A new differential subordination technique for a subclass of starlike functions

Lemma 1.1. [8] Let $w \in \mathbb{C}$ such that $|w| > \sqrt{2}$ and $R_0 = e\sqrt{2/(e^2 - 1)} \simeq 1.5208$ . Then $$\left|\log\left(\frac{w^2}{2-w^2}\right)\right| \ge 2$$ , whenever $|w| \ge R_0$ .

Lemma 1.4. [12] Let $\psi \in \Psi(\Omega, q)$ with q(0) = 1. If $p \in \mathcal{H}$ satisfies $$\psi(p(z), zp'(z), z^2p''(z); z) \in \Omega,$$ then $p \prec q$ . Throughout this article, we consider the function $q(z) = \mathfrak{B}(z)$ and define the admissibility class $\Psi(\Omega, q)$ , where $\Omega \subset \mathbb{C}$ . Since $\mathcal{E}(\mathfrak{B}) = \emptyset$ , therefore $\mathfrak{B} \in \mathcal{Q}$ . For $\zeta = e^{i\theta}$ $(0 \le \theta < 2\pi)$ we have, $$q(\zeta) = \sqrt{\frac{2}{1 + e^{-2e^{i\theta}}}}, \quad \zeta q'(\zeta) = \frac{\sqrt{2}e^{i\theta}e^{-2e^{i\theta}}}{(1 + e^{-2e^{i\theta}})^{3/2}}$$ and <span id="page-2-5"></span> $$\operatorname{Re}\left(\frac{\zeta q''(\zeta)}{q'(\zeta)}\right) = \frac{\cos\theta - 2e^{4\cos\theta}\cos\theta + e^{2\cos\theta}(\cos(\theta - 2\sin\theta) - 2\cos(\theta + 2\sin\theta))}{1 + e^{4\cos\theta} + 2e^{2\cos\theta}\cos(2\sin\theta)} =: g(\theta). \quad (1.3)$$ Here, <span id="page-2-1"></span> $$|q(\zeta)| = \frac{\sqrt{2}}{(1 + e^{-4\cos\theta} + 2e^{-2\cos\theta}\cos(2\sin\theta))^{\frac{1}{4}}} =: \omega(\theta). \tag{1.4}$$ From [8], we obtain that the minimum value of $\omega(\theta)$ is $\omega(\pi) = \sqrt{2/(1+e^2)}$ and maximum is $\omega(\theta_0) =: \omega_{\theta_0} \simeq 1.438$ , where $\theta_0 \simeq 1.31364$ . By simple analysis we also obtain that min $g(\theta) = g(0)$ , and <span id="page-2-2"></span> $$|\zeta q'(\zeta)| = \frac{\sqrt{2}e^{-2\cos\theta}}{(1 + e^{-4\cos\theta} + 2e^{-2\cos\theta}\cos(2\sin\theta))^{3/4}} := d(\theta).$$ (1.5) Further, a computation shows that $d'(\theta) = 0$ if and only if $\theta \in \{0, \theta_1, \pi\}$ where $\theta_1 \simeq 1.639$ . We observe that $d(\theta)$ is increasing in $[0, \theta_1]$ and decreasing in $[\theta_1, \pi]$ . Therefore, $\max d(\theta) = d(1.639) =: d_{\theta_1} \simeq 0.905$ and $\min d(\theta) = \min\{d(0), d(\pi)\} = \min\{0.158, 0.304\} = d(0)$ . Moreover, $\Psi(\Omega, \mathfrak{B})$ is defined to be the class of functions $\psi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ , which satisfies the following conditions: $$\psi(r, s, t; z) \not\in \Omega$$ , whenever $$r = \mathfrak{B}(\zeta) = \sqrt{\frac{2}{1 + e^{-2e^{i\theta}}}}; \quad s = m\zeta\mathfrak{B}'(\zeta) = \frac{\sqrt{2}me^{i\theta}e^{-2e^{i\theta}}}{(1 + e^{-2e^{i\theta}})^{3/4}}; \quad \text{Re}\left(1 + \frac{s}{t}\right) = m(1 + g(\theta)),$$ where $z \in \mathbb{D}$ , $0 \le \theta < 2\pi$ and $m \ge 1$ . Here, we start from first order differential subordination.

Lemma 2.1. Let -1 < B < 0 < A < 1 and $w \in \mathbb{C}$ such that |w| > 1, then $$\left| \frac{w-1}{A-Bw} \right| \ge 1$$ , whenever $|w| \ge \frac{1+A}{1+B}$ .

Lemma 2.2. Let $w \in \mathbb{C}$ such that |w| > 1, then $$|w^2 - 1| \ge 1$$ , whenever $|w| \ge \sqrt{2}$ .

Theorem 2.3. Let n be a positive integer, $\delta \in \{0,1\}$ , $R_0 = e\sqrt{2/(e^2-1)}$ , $\omega_{\theta_0} \simeq 1.438$ and $\beta \in \mathbb{C}$ such that $|\beta| \geq (R_0 + \omega_{\theta_0}^{\delta})(1 + e^2)^{3n/2}/(\sqrt{2}e)^n$ . Suppose $p(z) \in \mathcal{H}$ and satisfies the subordination $$(p(z))^{\delta} + \beta (zp'(z))^n \prec \mathfrak{B}(z),$$ then $p(z) \prec \mathfrak{B}(z)$ .

Theorem 2.5. Let n be a positive integer, $\delta \in \{0,1\}$ , $R_0 = e\sqrt{2/(e^2-1)}$ , $\omega_{\theta_0} \simeq 1.438$ and $\beta \in \mathbb{C}$ such that $|\beta| \geq (R_0 + \omega_{\theta_0}^{\delta})(1 + e^2)^{(3-n)/2}/(\sqrt{2}e)^{1-n}$ . Suppose $p(z) \in \mathcal{H}$ and satisfies the subordination $$(p(z))^{\delta} + \beta \frac{zp'(z)}{(p(z))^n} \prec \mathfrak{B}(z),$$ then $p(z) \prec \mathfrak{B}(z)$ .

Theorem 2.7. Let $R_0 = e\sqrt{2/(e^2 - 1)}$ , $\omega_{\theta_0} \simeq 1.438$ and $\beta, \gamma \in \mathbb{C}$ such that $\beta \neq 0$ with $|\beta|\omega_{\theta_0} + |\gamma| \leq \sqrt{2}e/((1 + e^2)^{3/2}(R_0 + \omega_{\theta_0}))$ . Suppose $p(z) \in \mathcal{H}$ and satisfies the subordination $$p(z) + \frac{zp'(z)}{\beta p(z) + \gamma} \prec \mathfrak{B}(z),$$ then $p(z) \prec \mathfrak{B}(z)$ . Proof. Let $q(z) = \sqrt{2/(1 + e^{-2z})} =: \mathfrak{B}(z)$ and $\Omega = q(\mathbb{D}) = \Omega_{\mathfrak{B}}$ . Now consider $\psi : \mathbb{C}^2 \times \mathbb{D} \to \mathbb{C}$ be as $\psi(r, s; z) = r + s/(\beta r + \gamma)$ . For $\psi$ to lie in the admissible class $\Psi(\Omega_{\mathfrak{B}}, \mathfrak{B})$ , we must have $$\psi\left(\sqrt{\frac{2}{1+e^{-2e^{i\theta}}}}, \frac{\sqrt{2}me^{i\theta}e^{-2e^{i\theta}}}{(1+e^{-2e^{i\theta}})^{3/2}}; z\right) \notin \Omega_{\mathfrak{B}} \qquad (z \in \mathbb{D}, 0 \le \theta < 2\pi, m \ge 1),$$ which is equivalent to $$\left|\log\left(\frac{(r+s/(\beta r+\gamma))^2}{2-(r+s/(\beta r+\gamma))^2}\right)\right| \ge 2.$$ We now use the Lemma 1.1 to prove the result, for which, let us assume $w = r + s/(\beta r + \gamma)$ . From (1.4) and (1.5), we have $|r| = \omega(\theta) \le \omega_{\theta_0}$ and $|s|/m = d(\theta) \ge \min d(\theta) = d(0)$ . Thus $$|w| = \left| r + \frac{s}{\beta r + \gamma} \right|$$ $$\geq \frac{|s|}{|\beta||r| + |\gamma|} - |r|$$ $$\geq \frac{md(0)}{|\beta|\omega_{\theta_0} + |\gamma|} - \omega_{\theta_0}.$$ Since $m \ge 1$ and $|\beta|\omega_{\theta_0} + |\gamma| \le \sqrt{2}e/((1+e^2)^{3/2}(R_0 + \omega_{\theta_0}))$ , we have $$|w| \ge \frac{\sqrt{2}e}{(1+e^2)^{3/2}(|\beta|\omega_{\theta_0}+|\gamma|)} - \omega_{\theta_0} \ge R_0 > \sqrt{2},$$ therefore the result follows at once by Lemma 1.1. Corollary 2.8. Let $R_0 = e\sqrt{2/(e^2-1)}$ , $\omega_{\theta_0} \simeq 1.438$ and $\beta, \gamma \in \mathbb{C}$ such that $\beta \neq 0$ with $\omega_{\theta_0}|\beta|+|\gamma| \leq \sqrt{2}e/((1+e^2)^{3/2}(R_0+\omega_{\theta_0}))$ . Suppose f be a function in $\mathcal{A}$ such that $$\frac{zf'(z)}{f(z)}\left(1+\left(\beta\frac{zf'(z)}{f(z)}+\gamma\right)^{-1}\left(1+\frac{zf''(z)}{f'(z)}-\frac{zf'(z)}{f(z)}\right)\right)\prec\mathfrak{B}(z),$$ then $f \in \mathcal{S}_{\mathfrak{B}}^*$ .

Theorem 2.9. Let $n \in \mathbb{N}$ , $\delta \in \{0,1\}$ , $\beta \in \mathbb{C} \setminus \{0\}$ , $\omega_{\theta_0} \simeq 1.438$ and $-1 < B \leq 0 < A \leq 1$ . Suppose $p \in \mathcal{H}$ such that it satisfies $$(p(z))^{\delta} + \beta (zp'(z))^n \prec \frac{1 + Az}{1 + Bz}$$ where <span id="page-6-0"></span> $$|\beta| \ge \left(\frac{(1+e^2)^{3/2}}{\sqrt{2}e}\right)^n \left(\frac{1+A}{1+B} + \omega_{\theta_0}^{\delta}\right) \tag{2.2}$$ then $p(z) \prec \mathfrak{B}(z)$ . Proof. Let $q(z) = \sqrt{2/(1+e^{-2z})} =: \mathfrak{B}(z)$ and $\Omega = \{w \in \mathbb{C} : |(w-1)/(A-Bw)| < 1\}$ . Let $\psi : \mathbb{C}^2 \times \mathbb{D} \to \mathbb{C}$ be defined as $\psi(r,s;z) = r^{\delta} + \beta s^n$ where $r = \mathfrak{B}(\xi)$ , $s = m\xi\mathfrak{B}'(\xi) = m\sqrt{2}\xi e^{-2\xi}/(1+e^{-2\xi})^{3/2}$ for $z \in \mathbb{D}$ , $\xi \in \overline{\mathbb{D}}$ and $m \geq 1$ . For $\psi$ to be an admissible function and for $\psi \in \Psi[\Omega,\mathfrak{B}]$ we need to show $\psi(\mathbb{D}) \not\subseteq \Omega$ , which is equivalent to $$\left| \frac{r^{\delta} + \beta s^n - 1}{A - B(r^{\delta} + \beta s^n)} \right| \ge 1.$$ We now use the Lemma 2.1 to prove the result, for which, let us assume $w = r^{\delta} + \beta s^{n}$ . From (1.4) and (1.5), we have $|r| = \omega(\theta) \le \omega_{\theta_0}$ and $|s|/m = d(\theta) \ge \min d(\theta) = d(0)$ . Thus $$|w| = |r^{\delta} + \beta s^{n}|$$ $$\geq |\beta||s|^{n} - |r|^{\delta}$$ $$\geq |\beta|m^{n}(d(0))^{n} - \omega_{\theta_{0}}^{\delta}$$ Since $m \geq 1$ and using (2.2), we have $$|w| \ge |\beta| \left( \frac{\sqrt{2}e}{(1+e^2)^{3/2}} \right)^n - \omega_{\theta_0}^{\delta} \ge \frac{1+A}{1+B} > 1,$$ therefore the result follows at once from Lemma 2.1.

Corollary 2.10. Let $n \in \mathbb{N}$ , $\delta \in \{0,1\}$ and $\beta \in \mathbb{C} \setminus \{0\}$ such that $\beta$ satisfies inequality (2.2). Moreover, let $\omega_{\theta_0} \simeq 1.438$ and $-1 < B \le 0 < A \le 1$ . Suppose $f \in \mathcal{A}$ such that $$\left(\frac{zf'(z)}{f(z)}\right)^{\delta} + \beta \left(\frac{zf'(z)}{f(z)} \left(1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)}\right)\right)^{n} \prec \frac{1 + Az}{1 + Bz},$$ then $f \in \mathcal{S}_{\mathfrak{B}}^*$ .

Theorem 2.11. Let $n \in \mathbb{N}$ , $\delta \in \{0,1\}$ , $\beta \in \mathbb{C} \setminus \{0\}$ , $\omega_{\theta_0} \simeq 1.438$ and $-1 \leq B < 0 \leq A \leq 1$ . Suppose $p \in \mathcal{H}$ such that it satisfies $$(p(z))^{\delta} + \beta \frac{zp'(z)}{(p(z))^n} \prec \frac{1 + Az}{1 + Bz},$$ where <span id="page-6-2"></span> $$|\beta| \ge \left(\frac{(1+e^2)^{\frac{3-n}{2}}}{(\sqrt{2}e)^{1-n}}\right) \left(\frac{1+A}{1+B} + \omega_{\theta_0}^{\delta}\right).$$ (2.3) Then $p(z) \prec \mathfrak{B}(z)$ . Proof. Let $q(z) = \sqrt{2/(1 + e^{-2z})} =: \mathfrak{B}(z)$ and $\Omega = \{w \in \mathbb{C} : |(w-1)/(A - Bw)| < 1\}$ . Let $\psi : \mathbb{C}^2 \times \mathbb{D} \to \mathbb{C}$ be defined as $\psi(r, s; z) = r^{\delta} + \beta s/r^n$ where $r = \mathfrak{B}(\xi)$ , $s = m\xi\mathfrak{B}'(\xi) = m\sqrt{2}\xi e^{-2\xi}/(1 + e^{-2\xi})^{3/2}$ for $z \in \mathbb{D}$ , $\xi \in \overline{\mathbb{D}}$ and $m \geq 1$ . For $\psi$ to be an admissible function and for $\psi \in \Psi[\Omega, \mathfrak{B}]$ we need to show $\psi(\mathbb{D}) \not\subseteq \Omega$ , which is equivalent to $$\left| \frac{r^{\delta} + \beta \frac{s}{r^n} - 1}{A - B(r^{\delta} + \beta \frac{s}{r^n})} \right| \ge 1.$$ We now use the Lemma 2.1 to prove the result, for which, let us assume $w = r^{\delta} + \beta s/r^{n}$ . From (1.4), we have $|r| = \omega(\theta) \le \omega_{\theta_0}$ . Thus $$|w| = \left| r^{\delta} + \beta \frac{s}{r^n} \right|$$ $$\geq |\beta| \left( \frac{2^{\frac{1-n}{2}} m e^{-2\cos\theta}}{(1 + e^{-4\cos\theta} + 2e^{-2\cos\theta}\cos(2\sin\theta))^{\frac{3-n}{4}}} \right) - (\omega(\theta))^{\delta}$$ $$\geq |\beta| m \chi_n(\theta) - (\omega(\theta))^{\delta},$$ where $$\chi_n(\theta) := \frac{2^{\frac{1-n}{2}} e^{-2\cos\theta}}{(1 + e^{-4\cos\theta} + 2e^{-2\cos\theta}\cos(2\sin\theta))^{\frac{3-n}{4}}}.$$ After a simple computation we obtain that $\min \chi_n(\theta) = \chi_n(0)$ for all $n \geq 1$ . Since $m \geq 1$ and using (2.3), we have $$|w| \ge |\beta| \left( \frac{(\sqrt{2}e)^{1-n}}{(1+e^2)^{\frac{3-n}{2}}} \right) - \omega_{\theta_0}^{\delta} \ge \frac{1+A}{1+B} > 1,$$ therefore the result follows from Lemma 2.1.

Corollary 2.12. Let $n \in \mathbb{N}$ , $\delta \in \{0,1\}$ and $\beta \in \mathbb{C} \setminus \{0\}$ such that $\beta$ satisfies inequality (2.3). Moreover, let $\omega_{\theta_0} \simeq 1.438$ and $-1 < B \le 0 < A \le 1$ . Suppose f be a function in A such that $$\left(\frac{zf'(z)}{f(z)}\right)^{\delta} + \beta \left(\frac{zf'(z)}{f(z)}\right)^{1-n} \left(1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)}\right) \prec \frac{1 + Az}{1 + Bz},$$ then $f \in \mathcal{S}_{\mathfrak{B}}^*$ .

Theorem 2.13. Let $\beta, \gamma \in \mathbb{C}$ such that $\beta \neq 0$ , $\omega_{\theta_0} \simeq 1.438$ and $-1 < B \leq 0 < A \leq 1$ . Suppose $p \in \mathcal{H}$ such that it satisfies $$p(z) + \frac{zp'(z)}{\beta p(z) + \gamma} \prec \frac{1 + Az}{1 + Bz}$$ provided <span id="page-7-0"></span> $$|\beta|\omega_{\theta_0} + |\gamma| \le \left(\frac{(1+e^2)^{3/2}}{\sqrt{2}e} \left(\frac{1+A}{1+B} + \omega_{\theta_0}\right)\right)^{-1}$$ (2.4) where Then $p(z) \prec \mathfrak{B}(z)$ . Proof. Let $q(z) = \sqrt{2/(1+e^{-2z})} =: \mathfrak{B}(z)$ and $\Omega = \{w \in \mathbb{C} : |(w-1)/(A-Bw)| < 1\}$ . Let $\psi : \mathbb{C}^2 \times \mathbb{D} \to \mathbb{C}$ be defined as $\psi(r,s;z) = r + s/(\beta r + \gamma)$ where $r = \mathfrak{B}(\xi)$ , $s = m\xi\mathfrak{B}'(\xi) = m\sqrt{2\xi}e^{-2\xi}/(1+e^{-2\xi})^{3/2}$ for $z \in \mathbb{D}$ , $\xi \in \overline{\mathbb{D}}$ and $m \geq 1$ . For $\psi$ to be an admissible function and for $\psi \in \Psi[\Omega,\mathfrak{B}]$ we need to show $\psi(\mathbb{D}) \not\subseteq \Omega$ , which is equivalent to $$\left| \frac{r + \frac{s}{\beta r + \gamma}}{A - B(r + \frac{s}{\beta r + \gamma})} \right| \ge 1.$$ We now use the Lemma 2.1 to prove the result, for which, let us assume $w = r + s/(\beta r + \gamma)$ . From (1.4) and (1.5), we have $|r| = \omega(\theta) \le \omega_{\theta_0}$ and $|s|/m = d(\theta) \ge \min d(\theta) = d(0)$ . Thus $$|w| = \left| r + \frac{s}{\beta r + \gamma} \right|$$ $$\geq \frac{|s|}{|\beta||r| + |\gamma|} - |r|$$ $$\geq \frac{md(0)}{|\beta|\omega_{\theta_0} + |\gamma|} - \omega_{\theta_0}.$$ Since $m \geq 1$ and using (2.4), we have $$|w| \ge \frac{\sqrt{2}e}{(1+e^2)^{3/2}(|\beta|\omega_{\theta_0}+|\gamma|)} - \omega_{\theta_0} \ge \frac{1+A}{1+B} > 1,$$ therefore the result follows at once from Lemma 2.1. Corollary 2.14. Let $\beta, \gamma \in \mathbb{C}$ such that $\beta \neq 0$ and satisfies inequality (2.4). Moreover, let $\omega_{\theta_0} \simeq 1.438$ and $-1 < B \leq 0 < A \leq 1$ . Suppose f be a function in A such that $$\frac{zf'(z)}{f(z)}\left(1+\left(\beta\frac{zf'(z)}{f(z)}+\gamma\right)^{-1}\left(1+\frac{zf''(z)}{f'(z)}-\frac{zf'(z)}{f(z)}\right)\right)\prec\frac{1+Az}{1+Bz},$$ then $f \in \mathcal{S}_{\mathfrak{B}}^*$ .

Theorem 2.15. Let $n \in \mathbb{N}$ , $\delta \in \{0,1\}$ , $\beta \in \mathbb{C} \setminus \{0\}$ and $\omega_{\theta_0} \simeq 1.438$ . Suppose $p \in \mathcal{H}$ such that it satisfies $$(p(z))^{\delta} + \beta (zp'(z)^n \prec \sqrt{1+z})^{\delta}$$ where <span id="page-8-0"></span> $$|\beta| \ge \left(\frac{(1+e^2)^{3/2}}{\sqrt{2}e}\right)^n \left(\sqrt{2} + \omega_{\theta_0}^{\delta}\right),\tag{2.5}$$ then $p(z) \prec \mathfrak{B}(z)$ . Proof. Let $q(z) = \sqrt{2/(1+e^{-2z})} =: \mathfrak{B}(z)$ and $\Omega = \{w \in \mathbb{C} : |w^2 - 1| < 1\}$ . Let $\psi : \mathbb{C}^2 \times \mathbb{D} \to \mathbb{C}$ be defined as $\psi(r,s;z) = r^{\delta} + \beta s^n$ where $r = \mathfrak{B}(\xi)$ , $s = m\xi\mathfrak{B}'(\xi) = m\sqrt{2}\xi e^{-2\xi}/(1+e^{-2\xi})^{3/2}$ for $z \in \mathbb{D}$ , $\xi \in \overline{\mathbb{D}}$ and $m \geq 1$ . For $\psi$ to be an admissible function and for $\psi \in \Psi[\Omega,\mathfrak{B}]$ we need to show $\psi(\mathbb{D}) \not\subseteq \Omega$ , which is equivalent to $$\left| (r^{\delta} + \beta s^n)^2 - 1 \right| \ge 1.$$ We now use the Lemma 2.2 to prove the result, for which, let us assume $w = r^{\delta} + \beta s^n$ . From (1.4) and (1.5), we have $|r| = \omega(\theta) \le \omega_{\theta_0}$ and $|s|/m = d(\theta) \ge \min d(\theta) = d(0)$ . Thus $$|w| = |r^{\delta} + \beta s^{n}|$$ $$\geq |\beta||s|^{n} - |r|^{\delta}$$ $$\geq |\beta|m^{n}(d(0))^{n} - \omega_{\theta 0}^{\delta}.$$ Since $m \geq 1$ and using (2.5), we have $$|w| \ge |\beta| \left(\frac{\sqrt{2}e}{(1+e^2)^{3/2}}\right)^n - \omega_{\theta_0}^{\delta} \ge \sqrt{2} > 1,$$ therefore the result follows from Lemma 2.2. Corollary 2.16. Let $n \in \mathbb{N}$ , $\delta \in \{0,1\}$ , $\beta \in \mathbb{C} \setminus \{0\}$ such that it satisfies (2.5) and $\omega_{\theta_0} \simeq 1.438$ . Further if $f \in \mathcal{A}$ and satisfies $$\left(\frac{zf'(z)}{f(z)}\right)^{\delta} + \beta \left(\frac{zf'(z)}{f(z)} \left(1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)}\right)\right)^{n} \prec \sqrt{1+z}$$ Then $p(z) \in \mathcal{S}_{\mathfrak{B}}^*$ .

Theorem 2.17. Let $n \in \mathbb{N}$ , $\delta \in \{0,1\}$ , $\beta \in \mathbb{C} \setminus \{0\}$ and $\omega_{\theta_0} \simeq 1.438$ . Suppose $p \in \mathcal{H}$ such that it satisfies $$(p(z))^{\delta} + \beta \frac{zp'(z)}{(p(z))^n} \prec \sqrt{1+z}$$ where <span id="page-9-0"></span> $$|\beta| \ge \left(\frac{(1+e^2)^{\frac{3-n}{2}}}{(\sqrt{2}e)^{1-n}}\right) \left(\sqrt{2} + \omega_{\theta_0}^{\delta}\right) \tag{2.6}$$ then $p(z) \prec \mathfrak{B}(z)$ . Proof. Let $q(z) = \sqrt{2/(1 + e^{-2z})} =: \mathfrak{B}(z)$ and $\Omega = \{w \in \mathbb{C} : |w^2 - 1| < 1\}$ . Let $\psi : \mathbb{C}^2 \times \mathbb{D} \to \mathbb{C}$ be defined as $\psi(r, s; z) = r^{\delta} + \beta s/r^n$ where $r = \mathfrak{B}(\xi)$ , $s = m\xi\mathfrak{B}'(\xi) = m\sqrt{2\xi}e^{-2\xi}/(1 + e^{-2\xi})^{3/2}$ for $z \in \mathbb{D}$ , $\xi \in \overline{\mathbb{D}}$ and $m \geq 1$ . For $\psi$ to be an admissible function and for $\psi \in \Psi[\Omega, \mathfrak{B}]$ we need to show $\psi(\mathbb{D}) \not\subseteq \Omega$ , which is equivalent to $$\left| \left( r^{\delta} + \beta \frac{s}{r^n} \right)^2 - 1 \right| \ge 1.$$ We now use the Lemma 2.2 to prove the result, for which, let us assume $w = r^{\delta} + \beta s/r^{n}$ . From (1.4), we have $|r| = \omega(\theta) \leq \omega_{\theta_0}$ . Thus $$|w| = \left| r^{\delta} + \beta \frac{s}{r^n} \right|$$ $$\geq |\beta| \left( \frac{2^{\frac{1-n}{2}} m e^{-2\cos\theta}}{(1 + e^{-4\cos\theta} + 2e^{-2\cos\theta}\cos(2\sin\theta))^{\frac{3-n}{4}}} \right) - (\omega(\theta))^{\delta}$$ $$\geq |\beta| m \chi_n(\theta) - (\omega(\theta))^{\delta},$$ ı where $$\chi_n(\theta) := \frac{2^{\frac{1-n}{2}} e^{-2\cos\theta}}{(1 + e^{-4\cos\theta} + 2e^{-2\cos\theta}\cos(2\sin\theta))^{\frac{3-n}{4}}}.$$ After a simple computation we obtain that $\min \chi_n(\theta) = \chi_n(0)$ for all $n \geq 1$ . Since $m \geq 1$ and using (2.6), we have $$|w| \ge |\beta| \left( \frac{(\sqrt{2}e)^{1-n}}{(1+e^2)^{\frac{3-n}{2}}} \right) - \omega_{\theta_0}^{\delta} \ge \sqrt{2} > 1,$$ therefore the result follows from Lemma 2.2. Corollary 2.18. Let $n \in \mathbb{N}$ , $\delta \in \{0,1\}$ , $\omega_{\theta_0} \simeq 1.438$ and $\beta \in \mathbb{C} \setminus \{0\}$ such that $\beta$ satisfies inequality (2.6). Suppose f be a function in A such that $$\left(\frac{zf'(z)}{f(z)}\right)^{\delta} + \beta \left(\frac{zf'(z)}{f(z)}\right)^{1-n} \left(1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)}\right) \prec \sqrt{1+z},$$ then $f \in \mathcal{S}_{\mathfrak{B}}^*$ .

Theorem 2.19. Let $\beta, \gamma \in \mathbb{C}$ such that $\beta \neq 0$ and $\omega_{\theta_0} \simeq 1.438$ . Suppose $p \in \mathcal{H}$ such that it satisfies $$p(z) + \frac{zp'(z)}{\beta p(z) + \gamma} \prec \sqrt{1+z}$$ provided <span id="page-10-0"></span> $$|\beta|\omega_{\theta_0} + |\gamma| \le \left(\frac{(1+e^2)^{3/2}}{\sqrt{2}e} \left(\sqrt{2} + \omega_{\theta_0}\right)\right)^{-1}$$ (2.7) where Then $p(z) \prec \mathfrak{B}(z)$ . Proof. Let $q(z) = \sqrt{2/(1+e^{-2z})} =: \mathfrak{B}(z)$ and $\Omega = \{w \in \mathbb{C} : |w^2-1| < 1\}$ . Let $\psi : \mathbb{C}^2 \times \mathbb{D} \to \mathbb{C}$ be defined as $\psi(r, s; z) = r + s/(\beta r + \gamma)$ where $r = \mathfrak{B}(\xi)$ , $s = m\xi\mathfrak{B}'(\xi) = m\sqrt{2}\xi e^{-2\xi}/(1 + e^{-2\xi})^{3/2}$ for $z \in \mathbb{D}$ , $\xi \in \overline{\mathbb{D}}$ and $m \geq 1$ . For $\psi$ to be an admissible function and for $\psi \in \Psi[\Omega, \mathfrak{B}]$ we need to show $\psi(\mathbb{D}) \not\subseteq \Omega$ , i.e. $$\left| \left( r + \frac{s}{\beta r + \gamma} \right)^2 - 1 \right| \ge 1.$$ We now use the Lemma 2.2 to prove the result, for which, let us assume $w = r + s/(\beta r + \gamma)$ . From (1.4) and (1.5), we have $|s|/m = d(\theta) \ge \min d(\theta) = d(0)$ and $|r| = \omega(\theta) \le \omega_{\theta_0}$ . Thus $$|w| = \left| r + \frac{s}{\beta r + \gamma} \right|$$ $$\geq \frac{|s|}{|\beta||r| + |\gamma|} - |r|$$ $$\geq \frac{md(0)}{|\beta|\omega_{\theta_0} + |\gamma|} - \omega_{\theta_0}.$$ Since $m \ge 1$ and using (2.7), we have $$|w| \ge \frac{\sqrt{2}e}{(1+e^2)^{3/2}(|\beta|\omega_{\theta_0}+|\gamma|)} - \omega_{\theta_0} \ge \sqrt{2} > 1,$$ therefore, by Lemma 2.2, the result holds.

Corollary 2.20. Let $\omega_{\theta_0} \simeq 1.438$ and $\beta, \gamma \in \mathbb{C}$ such that $\beta \neq 0$ and satisfies inequality (2.7). Suppose f be a function in A such that $$\frac{zf'(z)}{f(z)}\left(1+\left(\beta\frac{zf'(z)}{f(z)}+\gamma\right)^{-1}\left(1+\frac{zf''(z)}{f'(z)}-\frac{zf'(z)}{f(z)}\right)\right) \prec \sqrt{1+z},$$ then $f \in \mathcal{S}_{\mathfrak{B}}^*$ .

Theorem 3.1. Let $\delta \in \{0,1\}$ , $R_0 = e\sqrt{2/(e^2-1)}$ , $\omega_{\theta_0} \simeq 1.438$ and $\beta, \gamma \in \mathbb{C} \setminus \{0\}$ . If $p(z) \in \mathcal{H}$ such that $$(p(z))^{\delta} + \gamma z p'(z) + \beta z^2 p''(z) \prec \mathfrak{B}(z),$$ provided <span id="page-11-0"></span> $$\sqrt{2}e(\gamma(1+e^2)^2 + \beta(1-2e^4 - e^2)) \ge (R_0 + \omega_{\theta_0}^{\delta})(e^2 + 1)^{7/2}$$ (3.1) then $p(z) \prec \mathfrak{B}(z)$ .

Theorem 3.2. Let $\delta \in \{0,1\}$ , $\omega_{\theta_0} \simeq 1.438$ and $\beta, \gamma \in \mathbb{C} \setminus \{0\}$ . If $p(z) \in \mathcal{H}$ such that $$(p(z))^{\delta} + \gamma z p'(z) + \beta z^2 p''(z) \prec \frac{1 + Az}{1 + Bz}, \quad -1 < B \le 0 < A \le 1,$$ provided <span id="page-12-0"></span> $$\sqrt{2}e(\gamma(1+e^2)^2 + \beta(1-2e^4 - e^2)) \ge \left(\frac{1+A}{1+B} + \omega_{\theta_0}^{\delta}\right)(e^2+1)^{7/2}$$ (3.2) then $p(z) \prec \mathfrak{B}(z)$ .

Theorem 3.3. Let $\delta \in \{0,1\}$ , $\omega_{\theta_0} \simeq 1.438$ and $\beta, \gamma \in \mathbb{C} \setminus \{0\}$ . If $p(z) \in \mathcal{H}$ such that $$(p(z))^{\delta} + \gamma z p'(z) + \beta z^2 p''(z) \prec \sqrt{1+z}$$ provided <span id="page-13-0"></span> $$\sqrt{2}e(\gamma(1+e^2)^2 + \beta(1-2e^4 - e^2)) \ge \left(\sqrt{2} + \omega_{\theta_0}^{\delta}\right)(e^2 + 1)^{7/2}$$ (3.3) then $p(z) \prec \mathfrak{B}(z)$ .

Definition 1.2. [12] Let $\mathcal{Q}$ denotes the set of functions q that are analytic and univalent on $\overline{\mathbb{D}} \setminus \mathcal{E}(q)$ , where $$\mathcal{E}(q) = \left\{ \zeta \in \partial \mathbb{D} : \lim_{z \to \zeta} q(z) = \infty \right\}$$ such that $q'(\zeta) \neq 0$ for $\zeta \in \partial \mathbb{D} \backslash \mathcal{E}(q)$ .

Definition 1.3. [12] Let $\Omega$ be a set in $\mathbb{C}$ , $q \in \mathcal{Q}$ and n be a positive integer, the class of admissible functions $\Psi_n(\Omega, q)$ , consists of those functions $\psi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ that satisfy the admissibility condition $\psi(r, s, t; z) \notin \Omega$ , whenever $r = q(\zeta)$ , $s = m\zeta q'(\zeta)$ , $$\operatorname{Re}\left(\frac{t}{s}+1\right) \ge m\operatorname{Re}\left(\frac{\zeta q''(\zeta)}{q'(\zeta)}+1\right),$$ $z \in \mathbb{D}, \zeta \in \partial \mathbb{D} \setminus \mathcal{E}(q)$ and $m \geq n$ . In particular, we denote $\Psi_1(\Omega, q)$ as $\Psi(\Omega, q)$ .