Faber polynomial coefficient estimation of subclass of bi-subordinate univalent functions

Lemma 2.2. [18] Let u(z) be analytic function in the unit disc $\mathbb{U}$ with u(0) = 0 and |u(z)| < 1 for all $z \in U$ with the power series expansion $$u(z) = \sum_{n=1}^{\infty} c_n z^n ,$$ then $|c_n| \le 1$ for all n = 1, 2, 3, ... Furthermore, $|c_n| = 1$ for some n = 1, 2, 3, ... if and only if $$u(z) = e^{i\theta} z^n, \quad \theta \in \mathbb{R}.$$

Lemma 2.3. [7] Let the function $p(z) = 1 + \sum_{n=1}^{\infty} p_n z^n$ be so that $\Re(p(z)) > 0$ for $z \in U$ . Then for $-\infty < \alpha < \infty$ , <span id="page-4-0"></span> $$\left| p_2 - \alpha p_1^2 \right| \le \begin{cases} 2 - \alpha |p_1|^2 & ; \ \alpha < \frac{1}{2} \\ 2 - (1 - \alpha)|p_1|^2 & ; \ \alpha \ge \frac{1}{2} \end{cases}$$ (9) Let $\varphi(z) = \sum_{n=1}^{\infty} a_n z^n$ be a Schwarz function so that $|\varphi(z)| < 1$ , $z \in U$ . Set $p(z) = \frac{1+\varphi(z)}{1-\varphi(z)}$ where $p(z) = 1 + \sum_{n=1}^{\infty} p_n z^n$ is so that $\Re(p(z)) > 0$ for $z \in U$ . Comparing the corresponding coefficients of powers of z yields $p_1 = 2\varphi_1$ and $p_2 = 2(\varphi_2 + \varphi_1^2)$ . Now, substituting for $p_1$ and $p_2$ and letting $q = 1 - 2\alpha$ in (9), we obtain $$\left| \varphi_2 + \eta \varphi_1^2 \right| \le \begin{cases} 1 - (1 - \eta)|\varphi_1|^2 & ; \ \eta > 0 \\ 1 - (1 + \eta)|\varphi_1|^2 & ; \ \eta < 0 \end{cases}$$ (10)

Theorem 2.4. Let f defined by (1) belong to the class $\mathcal{H}_{\Sigma}(\tau, \lambda, \delta; \varphi)$ and $a_k = 0 \ (2 \le k \le n - 1)$ , then $$|a_n| \le \frac{B_1|\tau|}{1 + (n-1)(\lambda + n\delta)} \qquad (n \ge 4). \tag{11}$$

Corollary 2.5. [6, Theorem 2] Let $f \in \mathcal{N}_{\Sigma}(\alpha, \lambda, \delta)$ and $a_k = 0 \ (2 \le k \le n - 1)$ , then $$|a_n| \le \frac{2(1-\alpha)}{1+(n-1)(\lambda+n\delta)} \qquad (n \ge 4).$$ Let us put $\lambda = 1$ in Corollary 2.6, we have

Corollary 2.6. [21, Theorem 1] Let us consider $f \in \mathcal{N}_{\Sigma}^{(\alpha,\lambda)}$ and $a_k = 0 \ (2 \le k \le n-1)$ , then $$|a_n| \le \frac{2(1-\alpha)}{n(1+\delta(n-1))}$$ $(n \ge 4)$ . Let us put $\delta = 0$ in Corollary 2.6, we obtain

Corollary 2.7. [14, Theorem 1] If $f \in \mathfrak{D}(\alpha, \lambda)$ and $a_k = 0 \ (2 \le k \le n - 1)$ , then $$|a_n| \le \frac{2(1-\alpha)}{1+\lambda(n-1)}$$ $(n \ge 4)$ .

Theorem 2.8. Let $f \in \mathcal{H}_{\Sigma}(\tau, \lambda, \delta; \varphi)$ and $B_1 \geq |B_2|$ , then <span id="page-6-7"></span> $$|a_{2}| \leq \begin{cases} \frac{B_{1}\sqrt{B_{1}}|\tau|}{\sqrt{B_{1}^{2}|\tau|(1+2\lambda+6\delta)+(B_{1}+B_{2})(1+\lambda+2\delta)^{2}}} & if \ B_{2} < 0, B_{1} + B_{2} \leq 0\\ \frac{B_{1}\sqrt{B_{1}}|\tau|}{\sqrt{B_{1}^{2}|\tau|(1+2\lambda+6\delta)+(B_{1}-B_{2})(1+\lambda+2\delta)^{2}}} & if \ B_{2} > 0, B_{1} - B_{2} \leq 0 \end{cases}$$ $$(25)$$ <span id="page-6-9"></span> $$|a_3| \le \begin{cases} \frac{B_1|\tau|}{1+2\lambda+6\delta} & ; B_1 > |B_2| \\ \frac{|B_2\tau|}{1+2\lambda+6\delta} & ; B_1 < |B_2| \end{cases} , \tag{26}$$ and <span id="page-6-10"></span> $$|a_3 - 2a_2^2| \le \begin{cases} \frac{B_1|\tau|}{1 + 2\lambda + 6\delta} & ; B_1 > |B_2| \\ \frac{|B_2\tau|}{1 + 2\lambda + 6\delta} & ; B_1 < |B_2| \end{cases}$$ (27)

Corollary 2.9. [22, Theorem 1] Let $f \in \Sigma(\tau, \delta, \varphi)$ , then $$|a_2| \le \begin{cases} \frac{B_1 \sqrt{B_1 |\tau|}}{\sqrt{3B_1^2 |\tau|(1+2\delta)+4(B_1+B_2)(1+\delta)^2}} & B_2 < 0 \text{ and } B_1 + B_2 \ge 0\\ \frac{B_1 \sqrt{B_1 |\tau|}}{\sqrt{3B_1^2 |\tau|(1+2\delta)+4(B_1-B_2)(1+\delta)^2}} & B_2 > 0 \text{ and } B_1 - B_2 \ge 0 \end{cases},$$ $$|a_3| \le \begin{cases} \frac{B_1 |\tau|}{3(1+2\delta)} & B_1 > |B_2|\\ \frac{|B_2 \tau|}{3(1+2\delta)} & B_1 < |B_2| \end{cases}.$$ Let us put $\varphi(z) = \left(\frac{1+z}{1-z}\right)^{\alpha}$ , $B_1 = 2\alpha$ and $B_2 = 2\alpha^2$ , and $\tau = 1$ in Corollary 2.9 we have

Corollary 2.10. [11, Theorem 2.2] Let $f \in \mathcal{H}_{\Sigma}(\alpha, \delta)$ , then $$|a_2| \le \frac{2\alpha}{\sqrt{2(2+\alpha) + 4\delta(\alpha + \delta - \alpha\delta + 2)}},$$ $$|a_3| \le \frac{2\alpha}{3(1+2\delta)}.$$ By putting $\tau = 1 - \gamma$ and $\varphi(z) = \frac{1+z}{1-z}$ , $B_1 = B_2 = 2$ , in Corollary 2.9, we obtain

Corollary 2.11. [11, Theorem 3.2] Let $f \in \mathcal{H}_{\Sigma}(\gamma, \delta)$ , then $$|a_2| \le \sqrt{\frac{2(1-\gamma)}{3(1+2\delta)}},$$ $$|a_3| \leq \frac{2(1-\gamma)}{3(1+2\delta)}.$$ In case of $\tau=1$ , $\delta=0$ and $\varphi(z)=\left(\frac{1+z}{1-z}\right)^{\alpha}$ , $B_1=2\alpha$ , $B_2=2\alpha^2$ , in Theorem 2.8, we have

Corollary 2.12. [10, Theorem 2.2] Let $f \in \mathcal{B}_{\Sigma}(\alpha, \lambda)$ , then $$|a_2| \le \frac{2\alpha}{\sqrt{(1+\lambda)^2 + \alpha(1+2\lambda - \lambda^2)}},$$ $$|a_3| \le \frac{2\alpha}{1+2\lambda}.$$ Let us put $\tau = 1 - \gamma$ and $\varphi(z) = \frac{1+z}{1-z}$ , $B_1 = B_2 = 2$ , in Theorem 2.8, we obtain

Corollary 2.13. [6, Theorem 5] Let $0 \le \alpha < 1$ and $f \in \mathcal{N}_{\Sigma}(\gamma, \lambda, \delta)$ , then $$|a_2| \le \sqrt{\frac{2(1-\gamma)}{1+2\lambda+6\delta'}}$$ $$|a_3| \le \frac{2(1-\gamma)}{1+2\lambda+6\delta'}$$ and $$|a_3 - 2a_2^2| \le \frac{2(1-\gamma)}{1+2\lambda+6\delta}$$ . By putting $\delta = 0$ in Corollary 2.13, gets

Corollary 2.14. [10, Theorem 3.2] If f belong to $\mathcal{B}_{\Sigma}(\gamma, \lambda)$ and $0 \le \gamma < 1$ , then $$|a_2| \le \sqrt{\frac{2(1-\gamma)}{1+2\lambda}},$$ $$|a_3| \le \frac{2(1-\gamma)}{1+2\lambda},$$ Remark 2. Some results investigated in Corollaries from 2.9 to 2.14 represented an improvement of the estimate of $|a_3|$ of the earlier corresponding results.

Definition 2.1. Let $\lambda \geq 1$ , $\tau \in \mathbb{C}^* = \mathbb{C} - \{0\}$ , $0 \leq \delta \leq 1$ and $f, g \in \Sigma$ given by (1) and (3) respectively, then f is said to be in the class $\mathcal{H}_{\Sigma}(\tau, \lambda, \delta; \varphi)$ if $$1 + \frac{1}{\tau} \left( (1 - \lambda) \frac{f(z)}{z} + \lambda f'(z) + \delta z f''(z) - 1 \right) < \varphi(z), \tag{7}$$ and $$1 + \frac{1}{\tau} \left( (1 - \lambda) \frac{g(w)}{w} + \lambda g'(w) + \delta z g''(w) - 1 \right) < \varphi(w), \tag{8}$$ where $z, w \in U$ and $\varphi(z)$ is given by (2). Remark 1. For special choices of the parameters $\lambda$ , $\tau$ , $\delta$ and the function $\varphi(z)$ , the class $\mathcal{H}_{\Sigma}(\tau, \lambda, \delta; \varphi)$ reduced to the following subclasses: - 1. $\mathcal{H}_{\Sigma}(\tau, 1, \gamma; \varphi) = \Sigma(\tau, \gamma, \varphi)$ which introduced by A.E. Tudor [23] and recently studied by H.M. Srivastava and Deepak Bansal [22]. - 2. $\mathcal{H}_{\Sigma}(1,1,0;\varphi) = \mathcal{H}_{\sigma}(\varphi)$ which defined and studied by Rosihan M. Ali et al. [3]. - 3. $\mathcal{H}_{\Sigma}\left(1,1,\beta;\left(\frac{1+z}{1-z}\right)^{\alpha}\right) = \mathcal{H}_{\Sigma}(\alpha,\beta)$ which introduced by B.A. Frasin [11]. - 4. $\mathcal{H}_{\Sigma}\left(1,1,0;\left(\frac{1+z}{1-z}\right)^{\alpha}\right)=\mathcal{H}_{\Sigma}^{\alpha}$ which introduced by H.M. Srivastava et al. [20]. - 5. $\mathcal{H}_{\Sigma}\left(1,\lambda,0;\left(\frac{1+z}{1-z}\right)^{\alpha}\right)=\mathcal{B}_{\Sigma}(\alpha,\lambda)$ which is introduced by B.A. Frasin and M.K. Aouf [10], and recently studied by H.M. Srivastava et al. [21]. - 6. $\mathcal{H}_{\Sigma}\left(1-\gamma,1,\beta;\frac{1+z}{1-z}\right)=\mathcal{H}_{\Sigma}(\gamma,\beta)$ which introduced by B.A. Frasin [11]. - 7. $\mathcal{H}_{\Sigma}\left(1-\alpha,\lambda,\delta;\frac{1+z}{1-z}\right) = \mathcal{N}_{\Sigma}(\alpha,\lambda,\delta)$ which introduced by S. Bulut [6]. - 8. $\mathcal{H}_{\Sigma}\left(1-\beta,1,0;\frac{1+z}{1-z}\right)=\mathcal{H}_{\Sigma}(\beta)$ which introduced by H.M. Srivastava et al. [20]. - 9. $\mathcal{H}_{\Sigma}\left(1-\beta,\lambda,0;\frac{1+z}{1-z}\right)=\mathcal{B}_{\Sigma}(\beta,\lambda)$ which introduced by B.A. Frasin and M.A. Aouf [10] and recently studied by J.M. Jahangiri and S.G. Hamidi [14]. - 10. $\mathcal{H}_{\Sigma}\left(\tau,1,\gamma;\frac{1+Az}{1+Bz}\right) = \mathcal{R}_{\gamma,\sigma}^{\tau}(A,B)$ which introduced by A.E. Tudor [23].