Higher order differential subordinations for certain starlike functions

Lemma 1.2. [1] Let $z_0 \in \mathbb{D}$ and $r_0 = |z_0|$ . Let $f(z) = \sum_{k=n}^{\infty} a_k z^k$ be continuous on $\overline{\mathbb{D}}_{r_0}$ and analytic on $\mathbb{D} \cup \{z_0\}$ with $f(z) \neq 0$ and $n \geq 2$ . If $|f(z_0)| = \max\{|f(z)| : z \in \overline{\mathbb{D}}_{r_0}\}$ and $|f'(z_0)| = \max\{|f'(z)| : z \in \overline{\mathbb{D}}_{r_0}\}$ , then there exist real constants m, k and l such that $$\frac{z_0 f'(z_0)}{f(z_0)} = m, \quad 1 + \frac{z_0 f''(z_0)}{f'(z_0)} = k \quad and \quad 2 + \operatorname{Re}\left(\frac{z_0 f'''(z_0)}{f''(z_0)}\right) = l$$ where $l \geq k \geq m \geq n \geq 2$ .

Lemma 1.3. [15, Theorem 2.3b] Let $\phi \in \Psi_n[\Omega, q]$ with q(0) = a. If $p \in \mathcal{H}[a, n]$ satisfies $\phi(n(z), zn'(z), z^2n''(z); z) \in \Omega$ . then $p \prec q$ . Note that if $\Omega \subset \mathbb{C}$ is a simply connected domain, then there exists a conformal mapping h from $\mathbb{D}$ onto $\Omega = h(\mathbb{D})$ . Moreover, if the function $\phi(p(z), zp'(z), z^2p''(z); z)$ is analytic in $\mathbb{D}$ , then $\phi(p(z), zp'(z), z^2p''(z); z) \in \Omega$ can be expressed in terms of subordination as $$\phi(p(z), zp'(z), z^2p''(z); z) \prec h(z).$$ For the Ma-Minda functions to satisfy the criteria for third-order differential subordination, Kumar and Goel [9] modified the results of Antonino and Miller [1] in 2020. Moreover, they derived the first, second and third-order differential subordination implications for the class $\mathcal{S}_{SG}$ , using these modified results through admissibility conditions. Recently, Verma and Kumar [21] have obtained results pertaining to second and third-order differential subordination for the class $\mathcal{S}_e$ . Thus, this study is motivated to establish parameter-specific conditions enabling the validity of second and third-order subordination implications for the class $\mathcal{S}_s$ , as well as second-order subordination implications for the class $\mathcal{S}_s$ . The modified definition and lemma of the third-order differential subordination implication by Kumar and Goel [9], accompanied by some lemmas are presented below, which are required to deduce our results in the subsequent sections.

Lemma 1.5. [9] Let $p \in \mathcal{H}[a, n]$ with $m \ge n \ge 2$ , and let $q \in Q(a)$ such that it satisfies $$\left| \frac{zp'(z)}{q'(\zeta)} \right| \le m \quad for \quad z \in \mathbb{D} \quad and \quad \zeta \in \partial \mathbb{D} \setminus \mathbb{E}(q).$$ If $\Omega$ is a set in $\mathbb{C}$ , $\phi \in \Psi_n[\Omega, a]$ and $$\phi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z) \subset \Omega,$$ then $p \prec q$ .

Lemma 1.6. [4] Let $r_0 \approx 0.546302$ be the positive root of the equation $r^2 + 2\cot(1)r - 1 = 0$ . Then $$\left|\log\left(\frac{1+z}{1-z}\right)\right| \ge 1$$ on $|z| = R$ if and only if $R \ge r_0$ .

Lemma 1.7. [9, Lemma 4, Pg No. 192] For any complex number z, we have $$|\log(1+z)| \ge 1$$ if and only if $|z| \ge e-1$ . 2. The Class $$\mathcal{S}_s^*$$ We begin our study with the function $q(z) := 1 + \sin z$ and define the admissibility class $\Psi[\Omega, q]$ , where $\Omega \subset \mathbb{C}$ . We know that q(z) is analytic and univalent on $\overline{\mathbb{D}}$ , q(0) = 1 and it maps $\mathbb{D}$ onto the domain $\Omega_s := \{w \in \mathbb{C} : |\arcsin(w-1)| < 1\}$ . Since $\mathbb{E}(q) = \phi$ , for $\zeta \in \partial \mathbb{D} \setminus \mathbb{E}(q)$ if and only if $\zeta = e^{i\theta}$ for $\theta \in [0, 2\pi]$ . Now, consider <span id="page-2-0"></span> $$|q'(\zeta)| = \sqrt{\cosh^2(\sin\theta) - \sin^2(\cos\theta)} =: n_1(\theta)$$ (2.1) and $n_1(\theta)$ achieves its minimum value at $\theta = 0$ , denoted by <span id="page-3-2"></span> $$\nu_0 := n_1(0) = \sqrt{1 - \sin^2 1} \approx 0.540302.$$ (2.2) It is evident that $\min |q'(\zeta)| > 0$ , which implies that $q \in Q(1)$ , and consequently, the admissibility class $\Psi[\Omega,q]$ is well-defined. While considering $|\zeta|=1$ , we observe that $q(\zeta)\in q(\partial\mathbb{D})=\partial\Omega_s=\{w\in\mathbb{C}:|\arcsin(w-1)|=1\}$ . Consequently, $|\arcsin(q(\zeta)-1)|=1$ and $\arcsin(q(\zeta)-1)=e^{i\theta}$ for $\theta\in[0,2\pi]$ , which implies that $q(\zeta)=1+\sin\zeta$ . Moreover, $\zeta q'(\zeta)=e^{i\theta}\cos(e^{i\theta})$ and <span id="page-3-0"></span> $$\frac{\zeta q''(\zeta)}{q'(\zeta)} = e^{i\theta} \tan(e^{i\theta}). \tag{2.3}$$ By comparing the real parts on both sides of (2.3), we obtain <span id="page-3-1"></span> $$\operatorname{Re}\left(\frac{\zeta q''(\zeta)}{q'(\zeta)}\right) = \frac{-\cos\theta\sin(2\cos\theta) + \sin\theta\sinh(2\sin\theta)}{\cos(2\cos\theta) + \cosh(2\sin\theta)} =: n_2(\theta). \tag{2.4}$$ The function $n_2(\theta)$ as defined in (2.4), attains its minimum at $\theta = 0$ , denoted by <span id="page-3-3"></span> $$\nu_1 := n_2(0) = -\frac{\sin 2}{1 + \cos 2} \approx -1.55741.$$ (2.5) Moreover, the class $\Psi[\Omega, 1 + \sin z]$ is precisely defined as the class of all functions $\phi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ that satisfy the following conditions: $$\phi(r, s, t; z) \notin \Omega$$ for $z \in \mathbb{D}$ , $\theta \in [0, 2\pi]$ and $m \ge 1$ , whenever $$r = q(\zeta) = 1 + \sin(e^{i\theta}); \quad s = m\zeta q'(\zeta) = me^{i\theta}\cos(e^{i\theta}); \quad \operatorname{Re}\left(1 + \frac{t}{s}\right) \ge m(1 + n_2(\theta)).$$ (2.6) Furthermore, for $q(z) = 1 + \sin z$ , we have $$\zeta^2 \frac{q'''(\zeta)}{q'(\zeta)} = -e^{2i\theta}.$$ On comparing the real parts of both the sides, $$\operatorname{Re}\left(\zeta^2 \frac{q'''(\zeta)}{q'(\zeta)}\right) = -\cos 2\theta =: n_3(\theta).$$ The minimum value of $n_3(\theta)$ is -1, which is attained at $\theta = 0$ . Thus, if $\phi : \mathbb{C}^4 \times \mathbb{D} \to \mathbb{C}$ then $\phi \in \Psi[\Omega, 1 + \sin z]$ , provided $\phi$ satisfies the following conditions: $$\phi(r,s,t,u;z)\notin\Omega\quad\text{for}\quad z\in\mathbb{D},\quad\theta\in[0,2\pi]\quad\text{and}\quad k\geq m\geq 2$$ whenever $$r = q(\zeta) = 1 + \sin e^{i\theta}; \quad s = m\zeta q'(\zeta) = me^{i\theta} \cos(e^{i\theta});$$ $$\operatorname{Re}\left(1 + \frac{t}{s}\right) \ge m(1 + n_2(\theta)) \quad \text{and} \quad \operatorname{Re}\frac{u}{s} \ge m^2 n_3(\theta) + 3m(k - 1)n_2(\theta).$$ We now find conditions on the parameters $\beta_1$ and $\beta_2$ for different choices of h(z) namely $\sqrt{1+z}$ , (1+Cz)/(1+Dz), $2/(1+e^{-z})$ , $z+\sqrt{1+z^2}$ , $1+\sinh^{-1}z$ , $1+ze^z$ , $e^z$ and $1+\sin z$ , so that the following second order differential subordination implication holds: $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec h(z) \implies p(z) \prec 1 + \sin z$$ Note that the values of $\nu_0$ and $\nu_1$ are given in (2.2) and (2.5) respectively, which are widely used in the upcoming results. We begin with the following Theorem, for the case when $h(z) = \sqrt{1+z}$ .

Theorem 2.1. Let $\beta_1$ , $\beta_2 > 0$ and $\nu_0(\beta_1 + \beta_2\nu_1)(\nu_0(\beta_1 + \beta_2\nu_1) - 2) \ge 1$ . Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec \sqrt{1+z}$$ . Then $p(z) \prec 1 + \sin z$ . Proof. Suppose $h(z) = \sqrt{1+z}$ for $z \in \mathbb{D}$ . Then, $h(\mathbb{D}) = \{w \in \mathbb{C} : |w^2 - 1| < 1\} =: \Omega$ . Let $\phi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ be defined as $\phi(r, s, t; z) = 1 + \beta_1 s + \beta_2 t$ . It is clear that $\phi \in \Psi[\Omega, 1 + \sin z]$ provided $\phi(r, s, t, z) \notin \Omega$ for $z \in \mathbb{D}$ . Note that $$|(\phi(r, s, t; z))^{2} - 1| = |(1 + \beta_{1}s + \beta_{2}t)^{2} - 1|$$ $$\geq |\beta_{1}s + \beta_{2}t|(|\beta_{1}s + \beta_{2}t| - 2)$$ $$\geq |\beta_{1}s| \operatorname{Re}\left(1 + \frac{\beta_{2}}{\beta_{1}}\frac{t}{s}\right) \left(|\beta_{1}s| \operatorname{Re}\left(1 + \frac{\beta_{2}}{\beta_{1}}\frac{t}{s}\right) - 2\right)$$ $$\geq L(L - 2),$$ where $L = m\beta_1 n_1(\theta) \operatorname{Re}(1 + \beta_2(mn_2(\theta) + m - 1)/\beta_1)$ . Here $n_1(\theta)$ and $n_2(\theta)$ are given in (2.1) and (2.4), respectively. Since $m \ge 1$ , we have $$|(\phi(r, s, t; z))^{2} - 1| \ge \beta_{1} n_{1}(\theta) \operatorname{Re} \left( 1 + \frac{\beta_{2}}{\beta_{1}} n_{2}(\theta) \right) \left( \beta_{1} n_{1}(\theta) \operatorname{Re} \left( 1 + \frac{\beta_{2}}{\beta_{1}} n_{2}(\theta) \right) - 2 \right)$$ $$\ge \nu_{0}(\beta_{1} + \beta_{2} \nu_{1}) (\nu_{0}(\beta_{1} + \beta_{2} \nu_{1}) - 2)$$ $$\ge 1.$$ Therefore, $\phi(r, s, t; z) \notin \Omega$ and hence $\phi \in \Psi[\Omega, 1 + \sin z]$ . Now, the result follows at once by an application of Lemma 1.3. By considering the function p(z) = zf'(z)/f(z) in Theorem 2.1, we derive the following corollary: <span id="page-4-3"></span>Corollary 2.2. Suppose $\beta_1$ , $\beta_2 > 0$ and $f \in A$ . Let $$S_f(z) := 1 + \beta_1(S_2 - S_1^2 + S_1) + \beta_2(S_3 + 2S_2 + 2S_1^3 - 2S_1^2 - 3S_1S_2), \tag{2.7}$$ where <span id="page-4-2"></span> $$S_1 = \frac{zf'(z)}{f(z)}, \quad S_2 = \frac{z^2f''(z)}{f(z)} \quad and \quad S_3 = \frac{z^3f'''(z)}{f(z)}.$$ (2.8) Then $f \in \mathcal{S}_{s}^{*}$ provided $S_{f}(z) \prec \sqrt{1+z}$ and $\nu_{0}(\beta_{1}+\beta_{2}\nu_{1})(\nu_{0}(\beta_{1}+\beta_{2}\nu_{1})-2) \geq 1$ .

Theorem 2.3. Let $\beta_1$ , $\beta_2 > 0$ and $-1 < D < C \le 1$ with $\nu_0(\beta_1 + \beta_2 \nu_1)(1 - D^2) \ge (C - D)(1 + |D|)$ . Let p be analytic function in $\mathbb D$ with p(0) = 1 and <span id="page-4-1"></span> $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec \frac{1 + Cz}{1 + Dz}$$ Then $p(z) \prec 1 + \sin z$ . Proof. Let h(z) = (1+Cz)/(1+Dz) for $z \in \mathbb{D}$ , then we have $h(\mathbb{D}) = \{w \in \mathbb{C} : |w-(1-CD)/(1-D^2)| < (C-D)/(1-D^2)\} =: \Omega$ . Let $\phi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ be defined as $\phi(r,s,t;z) = 1 + \beta_1 s + \beta_2 t$ . It is clear that $\phi \in \Psi[\Omega, 1 + \sin z]$ only when $\phi(r, s, t; z) \notin \Omega$ for $z \in \mathbb{D}$ . Consider $$\left| \phi(r, s, t; z) - \frac{1 - CD}{1 - D^2} \right| = \left| 1 + \beta_1 s + \beta_2 t - \frac{1 - CD}{1 - D^2} \right|$$ $$\geq |\beta_1 s + \beta_2 t| - \frac{|D|(C - D)}{1 - D^2}$$ $$\geq |\beta_1 s| \operatorname{Re} \left( 1 + \frac{\beta_2}{\beta_1} \frac{t}{s} \right) - \frac{|D|(C - D)}{1 - D^2}$$ $$\geq m\beta_1 n_1(\theta) \operatorname{Re} \left( 1 + \frac{\beta_2}{\beta_1} (mn_2(\theta) + m - 1) \right) - \frac{|D|(C - D)}{1 - D^2}.$$ Using the fact that $m \geq 1$ and proceeding as in the proof of Theorem 2.1, we have $$\left| \phi(r, s, t; z) - \frac{1 - CD}{1 - D^2} \right| \ge \nu_0(\beta_1 + \beta_2 \nu_1) - \frac{|D|(C - D)}{1 - D^2}$$ $$\ge \frac{C - D}{1 - D^2}.$$ Thus, $\phi(r, s, t; z) \notin \Omega$ and hence $\phi \in \Psi[\Omega, 1 + \sin z]$ . Now, the result follows by an application of Lemma 1.3.

Theorem 2.4. Suppose $\beta_1$ , $\beta_2 > 0$ and $\nu_0(\beta_1 + \beta_2\nu_1) \ge r_0$ , where $r_0 \approx 0.546302$ is the positive root of the equation $r^2 + 2\cot(1)r - 1 = 0$ . Let p be analytic function in $\mathbb D$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec \frac{2}{1 + e^{-z}}.$$ Then $p(z) \prec 1 + \sin z$ . Proof. Assuming $h(z) = 2/(1+e^{-z})$ for $z \in \mathbb{D}$ . Then, $h(\mathbb{D}) = \{w \in \mathbb{C} : |\log(w/(2-w))| < 1\} =: \Omega$ . Let $\phi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ be defined as $\phi(r, s, t; z) = 1 + \beta_1 s + \beta_2 t$ . We know that $\phi \in \Psi[\Omega, 1 + \sin z]$ provided $\phi(r, s, t; z) \notin \Omega$ . First we consider $$|\beta_1 s + \beta_2 t| = |\beta_1|s| \left| 1 + \frac{\beta_2}{\beta_1} \frac{t}{s} \right|$$ $$\geq |\beta_1|s| \operatorname{Re} \left( 1 + \frac{\beta_2}{\beta_1} \frac{t}{s} \right)$$ $$\geq m\beta_1 n_1(\theta) \left( 1 + \frac{\beta_2}{\beta_1} (mn_2(\theta) + m - 1) \right).$$ As $m \geq 1$ , we obtain $$|\beta_1 s + \beta_2 t| \ge n_1(\theta)(\beta_1 + \beta_2 n_2(\theta))$$ $$\ge \nu_0(\beta_1 + \beta_2 \nu_1)$$ $$\ge r_0. \tag{2.9}$$ Now, we consider $$\left|\log\left(\frac{\phi(r,s,t;z)}{2-\phi(r,s,t;z)}\right)\right| = \left|\log\left(\frac{1+\beta_1s+\beta_2t}{1-(\beta_1s+\beta_2t)}\right)\right|.$$ Through Lemma 1.6 and (2.9), we have <span id="page-5-0"></span> $$\left|\log\left(\frac{1+\beta_1s+\beta_2t}{1-(\beta_1s+\beta_2t)}\right)\right| \ge 1,$$ which implies that $\phi \in \Psi[\Omega, 1 + \sin z]$ . Now, the result follows by an application of Lemma 1.3.

Theorem 2.5. Suppose $\beta_1$ , $\beta_2 > 0$ and $\nu_0(\beta_1 + \beta_2\nu_1) \ge \sqrt{2}$ . Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec z + \sqrt{1 + z^2}.$$ Then $p(z) \prec 1 + \sin z$ . Proof. Let $h(z) = z + \sqrt{1+z^2}$ for $z \in \mathbb{D}$ . Then, $h(\mathbb{D}) = \{w \in \mathbb{C} : |w^2 - 1| < 2|w|\} =: \Omega$ . Let $\phi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ be defined as $\phi(r, s, t; z) = 1 + \beta_1 s + \beta_2 t$ . For $\phi \in \Psi[\Omega, 1 + \sin z]$ , we must have $\phi(r, s, t; z) \notin \Omega$ . From the geometry of $z + \sqrt{1+z^2}$ (see Fig. 1), we note that $\Omega$ is constructed by the circles $$C_1: |z-1| = \sqrt{2}$$ and $C_2: |z+1| = \sqrt{2}$ .  <span id="page-6-0"></span>FIGURE 1. Graph of two circles, namely $C_1$ (blue boundary) and $C_2$ (orange boundary). While the shaded region represents $z + \sqrt{1+z^2}$ . It is obvious that $\Omega$ contains the disk enclosed by $C_1$ and excludes the portion of the disk enclosed by $C_1 \cap C_2$ . We have $$|\phi(r, s, t; z) - 1| = |\beta_1 s + \beta_2 t|.$$ Now, proceeding in the similar way as in the proof of Theorem 2.4, we have $$|\beta_1 s + \beta_2 t| \ge n_1(\theta)(\beta_1 + \beta_2 n_2(\theta))$$ $$\ge \nu_0(\beta_1 + \beta_2 \nu_1)$$ $$> \sqrt{2}.$$ The fact that $\phi(r, s, t; z)$ lies outside the circle $C_1$ suffices us to deduce that $\phi(r, s, t; z) \notin \Omega$ and hence $\phi \in \Psi[\Omega, 1 + \sin z]$ . Now, the result follows by an application of Lemma 1.3.

Theorem 2.6. Let $\beta_1$ , $\beta_2 > 0$ and $\nu_0(\beta_1 + \beta_2 \nu_1) \ge e$ . Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec 1 + z e^z.$$ Then $p(z) \prec 1 + \sin z$ .

Theorem 2.7. Suppose $\beta_1$ , $\beta_2 > 0$ and $2\nu_0(\beta_1 + \beta_2\nu_1) \ge \pi$ . Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec 1 + \sinh^{-1} z.$$ Then $p(z) \prec 1 + \sin z$ . Proof. Let $h(z) = 1 + \sinh^{-1} z$ for $z \in \mathbb{D}$ . Then, $h(\mathbb{D}) = \{w \in \mathbb{C} : |\sinh(w-1)| < 1\} =: \Omega_{\rho}$ . Let $\phi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ be defined as $\phi(r, s, t; z) = 1 + \beta_1 s + \beta_2 t$ . For $\phi \in \Psi[\Omega_{\rho}, 1 + \sin z]$ , we must have $\phi(r, s, t; z) \notin \Omega_{\rho}$ . Through [2, Remark 2.7], we note that the disk $\{\delta \in \mathbb{C} : |\delta - 1| < \pi/2\}$ is the smallest disk containing $\Omega_{\rho}$ . So, $$|\phi(r, s, t; z) - 1| = |\beta_1 s + \beta_2 t|.$$ Analogous to Theorem 2.4, we have $$|\beta_1 s + \beta_2 t| \ge n_1(\theta)(\beta_1 + \beta_2 n_2(\theta))$$ $$\ge \nu_0(\beta_1 + \beta_2 \nu_1)$$ $$\ge \frac{\pi}{2}.$$ Clearly, $\phi(r,s,t;z)$ does not belong to the disk $\{\delta \in \mathbb{C} : |\delta-1| < \pi/2\}$ which suffices us to conclude that $\phi(r,s,t;z) \notin \Omega_{\rho}$ . Therefore, $\phi \in \Psi[\Omega_{\rho},1+\sin z]$ and thus the result follows as an application of Lemma 1.3.

Theorem 2.8. Let $\beta_1$ , $\beta_2 > 0$ and $\nu_0(\beta_1 + \beta_2\nu_1) \ge e - 1$ . Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec e^z$$ . Then $p(z) \prec 1 + \sin z$ . Proof. Suppose $h(z) = e^z$ for $z \in \mathbb{D}$ . Then, $h(\mathbb{D}) = \{w \in \mathbb{C} : |\log w| < 1\} =: \Omega$ . Let $\phi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ be defined as $\phi(r, s, t; z) = 1 + \beta_1 s + \beta_2 t$ . For $\phi \in \Psi[\Omega, 1 + \sin z]$ , we must have $\phi(r, s, t; z) \notin \Omega$ . So, <span id="page-7-0"></span> $$|\phi(r, s, t; z) - 1| = |\beta_1 s + \beta_2 t|.$$ Proceeding similar to the proof of Theorem 2.4, we have $$|\beta_1 s + \beta_2 t| \ge n_1(\theta)(\beta_1 + \beta_2 n_2(\theta))$$ $$\ge \nu_0(\beta_1 + \beta_2 \nu_1)$$ $$\ge e - 1. \tag{2.10}$$ Further, we have $$|\log(\phi(r, s, t; z))| = |\log(1 + \beta_1 s + \beta_2 t)|.$$ Through Lemma 1.7 and (2.10), we have $$|\log(1+\beta_1 s + \beta_2 t)| \ge 1,$$ which implies that $\phi(r, s, t; z) \notin \Omega$ . Therefore, $\phi \in \Psi[\Omega, 1 + \sin z]$ and thus the result follows as an application of Lemma 1.3.

Theorem 2.9. Suppose $\beta_1$ , $\beta_2 > 0$ and $\nu_0(\beta_1 + \beta_2\nu_1) \ge \sinh 1$ . Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec 1 + \sin z.$$ Then $p(z) \prec 1 + \sin z$ . Proof. Suppose $h(z) = 1 + \sin z$ for $z \in \mathbb{D}$ . Then, $h(\mathbb{D}) = \{w \in \mathbb{C} : |\arcsin(w-1)| < 1\} =: \Omega_s$ . Let $\phi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ be defined as $\phi(r, s, t; z) = 1 + \beta_1 s + \beta_2 t$ . For $\phi \in \Psi[\Omega_s, 1 + \sin z]$ , we must have $\phi(r, s, t; z) \notin \Omega_s$ . Through [3, Lemma 3.3], we note that $\{\delta \in \mathbb{C} : |\delta - 1| < \sinh 1\}$ is the smallest disk containing $\Omega_s$ . So, $$|\phi(r, s, t; z) - 1| = |\beta_1 s + \beta_2 t|.$$ Analogous to Theorem 3.3, we have $$|\beta_1 s + \beta_2 t| \ge n_1(\theta)(\beta_1 + \beta_2 n_2(\theta))$$ $$\ge \nu_0(\beta_1 + \beta_2 \nu_1)$$ $$> \sinh 1.$$ Clearly, $\phi(r, s, t; z)$ lies outside the disk $\{\delta \in \mathbb{C} : |\delta - 1| < \sinh 1\}$ which is enough to conclude that $\phi(r, s, t; z) \notin \Omega_s$ . Therefore, $\phi \in \Psi[\Omega_s, 1 + \sin z]$ and the result follows as an application of Lemma 1.3. By taking p(z) = zf'(z)/f(z) in Theorems 2.7-2.12, we obtain the following corollary with $S_f$ as given by (2.7): Corollary 2.10. Suppose $\beta_1$ , $\beta_2 > 0$ and $f \in A$ . Then, $f \in \mathcal{S}_s^*$ if any of these conditions hold: - (i) $S_f(z) \prec (1+Cz)/(1+Dz)$ and $\nu_0(\beta_1+\beta_2\nu_1)(1-D^2) \geq (C-D)(1+|D|)$ , where $-1 < D < C \leq 1$ . - (ii) $S_f(z) \prec 2/(1 + e^{-z})$ and $\nu_0(\beta_1 + \beta_2 \nu_1) \ge r_0$ , where $r_0 \approx 0.546302$ is the positive root of the equation $r^2 + 2 \cot(1)r 1 = 0$ . - (iii) $S_f(z) \prec z + \sqrt{1+z^2}$ and $\nu_0(\beta_1 + \beta_2 \nu_1) \ge \sqrt{2}$ . - (iv) $S_f(z) \prec 1 + ze^z$ and $\nu_0(\beta_1 + \beta_2 \nu_1) \ge e$ . - (v) $S_f(z) \prec 1 + \sinh^{-1} z$ and $2\nu_0(\beta_1 + \beta_2\nu_1) \geq \pi$ . - (vi) $S_f(z) \prec e^z$ and $\nu_0(\beta_1 + \beta_2 \nu_1) \geq e 1$ . - (vii) $S_f(z) \prec 1 + \sin z$ and $\nu_0(\beta_1 + \beta_2 \nu_1) \geq \sinh 1$ . Next, we determine the sufficient conditions on the positive constants $\beta_1$ , $\beta_2$ and $\beta_3$ for various selections of h(z), notably including the case when $h(z) = 1 + \sin z$ itself. This study aims to establish third-order differential subordination implication, which is as follows: $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) + \beta_3 z^3 p'''(z) \prec h(z) \implies p(z) \prec 1 + \sin z.$$ The derived conditions involve the variables $\nu_0$ and $\nu_1$ , along with the constants m and k, which are explicitly defined in (2.2), (2.5) and Lemma 1.2, respectively. We begin with the Theorem below by choosing $h(z) = \sqrt{1+z}$ as follows:

Theorem 2.11. Suppose $\beta_1$ , $\beta_2$ , $\beta_3 > 0$ and satisfy $$\nu_0(\beta_1 + \beta_2\nu_1 + \beta_3(-m^2 + 3m(k-1)\nu_1))(\nu_0(\beta_1 + \beta_2\nu_1 + \beta_3(-m^2 + 3m(k-1)\nu_1)) - 2) \ge 1.$$ Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) + \beta_3 z^3 p'''(z) \prec \sqrt{1+z}$$ Then $p(z) \prec 1 + \sin z$ . Proof. Suppose $h(z) = \sqrt{1+z}$ for $z \in \mathbb{D}$ . Then, $h(\mathbb{D}) = \{w \in \mathbb{C} : |w^2 - 1| < 1\} =: \Omega$ . Let $\phi : \mathbb{C}^4 \times \mathbb{D} \to \mathbb{C}$ be defined as $\phi(r, s, t, u; z) = 1 + \beta_1 s + \beta_2 t + \beta_3 u$ . We know that $\phi \in \Psi[\Omega, 1 + \sin z]$ provided $\phi(r, s, t, u; z) \notin \Omega$ for $z \in \mathbb{D}$ . Consider $$|(\phi(r, s, t, u; z))^{2} - 1| = |(1 + \beta_{1}s + \beta_{2}t + \beta_{3}u)^{2} - 1|$$ $$\geq |\beta_{1}s + \beta_{2}t + \beta_{3}u|(|\beta_{1}s + \beta_{2}t + \beta_{3}u| - 2)$$ $$\geq |\beta_{1}s| \operatorname{Re}\left(1 + \frac{\beta_{2}}{\beta_{1}}\frac{t}{s} + \frac{\beta_{3}}{\beta_{1}}\frac{u}{s}\right) \left(|\beta_{1}s| \operatorname{Re}\left(1 + \frac{\beta_{2}}{\beta_{1}}\frac{t}{s} + \frac{\beta_{3}}{\beta_{1}}\frac{u}{s}\right) - 2\right)$$ $$\geq M(M - 2),$$ where $M = m\beta_1 n_1(\theta)(1 + \beta_2(mn_2(\theta) + m - 1)/\beta_1 + \beta_3(m^2n_3(\theta) + 3m(k - 1)n_2(\theta))/\beta_1)$ . As $m \ge 1$ and $\text{Re}(1 + n_2(\theta)) > 0$ , we get $$|(\phi(r, s, t, u; z))^{2} - 1| \ge \beta_{1} n_{1}(\theta) \left( 1 + \frac{\beta_{2}}{\beta_{1}} n_{2}(\theta) + \frac{\beta_{3}}{\beta_{1}} (m^{2} n_{3}(\theta) + 3m(k - 1) n_{2}(\theta)) \right) \left( \beta_{1} n_{1}(\theta) \left( 1 + \frac{\beta_{2}}{\beta_{1}} n_{2}(\theta) + \frac{\beta_{3}}{\beta_{1}} (m^{2} n_{3}(\theta) + 3m(k - 1) n_{2}(\theta)) \right) - 2 \right)$$ $$\ge \nu_{0}(\beta_{1} + \beta_{2} \nu_{1} + \beta_{3} (-m^{2} + 3m(k - 1) \nu_{1})) (\nu_{0}(\beta_{1} + \beta_{2} \nu_{1} + \beta_{3} (-m^{2} + 3m(k - 1) \nu_{1})) - 2)$$ $$\ge 1.$$ This implies that $\phi(r, s, t; z) \notin \Omega$ and thus $\phi \in \Psi[\Omega, 1 + \sin z]$ . The result follows as a consequence of Lemma 1.5. By considering p(z) = zf'(z)/f(z) in Theorem 2.11, we have the following: <span id="page-9-0"></span>Corollary 2.12. Suppose $\beta_1$ , $\beta_2$ , $\beta_3 > 0$ and $f \in A$ . Let $$\Theta_f(z) := 1 + \beta_1 S_1 + (\beta_1 + 2\beta_2)(S_2 - S_1^2) + (\beta_2 + 3\beta_3)(2S_1^3 - 3S_1S_2 + 3S_3) + \beta_3(S_4 - 3S_2^2 - 6S_1^4 - 4S_1S_3 + 12S_1^2S_2).$$ (2.11) where $S_1$ , $S_2$ and $S_3$ are mentioned in (2.8) and $S_4 := z^4 f^{(iv)}(z)/f(z)$ . Then, $f(z) \in \mathcal{S}_s^*$ provided $\Theta_f(z) \prec \sqrt{1+z}$ and $\nu_0(\beta_1 + \beta_2\nu_1 + \beta_3(-m^2 + 3m(k-1)\nu_1)) - 2\nu_0(\beta_1 + \beta_2\nu_1 + \beta_3(-m^2 + 3m(k-1)\nu_1)) \geq 1$ .

Theorem 2.13. Suppose $\beta_1$ , $\beta_2$ , $\beta_3 > 0$ and $-1 < D < C \le 1$ with $\nu_0(\beta_1 + \beta_2\nu_1 - m^2\beta_3 + 3m\beta_3(k-1)\nu_1)(1-D^2) \ge (1+|D|)(C-D)$ . Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta z^2 p''(z) + \beta_3 z^3 p'''(z) \prec \frac{1 + Cz}{1 + Dz}.$$ Then $p(z) \prec 1 + \sin z$ .

Theorem 2.14. Suppose $\beta_1$ , $\beta_2$ , $\beta_3 > 0$ and $\nu_0(\beta_1 + \beta_2\nu_1 - m^2\beta_3 + 3m\beta_3(k-1)\nu_1) \ge r_0$ , where $r_0 \approx 0.546302$ is the positive root of the equation $r^2 + 2\cot(1)r - 1 = 0$ . Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) + \beta_3 z^3 p'''(z) \prec \frac{2}{1 + e^{-z}}.$$ Then $p(z) \prec 1 + \sin z$ . Proof. Assume $h(z) = 2/(1 + e^{-z})$ for $z \in \mathbb{D}$ . Then, $h(\mathbb{D}) = \{w \in \mathbb{C} : |\log(w/(2-w))| < 1\} =: \Omega$ . Let $\phi : \mathbb{C}^4 \times \mathbb{D} \to \mathbb{C}$ be defined as $\phi(r, s, t, u; z) = 1 + \beta_1 s + \beta_2 t + \beta_3 u$ . For $\phi \in \Psi[\Omega, 1 + \sin z]$ , we must have $\phi(r, s, t, u; z) \notin \Omega$ . Consider $$|\beta_{1}s + \beta_{2}t + \beta_{3}u| = |\beta_{1}|s| \left| 1 + \frac{\beta_{2}}{\beta_{1}} \frac{t}{s} + \frac{\beta_{3}}{\beta_{1}} \frac{u}{s} \right|$$ $$\geq |\beta_{1}|s| \operatorname{Re} \left( 1 + \frac{\beta_{2}}{\beta_{1}} \frac{t}{s} + \frac{\beta_{3}}{\beta_{1}} \frac{u}{s} \right)$$ $$\geq m\beta_{1}n_{1}(\theta) \left( 1 + \frac{\beta_{2}}{\beta_{1}} (mn_{2}(\theta) + m - 1) + \frac{\beta_{3}}{\beta_{1}} (m^{2}n_{3}(\theta) + 3m(k - 1)n_{2}(\theta)) \right).$$ As $m \ge 1$ and $\text{Re}(1 + n_2(\theta)) > 0$ , we have $$|\beta_{1}s + \beta_{2}t + \beta_{3}u| \geq n_{1}(\theta) \left(\beta_{1} + \beta_{2}n_{2}(\theta) + \beta_{3}(m^{2}n_{3}(\theta) + 3m(k-1)n_{2}(\theta))\right)$$ $$\geq \nu_{0}(\beta_{1} + \beta_{2}\nu_{1} - m^{2}\beta_{3} + 3m\beta_{3}(k-1)\nu_{1})$$ $$\geq r_{0}.$$ (2.12) Now, we consider <span id="page-10-0"></span> $$\left| \log \left( \frac{\phi(r, s, t, u; z)}{2 - \phi(r, s, t, u; z)} \right) \right| = \left| \log \left( \frac{1 + \beta_1 s + \beta_2 t + \beta_3 u}{1 - (\beta_1 s + \beta_2 t + \beta_3 u)} \right) \right|.$$ In view of Lemma 1.6 and (2.12), we have $$\left| \log \left( \frac{1 + \beta_1 s + \beta_2 t + \beta_3 u}{1 - (\beta_1 s + \beta_2 t + \beta_3 u)} \right) \right| \ge 1,$$ which implies that $\phi \in \Psi[\Omega, 1+\sin z]$ . Thus, The result follows as an application of Lemma 1.5.

Theorem 2.15. Suppose $\beta_1$ , $\beta_2$ , $\beta_3 > 0$ and $\nu_0(\beta_1 + \beta_2\nu_1 - m^2\beta_3 + 3m\beta_3(k-1)\nu_1) \ge \sqrt{2}$ . Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) + \beta_3 z^3 p'''(z) \prec z + \sqrt{1 + z^2}.$$ Then $p(z) \prec 1 + \sin z$ . Proof. Take $h(z) = z + \sqrt{1+z^2}$ for $z \in \mathbb{D}$ . Then, $h(\mathbb{D}) = \{w \in \mathbb{C} : |w^2 - 1| < 2|w|\} =: \Omega$ . Let $\phi : \mathbb{C}^4 \times \mathbb{D} \to \mathbb{C}$ be defined as $\phi(r, s, t, u; z) = 1 + \beta_1 s + \beta_2 t + \beta_3 u$ . For $\phi \in \Psi[\Omega, 1 + \sin z]$ , we must have $\phi(r, s, t, u; z) \notin \Omega$ . From the geometry of $z + \sqrt{1+z^2}$ (see Fig. 1), we note that $\Omega$ is constructed by two circles $$C_1: |z-1| = \sqrt{2}$$ and $C_2: |z+1| = \sqrt{2}$ . It is obvious that $\Omega$ contains the disk enclosed by $C_1$ and excludes the portion of the disk enclosed by $C_1 \cap C_2$ . We have $$|\phi(r, s, t, u; z) - 1| = |\beta_1 s + \beta_2 t + \beta_3 u|.$$ Analogous to Theorem 2.14, we have $$|\beta_1 s + \beta_2 t + \beta_3 u| \ge n_1(\theta) \left( \beta_1 + \beta_2 n_2(\theta) + \beta_3 (m^2 n_3(\theta) + 3m(k-1)n_2(\theta)) \right)$$ $$\ge \nu_0(\beta_1 + \beta_2 \nu_1 - m^2 \beta_3 + 3m\beta_3 (k-1)\nu_1)$$ $$\ge \sqrt{2}.$$ Thus, we can say that $\phi(r, s, t, u; z)$ lies outside the circle $C_1$ which is sufficient to deduce that $\phi(r, s, t, u; z) \notin \Omega$ . Therefore, $\phi \in \Psi[\Omega, 1 + \sin z]$ and thus the result follows as a consequence of Lemma 1.5.

Theorem 2.16. Suppose $\beta_1$ , $\beta_2$ , $\beta_3 > 0$ and $\nu_0(\beta_1 + \beta_2\nu_1 - m^2\beta_3 + 3m\beta_3(k-1)\nu_1) \ge e$ . Let p be analytic function in $\mathbb D$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) + \beta_3 z^3 p'''(z) \prec 1 + z e^z.$$ Then $p(z) \prec 1 + \sin z$ .

Theorem 2.17. Suppose $\beta_1$ , $\beta_2$ , $\beta_3 > 0$ and $2\nu_0(\beta_1 + \beta_2\nu_1 - m^2\beta_3 + 3m\beta_3(k-1)\nu_1) \ge \pi$ . Let p be analytic function in $\mathbb D$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) + \beta_3 z^3 p'''(z) \prec 1 + \sinh^{-1} z.$$ Then $p(z) \prec 1 + \sin z$ . Proof. Assume $h(z) = 1 + \sinh^{-1} z$ for $z \in \mathbb{D}$ . Then, $h(\mathbb{D}) = \{w \in \mathbb{C} : |\sinh(w-1)| < 1\} =: \Omega_{\rho}$ . Let $\phi : \mathbb{C}^4 \times \mathbb{D} \to \mathbb{C}$ be defined as $\phi(r, s, t, u; z) = 1 + \beta_1 s + \beta_2 t + \beta_3 u$ . For $\phi \in \Psi[\Omega_{\rho}, 1 + \sin z]$ , we must have $\phi(r, s, t, u; z) \notin \Omega_{\rho}$ . Through [2, Remark 2.7], we note that the smallest disk containing $\Omega_{\rho}$ is $\{\delta \in \mathbb{C} : |\delta - 1| < \pi/2\}$ . Thus $$|\phi(r, s, t, u; z) - 1| = |\beta_1 s + \beta_2 t + \beta_3 u|.$$ Proceeding on the similar lines of Theorem 2.14, we have $$|\beta_1 s + \beta_2 t + \beta_3 u| \ge n_1(\theta) \left( \beta_1 + \beta_2 n_2(\theta) + \beta_3 (m^2 n_3(\theta) + 3m(k-1)n_2(\theta)) \right)$$ $$\ge \nu_0(\beta_1 + \beta_2 \nu_1 - m^2 \beta_3 + 3m\beta_3 (k-1)\nu_1)$$ $$\ge \frac{\pi}{2}.$$ Clearly, $\phi(r, s, t, u; z)$ lies outside the disk $\{\delta \in \mathbb{C} : |\delta - 1| < \pi/2\}$ which suffices to prove that $\phi(r, s, t, u; z) \notin \Omega_{\rho}$ . Therefore, $\phi \in \Psi[\Omega_{\rho}, 1 + \sin z]$ and thus the result follows as an application of Lemma 1.5.

Theorem 2.18. Suppose $\beta_1$ , $\beta_2$ , $\beta_3 > 0$ and $\nu_0(\beta_1 + \beta_2\nu_1 - m^2\beta_3 + 3m\beta_3(k-1)\nu_1) \ge e-1$ . Let p be analytic function in $\mathbb D$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) + \beta_3 z^3 p'''(z) \prec e^z.$$ Then $p(z) \prec 1 + \sin z$ . Proof. Take $h(z) = e^z$ for $z \in \mathbb{D}$ . Then, $h(\mathbb{D}) = \{w \in \mathbb{C} : |\log w| < 1\} =: \Omega$ . Let $\phi : \mathbb{C}^4 \times \mathbb{D} \to \mathbb{C}$ be defined as $\phi(r, s, t, u; z) = 1 + \beta_1 s + \beta_2 t + \beta_3 u$ . For $\phi \in \Psi[\Omega, 1 + \sin z]$ , we must have $\phi(r, s, t, u; z) \notin \Omega$ . So, $$|\phi(r, s, t, u; z) - 1| = |\beta_1 s + \beta_2 t + \beta_3 u|.$$ Proceeding on the similar lines of proof of Theorem 2.14, we get $$|\beta_1 s + \beta_2 t + \beta_3 u| \ge n_1(\theta) \left( \beta_1 + \beta_2 n_2(\theta) + \beta_3 (m^2 n_3(\theta) + 3m(k-1)n_2(\theta)) \right)$$ $$\ge \nu_0(\beta_1 + \beta_2 \nu_1 - m^2 \beta_3 + 3m\beta_3 (k-1)\nu_1)$$ $$\ge e - 1. \tag{2.13}$$ Next, we consider $$|\log(\phi(r, s, t, u; z))| = |\log(1 + \beta_1 s + \beta_2 t + \beta_3 u)|.$$ Through Lemma 1.7 and (2.13), we have <span id="page-12-0"></span> $$|\log(1 + \beta_1 s + \beta_2 t + \beta_3 u)| \ge 1,$$ which implies that $\phi(r, s, t, u; z) \notin \Omega$ . Therefore, $\phi \in \Psi[\Omega, 1 + \sin z]$ and the result follows as a consequence of Lemma 1.5.

Theorem 2.19. Suppose $\beta_1$ , $\beta_2$ , $\beta_3 > 0$ and $\nu_0(\beta_1 + \beta_2\nu_1 - m^2\beta_3 + 3m\beta_3(k-1)\nu_1) \ge \sinh 1$ . Let p be analytic function in $\mathbb D$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) + \beta_3 z^3 p'''(z) \prec 1 + \sin z.$$ Then $p(z) \prec 1 + \sin z$ .

Theorem 3.1. Suppose $\beta_1$ , $\beta_2 > 0$ and $(2\beta_1 - \beta_2)(2\beta_1 - \beta_2 - 4\sqrt{2}) \ge 8$ . Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec \sqrt{1+z}$$ . Then $p(z) \prec 1 + \sinh^{-1} z$ . Proof. Consider $h(z) = \sqrt{1+z}$ for $z \in \mathbb{D}$ . Then, $h(\mathbb{D}) = \{w \in \mathbb{C} : |w^2 - 1| < 1\} =: \Omega$ . Let $\phi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ be defined as $\phi(r, s, t; z) = 1 + \beta_1 s + \beta_2 t$ . It is clear that $\phi \in \Psi[\Omega, 1 + \sinh^{-1} z]$ provided $\phi(r, s, t; z) \notin \Omega$ for $z \in \mathbb{D}$ . Note that $$\begin{aligned} |(\phi(r, s, t; z))^{2} - 1| &= |(1 + \beta_{1}s + \beta_{2}t)^{2} - 1| \\ &\geq |\beta_{1}s + \beta_{2}t|(|\beta_{1}s + \beta_{2}t| - 2) \\ &\geq |\beta_{1}s| \operatorname{Re}\left(1 + \frac{\beta_{2}}{\beta_{1}} \frac{t}{s}\right) \left(|\beta_{1}s| \operatorname{Re}\left(1 + \frac{\beta_{2}}{\beta_{1}} \frac{t}{s}\right) - 2\right) \\ &\geq m\beta_{1}n_{4}(\theta) \operatorname{Re}\left(1 + \frac{\beta_{2}}{\beta_{1}}(mn_{5}(\theta) + m - 1)\right) \left(m\beta_{1}n_{4}(\theta) \operatorname{Re}\left(1 + \frac{\beta_{2}}{\beta_{1}}(mn_{5}(\theta) + m - 1)\right) - 2\right). \end{aligned}$$ Here $n_4(\theta)$ and $n_5(\theta)$ are given in (3.1) and (3.3). Since $m \ge 1$ , therefore $$|(\phi(r,s,t;z))^{2}-1| \geq \beta_{1}n_{4}(\theta)\operatorname{Re}\left(1+\frac{\beta_{2}}{\beta_{1}}n_{5}(\theta)\right)\left(\beta_{1}n_{4}(\theta)\operatorname{Re}\left(1+\frac{\beta_{2}}{\beta_{1}}n_{5}(\theta)\right)-2\right)$$ $$\geq \frac{1}{\sqrt{2}}\left(\beta_{1}-\frac{\beta_{2}}{2}\right)\left(\frac{1}{\sqrt{2}}\left(\beta_{1}-\frac{\beta_{2}}{2}\right)-2\right)$$ $$=\frac{(2\beta_{1}-\beta_{2})(2\beta_{1}-\beta_{2}-4\sqrt{2})}{8}$$ $$\geq 1.$$ Therefore, $\phi(r, s, t; z) \notin \Omega$ and hence $\phi \in \Psi[\Omega, 1 + \sinh^{-1} z]$ . Thus, result follows as an application of Lemma 1.3.

Theorem 3.2. Suppose $\beta_1$ , $\beta_2 > 0$ and $-1 < D < C \le 1$ with $(2\beta_1 - \beta_2)(1 - D^2) \ge 2\sqrt{2}(C - D)(1 + |D|)$ . Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec \frac{1 + Cz}{1 + Dz}$$ Then $p(z) \prec 1 + \sinh^{-1} z$ . Proof. Assume h(z) = (1 + Cz)/(1 + Dz) for $z \in \mathbb{D}$ . Then, $h(\mathbb{D}) = \{w \in \mathbb{C} : |w - (1 - CD)/(1 - D^2)| < (C - D)/(1 - D^2)\} =: \Omega$ . Let $\phi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ be defined as $\phi(r, s, t; z) = 1 + \beta_1 s + \beta_2 t$ . It is clear that $\phi \in \Psi[\Omega, 1 + \sinh^{-1} z]$ if $\phi(r, s, t; z) \notin \Omega$ for $z \in \mathbb{D}$ . Consider $$\begin{split} \left| \phi(r, s, t; z) - \frac{1 - CD}{1 - D^2} \right| &= \left| 1 + \beta_1 s + \beta t - \frac{1 - CD}{1 - D^2} \right| \\ &\geq \left| \beta_1 s + \beta_2 t \right| - \frac{|D|(C - D)}{1 - D^2} \\ &\geq \left| \beta_1 s \right| \operatorname{Re} \left( 1 + \frac{\beta_2}{\beta_1} \frac{t}{s} \right) - \frac{|D|(C - D)}{1 - D^2} \\ &\geq m \beta_1 n_4(\theta) \operatorname{Re} \left( 1 + \frac{\beta_2}{\beta_1} (m n_5(\theta) + m - 1) \right) - \frac{|D|(C - D)}{1 - D^2}. \end{split}$$ Using the fact that $m \geq 1$ and proceeding on the similar lines of the proof of Theorem 3.1, we have $$\left| \phi(r, s, t; z) - \frac{1 - CD}{1 - D^2} \right| \ge \frac{1}{\sqrt{2}} \left( \beta_1 - \frac{\beta_2}{2} \right) - \frac{|D|(C - D)}{1 - D^2} \\ \ge \frac{C - D}{1 - D^2}.$$ Thus, $\phi \in \Psi[\Omega, 1 + \sinh^{-1} z]$ and the result follows as an application of Lemma 1.3.

Theorem 3.3. Suppose $\beta_1$ , $\beta_2 > 0$ and $2\beta_1 - \beta_2 \ge 2\sqrt{2}r_0$ , where $r_0 \approx 0.546302$ is the positive root of the equation $r^2 + 2\cot(1)r - 1 = 0$ . Let p be analytic function in $\mathbb D$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec \frac{2}{1 + e^{-z}}.$$ Then $p(z) \prec 1 + \sinh^{-1} z$ . Proof. Suppose $h(z)=2/(1+e^{-z})$ for $z\in\mathbb{D}$ . Then, $h(\mathbb{D})=\{w\in\mathbb{C}:|\log(w/(2-w))|<1\}=:\Omega$ . Let $\phi:\mathbb{C}^3\times\mathbb{D}\to\mathbb{C}$ be defined as $\phi(r,s,t;z)=1+\beta_1s+\beta_2t$ . We know that $\phi\in\Psi[\Omega,1+\sinh^{-1}z]$ only when $\phi(r,s,t;z)\notin\Omega$ . First we consider, $$|\beta_1 s + \beta_2 t| = |\beta_1 s| \left| 1 + \frac{\beta_2}{\beta_1} \frac{t}{s} \right|$$ $$\geq |\beta_1 s| \operatorname{Re} \left( 1 + \frac{\beta_2}{\beta_1} \frac{t}{s} \right)$$ $$\geq m\beta_1 n_4(\theta) \left( 1 + \frac{\beta_2}{\beta_1} (mn_5(\theta) + m - 1) \right).$$ As $m \geq 1$ , we obtain <span id="page-15-1"></span> $$|\beta_1 s + \beta_2 t| \ge n_4(\theta)(\beta_1 + \beta_2 n_5(\theta))$$ $$\ge \frac{1}{\sqrt{2}} \left(\beta_1 - \frac{1}{2}\beta_2\right)$$ $$\ge r_0. \tag{3.5}$$ Now, we consider $$\left|\log\left(\frac{\phi(r,s,t;z)}{2-\phi(r,s,t;z)}\right)\right| = \left|\log\left(\frac{1+\beta_1s+\beta_2t}{1-(\beta_1s+\beta_2t)}\right)\right|.$$ Through Lemma 1.6 and (3.5), we have $$\left| \log \left( \frac{1 + \beta_1 s + \beta_2 t}{1 - (\beta_1 s + \beta_2 t)} \right) \right| \ge 1,$$ which implies that $\phi \in \Psi[\Omega, 1 + \sinh^{-1} z]$ . Therefore, result follows as an application of Lemma 1.3.

Theorem 3.4. Suppose $\beta_1$ , $\beta_2 > 0$ and $2\beta_1 - \beta_2 \ge 4$ . Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec z + \sqrt{1 + z^2}.$$ Then $p(z) \prec 1 + \sinh^{-1} z$ . Proof. Take $h(z)=z+\sqrt{1+z^2}$ for $z\in\mathbb{D}$ . Then, $h(\mathbb{D})=\{w\in\mathbb{C}:|w^2-1|<2|w|\}=:\Omega$ . Let $\phi:\mathbb{C}^3\times\mathbb{D}\to\mathbb{C}$ be defined as $\phi(r,s,t;z)=1+\beta_1s+\beta_2t$ . For $\phi\in\Psi[\Omega,1+\sinh^{-1}z]$ , we must have $\phi(r,s,t;z)\notin\Omega$ . From the geometry of $z+\sqrt{1+z^2}$ (see Fig. 1), we note that $\Omega$ is constructed by the circles $$C_1: |z-1| = \sqrt{2}$$ and $C_2: |z+1| = \sqrt{2}$ . It is obvious that $\Omega$ contains the disk enclosed by $C_1$ and excludes the portion of the disk enclosed by $C_1 \cap C_2$ . We have $$|\phi(r, s, t; z) - 1| = |\beta_1 s + \beta_2 t|.$$ Similar to the proof of Theorem 3.3, we have $$|\beta_1 s + \beta_2 t| \ge n_4(\theta)(\beta_1 + \beta_2 n_5(\theta))$$ $$\ge \frac{1}{\sqrt{2}} \left(\beta_1 - \frac{1}{2}\beta_2\right)$$ $$\ge \sqrt{2}.$$ The fact that $\phi(r, s, t; z)$ lies outside the circle $C_1$ suffices us to deduce that $\phi(r, s, t; z) \notin \Omega$ . Consequently, $\phi \in \Psi[\Omega, 1 + \sinh^{-1} z]$ and thus the result follows as an application of Lemma 1.3.

Theorem 3.5. Suppose $\beta_1$ , $\beta_2 > 0$ and $2\beta_1 - \beta_2 \ge 2\sqrt{2}\sinh 1$ . Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec 1 + \sin z.$$ Then $p(z) \prec 1 + \sinh^{-1} z$ . Proof. Consider $h(z) = 1 + \sin z$ for $z \in \mathbb{D}$ . Then, $h(\mathbb{D}) = \{w \in \mathbb{C} : |\arcsin(w-1)| < 1\} =: \Omega_s$ . Let $\phi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ be defined as $\phi(r, s, t; z) = 1 + \beta_1 s + \beta_2 t$ . For $\phi \in \Psi[\Omega_s, 1 + \sinh^{-1} z]$ , we must have $\phi(r, s, t; z) \notin \Omega_s$ . From [3, Lemma 3.3], we note that the smallest disk containing $\Omega_s$ is $\{\delta \in \mathbb{C} : |\delta - 1| < \sinh 1\}$ . So, $$|\phi(r, s, t; z) - 1| = |\beta_1 s + \beta_2 t|.$$ Proceeding on the same lines of the proof of Theorem 3.3, we have $$|\beta_1 s + \beta_2 t| \ge n_4(\theta)(\beta_1 + \beta_2 n_5(\theta))$$ $$\ge \frac{1}{\sqrt{2}} \left(\beta_1 - \frac{1}{2}\beta_2\right)$$ $$\ge \sinh 1.$$ Clearly, $\phi(r, s, t; z)$ lies outside the disk $\{\delta \in \mathbb{C} : |\delta - 1| < \sinh 1\}$ , which is enough to conclude that $\phi(r, s, t; z) \notin \Omega_s$ . Therefore, $\phi \in \Psi[\Omega_s, 1 + \sinh^{-1} z]$ and the result follows as an application of Lemma 1.3.

Theorem 3.6. Let $\beta_1$ , $\beta_2 > 0$ and $2\beta_1 - \beta_2 \ge 2\sqrt{2}e$ . Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec 1 + z e^z$$ . Then $p(z) \prec 1 + \sinh^{-1} z$ .

Theorem 3.7. Suppose $\beta_1$ , $\beta_2 > 0$ and $2\beta_1 - \beta_2 \ge 2\sqrt{2}(e-1)$ . Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec e^z$$ . Then $p(z) \prec 1 + \sinh^{-1} z$ . Proof. Take $h(z) = e^z$ for $z \in \mathbb{D}$ . Then, $h(\mathbb{D}) = \{w \in \mathbb{C} : |\log w| < 1\} =: \Omega$ . Let $\phi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ be defined as $\phi(r, s, t; z) = 1 + \beta_1 s + \beta_2 t$ . For $\phi \in \Psi[\Omega, 1 + \sinh^{-1} z]$ , we must have $\phi(r, s, t; z) \notin \Omega$ . So, $$|\phi(r, s, t; z) - 1| = |\beta_1 s + \beta_2 t|.$$ Similar to the proof of Theorem 3.3, we have $$|\beta_1 s + \beta t| \ge n_4(\theta)(\beta_1 + \beta_2 n_5(\theta))$$ $$\ge \frac{1}{\sqrt{2}} \left(\beta_1 - \frac{1}{2}\beta_2\right)$$ $$\ge e - 1. \tag{3.6}$$ Further, we have $$|\log(\phi(r, s, t; z))| = |\log(1 + \beta_1 s + \beta_2 t)|.$$ Through Lemma 1.7 and (3.6), we have <span id="page-17-0"></span> $$|\log(1+\beta_1s+\beta_2t)| \ge 1,$$ which implies that $\phi(r, s, t; z) \notin \Omega$ . Therefore, $\phi \in \Psi[\Omega, 1 + \sinh^{-1} z]$ and thus the result follows as an application of Lemma 1.3.

Theorem 3.8. Suppose $\beta_1$ , $\beta_2 > 0$ and $2\beta_1 - \beta_2 \ge \sqrt{2}\pi$ . Let p be analytic function in $\mathbb{D}$ with p(0) = 1 and $$1 + \beta_1 z p'(z) + \beta_2 z^2 p''(z) \prec 1 + \sinh^{-1} z.$$ Then $p(z) \prec 1 + \sinh^{-1} z$ .

Definition 1.1. [1] Let $\phi(r, s, t, u; z) : \mathbb{C}^4 \times \mathbb{D} \to \mathbb{C}$ and h(z) be a univalent function in $\mathbb{D}$ , if p is an analytic function in $\mathbb{D}$ satisfying the third-order differential subordination <span id="page-1-0"></span> $$\phi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z) \prec h(z)$$ (1.2) then p is called the solution of the differential subordination. The univalent function q is said to be a dominant of the solutions of the differential subordination if $p \prec q$ for all p satisfying (1.2). A dominant $\bar{q}$ that satisfies $\bar{q} \prec q$ for all dominants q of (1.2) is said to be the best dominant of (1.2), which is unique upto the rotations of $\mathbb{D}$ . Moreover, suppose Q be the set of analytic and univalent functions $q \in \overline{\mathbb{D}} \setminus \mathbb{E}(q)$ , where $$\mathbb{E}(q) = \{ \zeta \in \partial \mathbb{D} : \lim_{z \to \zeta} q(z) = \infty \}$$ such that $q'(\zeta) \neq 0$ for $\zeta \in \partial \mathbb{D} \setminus \mathbb{E}(q)$ . The subclass of Q for which q(0) = a is denoted by Q(a).

Definition 1.4. [9] Let $\Omega$ be a set in $\mathbb{C}$ , $q \in Q$ and $k \geq m \geq n \geq 2$ . The class of admissible operators $\Psi_n[\Omega, q]$ consists of those $\phi : \mathbb{C}^4 \times \mathbb{D} \to \mathbb{C}$ that satisfy the admissibility conditions $$\phi(r, s, t, u; z) \notin \Omega$$ whenever $z \in \mathbb{D}$ , $$r = q(\zeta), \quad s = m\zeta q'(\zeta), \quad \operatorname{Re}\left(1 + \frac{t}{s}\right) \ge m\left(1 + \operatorname{Re}\frac{\zeta q''(\zeta)}{q'(\zeta)}\right)$$ and $$\operatorname{Re} \frac{u}{s} \ge m^2 \operatorname{Re} \frac{\zeta^2 q'''(\zeta)}{q'(\zeta)} + 3m(k-1) \operatorname{Re} \frac{\zeta q''(\zeta)}{q'(\zeta)} \quad \text{for} \quad \zeta \in \partial \mathbb{D} \setminus \mathbb{E}(q).$$