Differential Subordination of Certain Class of Starlike Functions

Theorem 3.1. Suppose $\alpha_1$ , $\alpha_2 > 0$ and $(\alpha_1 - \alpha_2)(1 - D^2) \ge e(C - D)(1 + |D|)$ , where $-1 < D < C \le 1$ . Consider the analytic function p in $\mathbb{D}$ such that p(0) = 1 and $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) \prec \frac{1 + Cz}{1 + Dz}$$ implies $p(z) \prec e^z$ .

Theorem 3.3. Suppose $\alpha_1$ , $\alpha_2 > 0$ and ${\alpha_1}^2 - 2\alpha_1\alpha_2 + {\alpha_2}^2 - 2e\alpha_1 + 2e\alpha_2 \ge e^2$ . Consider the analytic function p in $\mathbb D$ such that p(0) = 1 and <span id="page-4-0"></span> $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) \prec \sqrt{1+z}$$ implies $p(z) \prec e^z$ . Proof. Suppose $h(z) = \sqrt{1+z}$ for $z \in \mathbb{D}$ and $h(\mathbb{D}) = \Omega := \{\delta \in \mathbb{C} : |\delta^2 - 1| < 1\}$ . Let us define $\xi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ as $\xi(\eta_1, \eta_2, \eta_3; z) = 1 + \alpha_1 \eta_2 + \alpha_2 \eta_3$ . It is clear that $\xi \in \Psi[\Omega, \Delta_e]$ if $\xi(r, s, t, z) \notin \Omega$ for $z \in \mathbb{D}$ . Note that $$|(\xi(r, s, t; z))^{2} - 1| = |(1 + \alpha_{1}s + \alpha_{2}t)^{2} - 1|$$ $$\geq |\alpha_{1}s + \alpha_{2}t|(|\alpha_{1}s + \alpha_{2}t| - 2)$$ $$\geq |\alpha_{1}s| \operatorname{Re}\left(1 + \frac{\alpha_{2}}{\alpha_{1}}\frac{t}{s}\right) \left(|\alpha_{1}s| \operatorname{Re}\left(1 + \frac{\alpha_{2}}{\alpha_{1}}\frac{t}{s}\right) - 2\right)$$ Using the fact that $m \geq 1$ and analogous to Theorem 3.1, we have $$|(\xi(r,s,t;z))^2 - 1| \ge \frac{1}{e} \left(\alpha_1 - \alpha_2\right) \left(\frac{1}{e} \left(\alpha_1 - \alpha_2\right) - 2\right)$$ $$= \frac{\alpha_1^2 - 2\alpha_1\alpha_2 + \alpha_2^2 - 2e\alpha_1 + 2e\alpha_2}{e^2}$$ $$\ge 1.$$ Therefore, $\xi \in \Psi[\Omega, \Delta_e]$ and thus $p(z) \prec e^z$ as a consequence of Lemma E.

Corollary 3.4. By considering p(z) = zf'(z)/f(z) in above theorem and through (3.6), we deduce that $f(z) \in \mathcal{S}_e^*$ if $Y_f(z) \prec \sqrt{1+z}$ and $\alpha_1^2 - 2\alpha_1\alpha_2 + \alpha_2^2 - 2e\alpha_1 + 2e\alpha_2 \geq e^2$ . Next, we consider $h(z) = 2/(1 + e^{-z})$ and $z + \sqrt{1 + z^2}$ in the coming results.

Theorem 3.5. Suppose $\alpha_1$ , $\alpha_2 > 0$ and $\alpha_1 - \alpha_2 \ge er_0$ , where $r_0 \approx 0.546302$ is the positive root of the equation $r^2 + 2\cot(1)r - 1 = 0$ . Consider the analytic function p in $\mathbb{D}$ such that p(0) = 1 and $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) \prec \frac{2}{1 + e^{-z}}$$ implies $p(z) \prec e^z$ .

Theorem 3.6. Suppose $\alpha_1$ , $\alpha_2 > 0$ and $\alpha_1 - \alpha_2 \ge \sqrt{2}e$ . Let p be an analytic function in $\mathbb{D}$ such that p(0) = 1 and $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) \prec z + \sqrt{1 + z^2}$$ implies $p(z) \prec e^z$ . Proof. Suppose $h(z) = z + \sqrt{1+z^2}$ for $z \in \mathbb{D}$ and $h(\mathbb{D}) = \Omega := \{\delta \in \mathbb{C} : |\delta^2 - 1| < 2|\delta|\}$ . We define $\xi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ as $\xi(\eta_1, \eta_2, \eta_3; z) = 1 + \alpha_1 \eta_2 + \alpha_2 \eta_3$ . For $\xi \in \Psi[\Omega, \Delta_e]$ , we must have $\xi(r, s, t; z) \notin \Omega$ . From the graph of $z + \sqrt{1+z^2}$ (see Fig. 1), we note that $\Omega$ is constructed by the circles $C_1$ and $C_2$ , given by $$C_1: |z-1| = \sqrt{2}$$ 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 $C_2 \cap C_1$ . We have $$|\xi(r, s, t; z) - 1| = |\alpha_1 s + \alpha_2 t|.$$  <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 (solid purple) represents $z + \sqrt{1+z^2}$ . Analogous to Theorem 3.5, we have $$|\alpha_1 s + \alpha_2 t| \ge b(\theta)(\alpha_1 + \alpha_2 l(\theta))$$ $$\ge \frac{1}{e}(\alpha_1 - \alpha_2)$$ $$\ge \sqrt{2}.$$ The fact that $\xi(r, s, t; z) \notin C_1$ suffices us to deduce that $\xi(r, s, t; z) \notin \Omega$ . Consequently, $\xi \in \Psi[\Omega, \Delta_e]$ and thus $p(z) \prec e^z$ by using Lemma E. Now, if we choose h(z) to be $1 + \sin z$ and $1 + ze^z$ , we derive the conditions on $\alpha_1$ and $\alpha_2$ as follows:

Theorem 3.7. Suppose $\alpha_1$ , $\alpha_2 > 0$ and $\alpha_1 - \alpha_2 \ge e \sinh 1$ . Let p be an analytic function in $\mathbb{D}$ such that p(0) = 1 and $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) \prec 1 + \sin z$$ implies $p(z) \prec e^z$ . Proof. Suppose $h(z) = 1 + \sin z$ for $z \in \mathbb{D}$ and $h(\mathbb{D}) = \Omega := \{\delta \in \mathbb{C} : |\arcsin(\delta - 1)| < 1\}$ . Define $\xi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ as $\xi(\eta_1, \eta_2, \eta_3; z) = 1 + \alpha_1 \eta_2 + \alpha_2 \eta_3$ . For $\xi \in \Psi[\Omega, \Delta_e]$ , we must have $\xi(r, s, t; z) \notin \Omega$ . Through [3, Lemma 3.3], we note that the smallest disk containing $\Omega$ is $\{w \in \mathbb{C} : |w - 1| < \sinh 1\}$ . So, $$|\xi(r, s, t; z) - 1| = |\alpha_1 s + \alpha_2 t|.$$ Analogous to Theorem 3.5, we have $$|\alpha_1 s + \alpha_2 t| \ge b(\theta)(\alpha_1 + \alpha_2 l(\theta))$$ $$\ge \frac{1}{e}(\alpha_1 - \alpha_2)$$ $$\ge \sinh 1.$$ Clearly, $\xi(r, s, t; z) \notin \{w : |w - 1| < \sinh 1\}$ which is enough to conclude that $\xi(r, s, t; z) \notin \Omega$ . Therefore, $\xi \in \Psi[\Omega, \Delta_e]$ and thus $p(z) \prec e^z$ through Lemma E.

Theorem 3.8. Suppose $\alpha_1$ , $\alpha_2 > 0$ and $\alpha_1 - \alpha_2 \ge e^2$ . Consider the analytic function p in $\mathbb{D}$ such that p(0) = 1 and $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) \prec 1 + z e^z$$ implies $p(z) \prec e^z$ . Proof. Suppose $h(z) = 1 + ze^z$ for $z \in \mathbb{D}$ and $\Omega = h(\mathbb{D})$ . Let us define $\xi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ as $\xi(\eta_1, \eta_2, \eta_3; z) = 1 + \alpha_1 \eta_2 + \alpha_2 \eta_3$ . For $\xi \in \Psi[\Omega, \Delta_e]$ , we need to have $\xi(r, s, t; z) \notin \Omega$ . Through [9, Lemma 3.3], we observe that the smallest disk containing $\Omega$ is $\{w \in \mathbb{C} : |w - 1| < e\}$ . Thus $$|\xi(r, s, t; z) - 1| = |\alpha_1 s + \alpha_2 t|.$$ Analogous to Theorem 3.5, we have $$|\alpha_1 s + \alpha_2 t| \ge b(\theta)(\alpha_1 + \alpha_2 l(\theta))$$ $$\ge \frac{1}{e}(\alpha_1 - \alpha_2)$$ $$\ge e.$$ Clearly, $\xi(r, s, t; z) \notin \{w \in \mathbb{C} : |w - 1| < e\}$ which suffices us to conclude that $\xi(r, s, t; z) \notin \Omega$ . Therefore, $\xi \in \Psi[\Omega, \Delta_e]$ and thus $p(z) \prec e^z$ by using Lemma E. Now, we conclude this section by choosing h(z) to be $1 + \sinh^{-1} z$ and $e^z$ itself, followed by its subsequent corollary.

Theorem 3.9. Suppose $\alpha_1$ , $\alpha_2 > 0$ and $2(\alpha_1 - \alpha_2) \ge \pi e$ . Consider the analytic function p in $\mathbb{D}$ such that p(0) = 1 and $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) \prec 1 + \sinh^{-1} z$$ implies $p(z) \prec e^z$ . Proof. Suppose $h(z) = 1 + \sinh^{-1} z$ for $z \in \mathbb{D}$ and $h(\mathbb{D}) = \Omega := \{\delta \in \mathbb{C} : |\sinh(\delta - 1)| < 1\}$ . We define $\xi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ as $\xi(\eta_1, \eta_2, \eta_3; z) = 1 + \alpha_1 \eta_2 + \alpha_2 \eta_3$ . For $\xi \in \Psi[\Omega, \Delta_e]$ , we must have $\xi(r, s, t; z) \notin \Omega$ . Through [2, Remark 2.7], we note that the disk $\{w \in \mathbb{C} : |w - 1| < \pi/2\}$ is the smallest disk containing $\Omega$ . So, $$|\xi(r, s, t; z) - 1| = |\alpha_1 s + \alpha_2 t|.$$ Analogous to Theorem 3.5, we have $$|\alpha_1 s + \alpha_2 t| \ge b(\theta)(\alpha_1 + \alpha_2 l(\theta))$$ $$\ge \frac{1}{e}(\alpha_1 - \alpha_2)$$ $$\ge \frac{\pi}{2}.$$ Clearly, $\xi(r, s, t; z)$ does not belong to the disk $\{w \in \mathbb{C} : |w - 1| < \pi/2\}$ which suffices us to conclude that $\xi(r, s, t; z) \notin \Omega$ . Therefore, $\xi \in \Psi[\Omega, \Delta_e]$ and thus $p(z) \prec e^z$ by using Lemma E.

Theorem 3.10. Suppose $\alpha_1$ , $\alpha_2 > 0$ and $\alpha_1 - \alpha_2 \ge e(e-1)$ . Consider the analytic function p in $\mathbb{D}$ such that p(0) = 1 and $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) \prec e^z$$ implies $p(z) \prec e^z$ . Proof. Suppose $h(z) = e^z$ for $z \in \mathbb{D}$ and $\Omega = \Delta_e$ . Define $\xi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ as $\xi(\eta_1, \eta_2, \eta_3; z) = 1 + \alpha_1 \eta_2 + \alpha_2 \eta_3$ . For $\xi \in \Psi[\Delta_e, \Delta_e]$ , we must have $\xi(r, s, t; z) \notin \Delta_e$ . So, $$|\xi(r, s, t; z) - 1| = |\alpha_1 s + \alpha_2 t|.$$ Analogous to Theorem 3.5, we have <span id="page-7-0"></span> $$|\alpha_1 s + \alpha_2 t| \ge b(\theta)(\alpha_1 + \alpha_2 l(\theta))$$ $$\ge \frac{1}{e}(\alpha_1 - \alpha_2)$$ $$\ge e - 1. \tag{3.8}$$ Further, we have $$|\log(\xi(r, s, t; z))| = |\log(1 + \alpha_1 s + \alpha_2 t)|.$$ Through Lemma C and (3.8), we have $$|\log(1 + \alpha_1 s + \alpha_2 t)| \ge 1,$$ which implies that $\xi(r,s,t;z) \notin \Delta_e$ . Therefore, $\xi \in \Psi[\Delta_e,\Delta_e]$ and thus $p(z) \prec e^z$ through Lemma Theorems 4.5-4.10 yield the following conclusion if p(z) = zf'(z)/f(z) is considered as in (3.6). Corollary 3.11. Suppose $\alpha_1$ , $\alpha_2 > 0$ and $f \in \mathcal{A}$ . Then, $f \in \mathcal{S}_e^*$ if any of the conditions hold: - (i) $Y_f(z) \prec 2/(1+e^{-z})$ and $\alpha_1 \alpha_2 \geq er_0$ , where $r_0 \approx 0.546302$ is the positive root of the equation $r^2 + 2 \cot(1)r - 1 = 0$ . - (ii) $Y_f(z) \prec z + \sqrt{1+z^2}$ and $\alpha_1 \alpha_2 \ge \sqrt{2}e$ . - (iii) $Y_f(z) \prec 1 + \sin z$ and $\alpha_1 \alpha_2 \ge e \sinh 1$ . - (iv) $Y_f(z) \prec 1 + ze^z$ and $\alpha_1 \alpha_2 \ge e^2$ . - (v) $Y_f(z) \prec 1 + \sinh^{-1} z \text{ and } 2(\alpha_1 \alpha_2) \geq \pi e.$ (vi) $Y_f(z) \prec e^z \text{ and } \alpha_1 \alpha_2 \geq e(e-1).$ Lastly, we conclude this article by determining the sufficient conditions on $\alpha_1$ , $\alpha_2$ and $\alpha_3 \in \mathbb{R}^+$ involving the constants m and k (defined in Lemma A) to satisfy the third order differential subordination implication, given by $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) + \alpha_3 z^3 p'''(z) \prec h(z) \implies p(z) \prec e^z.$$ We obtain the following result upon considering h(z) = (1 + Cz)/(1 + Dz).

Theorem 3.12. Suppose $\alpha_1$ , $\alpha_2$ , $\alpha_3 > 0$ and $(\alpha_1 - \alpha_2 - m^2\alpha_3 - 3m(k-1)\alpha_3)(1-D^2) \ge$ e(1+|D|)(C-D) where $-1 < D < C \le 1$ . Consider the analytic function p in $\mathbb D$ such that $p(0) = 1 \ and$ $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) + \alpha_3 z^3 p'''(z) \prec \frac{1 + Cz}{1 + Dz}$$ implies $p(z) \prec e^z$ .

Theorem 3.14. Suppose $\alpha_1$ , $\alpha_2$ and $\alpha_3$ be positive real numbers and $(\alpha_1 - \alpha_2 - m^2\alpha_3 - 3m(k - 1)\alpha_3)(\alpha_1 - \alpha_2 - m^2\alpha_3 - 3m(k - 1)\alpha_3 - 2e) \ge e^2$ . Consider the analytic function p in $\mathbb{D}$ such that p(0) = 1 and <span id="page-9-0"></span> $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) + \alpha_3 z^3 p'''(z) \prec \sqrt{1+z}$$ implies $p(z) \prec e^z$ . Proof. Suppose $h(z) = \sqrt{1+z}$ and $h(\mathbb{D}) = \Omega := \{\delta \in \mathbb{C} : |\delta^2 - 1| < 1\}$ . Consider $\xi : \mathbb{C}^4 \times \mathbb{D} \to \mathbb{C}$ defined as $\xi(\eta_1, \eta_2, \eta_3, \eta_4; z) = 1 + \alpha_1 \eta_2 + \alpha_2 \eta_3 + \alpha_3 \eta_4$ . We know that $\xi \in \Psi[\Omega, \Delta_e]$ if $\xi(r, s, t, u; z) \notin \Omega$ for $z \in \mathbb{D}$ . Consider $$\begin{aligned} |(\xi(r,s,t,u;z))^2 - 1| &= |(1 + \alpha_1 s + \alpha_2 t + \alpha_3 u)^2 - 1| \\ &\geq |\alpha_1 s + \alpha_2 t + \alpha_3 u| (|\alpha_1 s + \alpha_2 t + \alpha_3 u| - 2) \\ &\geq |\alpha_1 s| \operatorname{Re} \left( 1 + \frac{\alpha_2}{\alpha_1} \frac{t}{s} + \frac{\alpha_3}{\alpha_1} \frac{u}{s} \right) \left( |\alpha_1 s| \operatorname{Re} \left( 1 + \frac{\alpha_2}{\alpha_1} \frac{t}{s} + \frac{\alpha_3}{\alpha_1} \frac{u}{s} \right) - 2 \right). \end{aligned}$$ Analogous to the proof of Theorem 3.12 and the fact that $m \geq 1$ , we obtain $$|(\xi(r,s,t,u;z))^{2}-1| \geq \alpha_{1}b(\theta)\left(1+\frac{\alpha_{2}}{\alpha_{1}}l(\theta)+\frac{\alpha_{3}}{\alpha_{1}}(m^{2}h(\theta)+3m(k-1)l(\theta))\right) \times$$ $$\left(\alpha_{1}b(\theta)\left(1+\frac{\alpha_{2}}{\alpha_{1}}l(\theta)+\frac{\alpha_{3}}{\alpha_{1}}(m^{2}h(\theta)+3m(k-1)l(\theta))\right)-2\right)$$ $$\geq \frac{1}{e^{2}}\left(\alpha_{1}-\alpha_{2}-\alpha_{3}(m^{2}+3m(k-1))\right)\left(\alpha_{1}-\alpha_{2}-\alpha_{3}(m^{2}+3m(k-1))-2e\right)$$ $$\geq 1.$$ This implies that $\xi \in \Psi[\Omega, \Delta_e]$ . Thus, $p(z) \prec e^z$ as a consequence of Lemma F. Corollary 3.15. By considering p(z) = zf'(z)/f(z) in above theorem and through (3.9), we deduce that $f(z) \in \mathcal{S}_e^*$ if $\chi_f(z) \prec \sqrt{1+z}$ and $(\alpha_1 - \alpha_2 - m^2\alpha_3 - 3m\alpha_3(k-1))(\alpha_1 - \alpha_2 - m^2\alpha_3 - 3m(k-1)\alpha_3 - 2e) \geq e^2$ . The next two results deal with the case when $h(z) = 2/(1 + e^{-z})$ and $z + \sqrt{1 + z^2}$ .

Theorem 3.16. Suppose $\alpha_1$ , $\alpha_2$ , $\alpha_3 > 0$ and $\alpha_1 - \alpha_2 - m^2\alpha_3 - 3m(k-1)\alpha_3 \ge er_0$ , where $r_0 \approx 0.546302$ is the positive root of the equation $r^2 + 2\cot(1)r - 1 = 0$ . Consider the analytic function p in $\mathbb{D}$ such that p(0) = 1 and $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) + \alpha_3 z^3 p'''(z) \prec \frac{2}{1 + e^{-z}}$$ implies $p(z) \prec e^z$ . Proof. Suppose $h(z) = 2/(1 + e^{-z})$ and $h(\mathbb{D}) = \Omega := \{\delta \in \mathbb{C} : |\log(\delta/(2 - \delta))| < 1\}$ . Define $\xi : \mathbb{C}^4 \times \mathbb{D} \to \mathbb{C}$ as $\xi(\eta_1, \eta_2, \eta_3, \eta_4; z) = 1 + \alpha_1\eta_2 + \alpha_2\eta_3 + \alpha_3\eta_4$ . For $\xi \in \Psi[\Omega, \Delta_e]$ , it is required that $\xi(r, s, t, u; z) \notin \Omega$ . Consider $$|\alpha_1 s + \alpha_2 t + \alpha_3 u| = \alpha_1 |s| \left| 1 + \frac{\alpha_2}{\alpha_1} \frac{t}{s} + \frac{\alpha_3}{\alpha_1} \frac{u}{s} \right|$$ $$\geq \alpha_1 |s| \operatorname{Re} \left( 1 + \frac{\alpha_2}{\alpha_1} \frac{t}{s} + \frac{\alpha_3}{\alpha_1} \frac{u}{s} \right)$$ $$\geq m \alpha_1 b(\theta) \left( 1 + \frac{\alpha_2}{\alpha_1} (ml(\theta) + m - 1) + \frac{\alpha_3}{\alpha_1} (m^2 h(\theta) + 3m(k - 1)l(\theta)) \right).$$ As $m \ge 1$ and $\text{Re}(1 + l(\theta)) > 0$ , we have $$|\alpha_1 s + \alpha_2 t + \alpha_3 u| \ge b(\theta) \left( \alpha_1 + \alpha_2 l(\theta) + \alpha_3 (m^2 h(\theta) + 3m(k-1)l(\theta)) \right)$$ $$\ge \frac{1}{e} (\alpha_1 - \alpha_2 - m^2 \alpha_3 - 3m(k-1)\alpha_3)$$ $$\ge r_0. \tag{3.10}$$ Next, we consider $$\left| \log \left( \frac{\xi(r, s, t, u; z)}{2 - \xi(r, s, t, u; z)} \right) \right| = \left| \log \left( \frac{1 + \alpha_1 s + \alpha_2 t + \alpha_3 u}{1 - (\alpha_1 s + \alpha_2 t + \alpha_3 u)} \right) \right|.$$ Through Lemma D and (3.10), we have <span id="page-10-0"></span> $$\left| \log \left( \frac{1 + \alpha_1 s + \alpha_2 t + \alpha_3 u}{1 - (\alpha_1 s + \alpha_2 t + \alpha_3 u)} \right) \right| \ge 1.$$ Therefore, $\xi \in \Psi[\Omega, \Delta_e]$ and $p(z) \prec e^z$ as a consequence of Lemma F.

Theorem 3.17. Suppose $\alpha_1$ , $\alpha_2$ , $\alpha_3 > 0$ and $\alpha_1 - \alpha_2 - m^2\alpha_3 - 3m(k-1)\alpha_3 \ge \sqrt{2}e$ . Consider the analytic function p in $\mathbb{D}$ such that p(0) = 1 and $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) + \alpha_3 z^3 p'''(z) \prec z + \sqrt{1 + z^2}$$ implies $p(z) \prec e^z$ . Proof. Suppose $h(z) = z + \sqrt{1 + z^2}$ for $z \in \mathbb{D}$ and thus $h(\mathbb{D}) = \Omega := \{\delta \in \mathbb{C} : |\delta^2 - 1| < 2|\delta|\}$ . Define $\xi : \mathbb{C}^4 \times \mathbb{D} \to \mathbb{C}$ as $\xi(\eta_1, \eta_2, \eta_3, \eta_4; z) = 1 + \alpha_1 \eta_2 + \alpha_2 \eta_3 + \alpha_3 \eta_4$ . For $\xi \in \Psi[\Omega, \Delta_e]$ , we must have $\xi(r, s, t, u; z) \notin \Omega$ . From the graph of $z + \sqrt{1 + z^2}$ (see Fig. 1), we note that $\Omega$ is constructed by two circles $C_1$ and $C_2$ , given by $$C_1: |z-1| = \sqrt{2}$$ 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 $C_2 \cap C_1$ . We have $$|\xi(r, s, t, u; z) - 1| = |\alpha_1 s + \alpha_2 t + \alpha_3 u|.$$ Similar to the proof of Theorem 3.16, we have $$|\alpha_1 s + \alpha_2 t + \alpha_3 u| \ge b(\theta) \left( \alpha_1 + \alpha_2 l(\theta) + \alpha_3 (m^2 h(\theta) + 3m(k-1)l(\theta)) \right)$$ $$\ge \frac{1}{e} (\alpha_1 - \alpha_2 - m^2 \alpha_3 - 3m(k-1)\alpha_3)$$ $$\ge \sqrt{2}.$$ Thus, we can say that $\xi(r, s, t, u; z)$ does not belong to the circle $C_1$ which suffices us to deduce that $\xi(r, s, t, u; z) \notin \Omega$ . Therefore, $\xi \in \Psi[\Omega, \Delta_e]$ and thus $p(z) \prec e^z$ by using Lemma F. If we take h(z) to be $1 + \sin z$ and $1 + ze^z$ , the results are as follow:

Theorem 3.18. Suppose $\alpha_1$ , $\alpha_2$ , $\alpha_3 > 0$ and $\alpha_1 - \alpha_2 - m^2\alpha_3 - 3m(k-1)\alpha_3 \ge e \sinh 1$ . Consider the analytic function p in $\mathbb{D}$ such that p(0) = 1 and $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) + \alpha_3 z^3 p'''(z) \prec 1 + \sin z$$ implies $p(z) \prec e^z$ . Proof. Suppose $h(z) = 1 + \sin z$ and $h(\mathbb{D}) = \Omega := \{ \delta \in \mathbb{C} : |\arcsin(\delta - 1)| < 1 \}$ . Let us define $\xi : \mathbb{C}^4 \times \mathbb{D} \to \mathbb{C}$ as $\xi(\eta_1, \eta_2, \eta_3, \eta_4; z) = 1 + \alpha_1 \eta_2 + \alpha_2 \eta_3 + \alpha_3 \eta_4$ . For $\xi \in \Psi[\Omega, \Delta_e]$ , it is required that $\xi(r, s, t, u; z) \notin \Omega$ . Through [3, Lemma 3.3], we note that the smallest disk containing $\Omega$ is $\{w : |w - 1| < \sinh 1\}$ . So, $$|\xi(r, s, t, u; z) - 1| = |\alpha_1 s + \alpha_2 t + \alpha_3 u|.$$ Analogous to the proof of Theorem 3.16, we have $$|\alpha_1 s + \alpha_2 t + \alpha_3 u| \ge b(\theta) \left( \alpha_1 + \alpha_2 l(\theta) + \alpha_3 (m^2 h(\theta) + 3m(k-1)l(\theta)) \right)$$ $$\ge \frac{1}{e} (\alpha_1 - \alpha_2 - m^2 \alpha_3 - 3m(k-1)\alpha_3)$$ $$> \sinh 1.$$ Clearly, $\xi(r, s, t, u; z)$ does not belong to the disk $\{w : |w-1| < \sinh 1\}$ indicates that $\xi(r, s, t, u; z) \notin \Omega$ . Therefore, $\xi \in \Psi[\Omega, \Delta_e]$ and thus $p(z) \prec e^z$ by employing Lemma F.

Theorem 3.19. Suppose $\alpha_1$ , $\alpha_2$ , $\alpha_3 > 0$ and $\alpha_1 - \alpha_2 - m^2\alpha_3 - 3m(k-1)\alpha_3 \ge e^2$ . Consider the analytic function p in $\mathbb{D}$ such that p(0) = 1 and $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) + \alpha_3 z^3 p'''(z) \prec 1 + z e^z$$ implies $p(z) \prec e^z$ . Proof. Suppose $h(z) = 1 + ze^z$ and $\Omega = h(\mathbb{D})$ . We define $\xi : \mathbb{C}^4 \times \mathbb{D} \to \mathbb{C}$ as $\xi(\eta_1, \eta_2, \eta_3, \eta_4; z) = 1 + \alpha_1 \eta_2 + \alpha_2 \eta_3 + \alpha_3 \eta_4$ . For $\xi \in \Psi[\Omega, \Delta_e]$ , we need to have that $\xi(r, s, t, u; z) \notin \Omega$ . Through [9, Lemma 3.3], we note that the smallest disk containing $\Omega$ is $\{w \in \mathbb{C} : |w - 1| < e\}$ . Also, $$|\xi(r, s, t, u; z) - 1| = |\alpha_1 s + \alpha_2 t + \alpha_3 u|.$$ Analogous to the proof of Theorem 3.16, we have $$|\alpha_1 s + \alpha_2 t + \alpha_3 u| \ge b(\theta) \left( \alpha_1 + \alpha_2 l(\theta) + \alpha_3 (m^2 h(\theta) + 3m(k-1)l(\theta)) \right)$$ $$\ge \frac{1}{e} (\alpha_1 - \alpha_2 - m^2 \alpha_3 - 3m(k-1)\alpha_3)$$ $$\ge e.$$ Clearly, $\xi(r, s, t, u; z)$ does not belong to the disk $\{w \in \mathbb{C} : |w - 1| < e\}$ , is enough to conclude that $\xi(r, s, t, u; z) \notin \Omega$ . Therefore, $\xi \in \Psi[\Omega, \Delta_e]$ and thus $p(z) \prec e^z$ by using Lemma F. We complete this section by considering $h(z) = 1 + \sinh^{-1} z$ and $e^z$ itself which are followed by a resulting corollary.

Theorem 3.20. Suppose $\alpha_1$ , $\alpha_2$ , $\alpha_3 > 0$ and $2(\alpha_1 - \alpha_2 - m^2\alpha_3 - 3m(k-1)\alpha_3) \ge \pi e$ . Consider the analytic function p in $\mathbb{D}$ such that p(0) = 1 and $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) + \alpha_3 z^3 p'''(z) \prec 1 + \sinh^{-1} z$$ implies $p(z) \prec e^z$ . Proof. Suppose $h(z) = 1 + \sinh^{-1} z$ and $h(\mathbb{D}) = \Omega := \{\delta \in \mathbb{C} : |\sinh(\delta - 1)| < 1\}$ . Define $\xi : \mathbb{C}^4 \times \mathbb{D} \to \mathbb{C}$ as $\xi(\eta_1, \eta_2, \eta_3, \eta_4; z) = 1 + \alpha_1 \eta_2 + \alpha_2 \eta_3 + \alpha_3 \eta_4$ . For $\xi \in \Psi[\Omega, \Delta_e]$ , it is required that $\xi(r, s, t, u; z) \notin \Omega$ . Through [2, Remark 2.7], we note that the smallest disk containing $\Omega$ is $\{w \in \mathbb{C} : |w - 1| < \pi/2\}$ . So, $$|\xi(r, s, t, u; z) - 1| = |\alpha_1 s + \alpha_2 t + \alpha_3 u|.$$ Similar to the proof of Theorem 3.16, we have $$|\alpha_1 s + \alpha_2 t + \alpha_3 u| \ge b(\theta) \left( \alpha_1 + \alpha_2 l(\theta) + \alpha_3 (m^2 h(\theta) + 3m(k-1)l(\theta)) \right)$$ $$\ge \frac{1}{e} (\alpha_1 - \alpha_2 - m^2 \alpha_3 - 3m(k-1)\alpha_3)$$ $$\ge \frac{\pi}{2}.$$ Clearly, $\xi(r, s, t, u; z) \notin \{w \in \mathbb{C} : |w - 1| < \pi/2\}$ which suffices to prove that $\xi(r, s, t, u; z) \notin \Omega$ . Therefore, $\xi \in \Psi[\Omega, \Delta_e]$ and thus $p(z) \prec e^z$ by Lemma F.

Theorem 3.21. Suppose $\alpha_1$ , $\alpha_2$ , $\alpha_3 > 0$ and $\alpha_1 - \alpha_2 - m^2 \alpha_3 - 3m(k-1)\alpha_3 \ge e(e-1)$ . Consider the analytic function p in $\mathbb{D}$ such that p(0) = 1 and $$1 + \alpha_1 z p'(z) + \alpha_2 z^2 p''(z) + \alpha_3 z^3 p'''(z) \prec e^z$$ implies $p(z) \prec e^z$ .

Definition 1.1. [1] Let $\xi(r, s, t, u; z) : \mathbb{C}^4 \times \mathbb{D} \to \mathbb{C}$ 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-1-1"></span> $$\xi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z) \prec h(z)$$ (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 rotation. Furthermore, let Q denote 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). <span id="page-1-2"></span>Lemma A. [1] Let $z_0 \in \mathbb{D}$ and $r_0 = |z_0|$ . Let $f(z) = a_n z^n + a_{n+1} z^{n+1} + \cdots$ 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 $$\frac{z_0 f'(z_0)}{f(z_0)} = m, \quad 1 + \frac{z_0 f''(z_0)}{f'(z_0)} = k \quad \text{and} \quad 2 + \text{Re}\left(\frac{z_0 f'''(z_0)}{f''(z_0)}\right) = l \quad \text{where} \quad l \ge k \ge m \ge n \ge 2.$$

Definition 2.1. [8] Let $\Omega$ be a set in $\mathbb{C}$ , $q \in Q$ and $k \geq m \geq n \geq 2$ . The class of admissible operators $\xi_n[\Omega, q]$ consists of those $\xi : \mathbb{C}^4 \times \mathbb{D} \to \mathbb{C}$ that satisfy the admissibility condition $$\xi(r, s, t, u; z) \notin \Omega$$ whenever $z \in \mathbb{D}$ , $r = q(\zeta)$ , $s = m\zeta q'(\zeta)$ , $$\operatorname{Re}\left(1+\frac{t}{s}\right) \geq m\left(1+\operatorname{Re}\frac{\zeta q''(\zeta)}{q'(\zeta)}\right) \quad \text{and} \quad \operatorname{Re}\frac{u}{s} \geq m^2\operatorname{Re}\frac{\zeta^2 q'''(\zeta)}{q'(\zeta)} + 3m(k-1)\operatorname{Re}\frac{\zeta q''(\zeta)}{q'(\zeta)}$$ for $\zeta \in \partial \mathbb{D} \setminus \mathbb{E}(q)$ . <span id="page-2-3"></span>Lemma B. [8] Let $p \in \mathcal{H}[a,n]$ with $m \geq n \geq 2$ , and let $q \in Q(a)$ such that it satisfies $$\left| \frac{zp'(z)}{q'(\zeta)} \right| \le m,\tag{2.1}$$ for $z \in \mathbb{D}$ and $\zeta \in \partial \mathbb{D} \setminus \mathbb{E}(q)$ . If $\Omega$ is a set in $\mathbb{C}, \xi \in \Psi_n[\Omega, a]$ and $$\xi(p(z), zp'(z), z^2p''(z), z^3p'''(z); z) \subset \Omega,$$ then $p \prec q$ . Here, we provide two lemmas which are necessary for the proof of results in the coming sections. <span id="page-2-6"></span>Lemma C. [8, Lemma 4, Pg No. 192] For any complex number z, we have $$|\log(1+z)| \ge 1$$ if and only if $|z| \ge e - 1$ . <span id="page-2-5"></span>Lemma D. [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 $R \ge r_0$ .