpha) \subset \mathcal{S}_{\tau}$ , for $k \ge (\pi + 4(1 \alpha))/\pi$ . In particular, $k \mathcal{ST} \subset \mathcal{S}_{\tau}$ for $k \ge 1 + 4/\pi$ .

Theorem 3.2. Let $f \in \mathcal{S}_{\tau}^*$ . Then $f \in \mathcal{C}_{\gamma}$ in $|z| < r_{\gamma}$ , where $r_{\gamma} (\approx 0.387888)$ is the least positive root of the equation <span id="page-5-0"></span> $$(1 - \arctan r)(1 - r^4)(1 - \arctan r - \gamma) - r = 0, \quad (0 \le \gamma < 1). \tag{3.1}$$

Theorem 3.3. The $\mathcal{S}_{\tau}$ -radii for the classes $\mathcal{S}_L$ , $\mathcal{S}_C$ , $\mathcal{S}_e$ , $\mathcal{S}_{\wp}^*$ are as follows: - (1) $R_{S}(\mathcal{S}_L^) = \pi(8-\pi)/16$ . - (2) $R_{S_{\tau}}(\mathcal{S}_C^) = \sqrt{1 + 3\pi/8} 1.$ - (3) $R_{S_{\pi}^{}}(\mathcal{S}_{e}^{}) = \ln(1 + \pi/4).$ - (4) $R_{S_{\tau}}(\Delta^) = \frac{\pi(8+\pi)}{8(4+\pi)}.$ - (5) $R_{S_{\tau}}(\mathcal{S}_{\wp}^) = 0.484035$ All the above radii are sharp.

Theorem 3.4. The radii of $S_{\tau}$ for the classes $S_e^, S_{SG}^, S_C^, S_{\omega}^, \Delta$ , are as follows: - (1) $R_{\mathcal{S}_{z}^{}}(\mathcal{S}_{\tau}^{}) = \tan(1 1/e) \approx 0.732368.$ - (2) $R_{\mathcal{S}_{SG}}(\mathcal{S}_{\tau}^) = \tan((e-1)/(e+1)) \approx 0.498088.$ - (3) $R_{\mathcal{S}_{\sigma}}(\mathcal{S}_{\tau}^) = \tan(2/3) \approx 0.786843.$ - (4) $R_{\mathcal{S}_{\tau}}(\mathcal{S}_{\tau}^) = \tan(1/e) \approx 0.385426.$ - (5) $R_{\Delta}(\mathcal{S}_{\tau}^) = \tan(2 \sqrt{2}) \approx 0.66347.$ All the above radii are sharp.

Lemma 4.1. [13] Let $p \in \mathcal{P}$ be of the form $1 + \sum_{n=1}^{\infty} p_n z^n$ . Then <span id="page-9-3"></span> $$|p_1^4 - 3p_1^2p_2 + p_2^2 + 2p_1p_3 - p_4| \le 2 (4.1)$$ and <span id="page-9-4"></span> $$|p_3 - 2p_1p_2 + p_1^3| \le 2. (4.2)$$

Lemma 4.2. [14] Let $p \in \mathcal{P}$ be of the form $1 + \sum_{n=1}^{\infty} p_n z^n$ . Then $$|p_2 - \beta p_1^2| \le \begin{cases} 2 - 4\beta, & \beta \le 0; \\ 2, & 0 \le \beta \le 1; \\ 4\beta - 2, & \beta \ge 1 \end{cases}$$ when $\beta < 0$ or $\beta > 1$ , the equality holds if and only if p(z) = (1+z)/(1-z) or one of its rotations. If $0 < \beta < 1$ , then the inequality holds if and only if $p(z) = (1+z^2)/(1-z^2)$ or one of its rotations. If $\beta = 0$ , the equality holds if and only if $p(z) = (1+\eta)(1+z)/(2(1-z)) + (1-\eta)(1-z)/(2(1+z))(0 \le z)$ $\eta \leq 1$ ) or one of its rotations. If $\beta = 1$ , the equality holds if and only if p is the reciprocal of one of the functions such that the equality holds in case of $\beta = 0$ . Though the above upper bound is sharp for $0 < \beta < 1$ , still it can be improved as follows: <span id="page-9-5"></span> $$|p_2 - \beta p_1^2| + \beta |p_1|^2 \le 2 \quad (0 < \beta \le 1/2)$$ (4.3) and $$|p_2 - \beta p_1^2| + (1 - \beta)|p_1|^2 \le 2 \quad (1/2 < \beta \le 1).$$ Also, we recall that <span id="page-9-6"></span> $$\max_{0 \le t \le 4} (At^2 + Bt + C) = \begin{cases} C, & B \le 0, A \le \frac{-B}{4}; \\ 16A + 4B + C, & B \ge 0, A \ge \frac{-B}{8} & \text{or} \quad B \le 0, A \ge \frac{-B}{4}; \\ \frac{4AC - B^2}{4A}, & B > 0, A \le \frac{-B}{8}. \end{cases}$$ (4.4)

Theorem 4.3. Let $f(z) = z + \sum_{n=2}^{n=\infty} a_n z^n \in \mathcal{S}_{\tau}^*$ , then $|a_2| \leq 1, |a_3| \leq 1/2, |a_4| \leq 1/3$ , and $|a_5| \le 323/528 \approx 0.611742$ . All these bounds are sharp.

Lemma 4.4. [10, 13] Let $p \in \mathcal{P}$ of the form $1 + \sum_{n=1}^{\infty} p_n z^n$ . Then $$2p_2 = p_1^2 + \gamma(4 - p_1^2),$$ $$4p_3 = p_1^3 + 2p_1(4 - p_1^2)\gamma - p_1(4 - p_1^2)\gamma^2 + 2(4 - p_1^2)(1 - |\gamma|^2)\eta$$ and $$8p_4 = p_1^4 + (4 - p_1^2)\gamma(p_1^2(\gamma^2 - 3\gamma + 3) + 4\gamma) - 4(4 - p_1^2)(1 - |\gamma|^2)(p_1(\gamma - 1)\eta + \bar{\gamma}\eta^2 - (1 - |\eta|^2)\rho),$$ for some $\gamma$ , $\eta$ and $\rho$ such that $|\gamma| < 1$ , $|\eta| < 1$ and $|\rho| < 1$ . Theorem 4.5. Let $f \in \mathcal{S}_{\tau}^*$ . Then $$|a_2a_3 - a_4| \le \frac{1}{3}.$$ The result is sharp. Let the function $f: \mathbb{D} \to \mathbb{C}$ be defined as <span id="page-11-0"></span> $$f(z) = z \exp\left(\int_0^z \frac{\arctan(t^3)}{t} dt\right) = z + \frac{z^4}{3} + \frac{z^7}{18} + \cdots,$$ (4.10) with f(0) = 0 and f'(0) = 1. The function defined in equation (4.10) acts as an extremal function for the bounds of $|a_2a_3 - a_4|$ for the values of $a_2 = a_3 = 0$ and $a_4 = 1/3$ . We now establish the bound for $H_2(2)$ in the following result by skipping the proof. Theorem 4.6. Let $f \in \mathcal{S}_{\tau}^*$ . Then $$|H_2(2)| \le \frac{1}{4}.$$ The result is sharp. For the sharpness of the above result, we consider the function $f: \mathbb{D} \to \mathbb{C}$ be defined as $$f(z) = z \exp\left(\int_0^z \frac{\arctan(t^2)}{t} dt\right) = z + \frac{z^3}{2} + \frac{z^5}{8} + \cdots,$$ with f(0) = 0 and f'(0) = 1. Obtaining the upper bound for the third order Hankel determinant is a challenging task, especially if we aim to find a sharp result as noted in [1,8]. Thus, finding the sharp bound for the third order Hankel determinant in [8] was an open problem and has recently been solved in [26]. Motivated by the works in [1,26,27], we now establish the sharp upper bound for the third order Hankel determinant for functions belonging to the class $\mathcal{S}_{\tau}^*$ . Theorem 4.7. Let $f \in \mathcal{S}_{\tau}^*$ . Then <span id="page-12-1"></span> $$|H_3(1)| \le \frac{1}{9}.\tag{4.11}$$ The result is sharp.

Definition 2.8. (Integral representation) A function $f \in \mathcal{S}_{\tau}^*$ if and only if an analytic function $\psi(z) \prec \tau(z)$ exists such that <span id="page-3-0"></span> $$f(z) = z \exp \int_0^z \frac{\psi(t) - 1}{t} dt.$$ (2.3) By choosing $\psi(t) = \tau(t)$ , in the expression given in equation (2.3), we obtain the function, $$\tilde{\tau}(z) = z \exp \int_0^z \frac{\arctan t}{t} dt \in \mathcal{S}_{\tau}^*.$$ Given Table 1 shows some $\psi_i(i=1,2,3)$ functions which belong to the domain $\Omega_{\tau}$ . Table 1. Functions associated with the class $\mathcal{S}_{\tau}^*$ <span id="page-3-1"></span> | i | $\psi_i(z)$ | $f_i(z)$ | |---|------------------------------|-----------------------------------------------| | 1 | 1 + z/2 | $z \exp(z/2)$ | | 2 | $1 + \frac{z \exp(z/17)}{2}$ | $z \exp\left(\frac{17}{2}(e^{z/17}-1)\right)$ | | 3 | $1 + \frac{z\sin z}{4}$ | $z \exp\left(\frac{1-\cos z}{4}\right)$ | And since $\tau(z)$ is univalent in $\mathbb{D}$ , $\psi_i(\mathbb{D}) \subset \tau(\mathbb{D})$ and $\psi_i(0) = \tau(0)$ are enough to prove that $\psi_i(z) \prec \tau(z)$ . Thus, the functions $f_i(z)$ belong to the class $\mathcal{S}_{\tau}^*$ using the integral representation given in equation (2.3). In particular, $\psi(z) = \tau(z)$ gives a function <span id="page-3-2"></span> $$\tilde{\tau}(z) = z \exp \int_0^z \frac{\arctan(t)}{t} dt = z + z^2 + \frac{z^3}{2} + \frac{z^4}{18} - \frac{5}{72} z^5 - \frac{13}{1800} z^6 + \cdots, \tag{2.4}$$ which plays the role of an extremal function for many problems involving $\mathcal{S}_{\tau}^*$ . Note that $\tau(z)$ meets all the conditions of Ma-Minda function and therefore yields the following simple results: