Faber polynomial coefficient estimates for a class of analytic bi-univalent functions

Theorem 3.1. For $\lambda \geq 1$ , $\mu \geq 0$ , $\delta \geq 0$ and $0 \leq \alpha < 1$ , let the function $f \in \mathfrak{B}^{\mu}_{\Sigma}(\alpha, \lambda, \delta)$ be given by (1.1). If $a_k = 0$ $(2 \leq k \leq n - 1)$ , then $$|a_n| \le \frac{2(1-\alpha)}{\mu + (n-1)\lambda + n(n-1)\xi\delta} \quad (n \ge 4).$$

Corollary 3.2. For $\lambda \geq 1$ , $\delta \geq 0$ and $0 \leq \alpha < 1$ , let the function $f \in \mathfrak{B}_{\Sigma}(\alpha, \lambda, \delta)$ be given by (1.1). If $a_k = 0$ ( $2 \leq k \leq n-1$ ), then $$|a_n| \le \frac{2(1-\alpha)}{1+(n-1)\lambda+n(n-1)\xi\delta} \quad (n \ge 4).$$

Theorem 3.3. For $\lambda \geq 1$ , $\mu \geq 0$ , $\delta \geq 0$ and $0 \leq \alpha < 1$ , let the function $f \in \mathfrak{B}^{\mu}_{\Sigma}(\alpha, \lambda, \delta)$ be given by (1.1). Then one has the following $$|a_2| \le \begin{cases} \sqrt{\frac{4(1-\alpha)}{(\mu+2\lambda+6\xi\delta)(\mu+1)}}, & 0 \le \alpha \le \frac{\mu+2\lambda-\lambda^2}{(\mu+2\lambda+6\xi\delta)(\mu+1)} \\ \frac{2(1-\alpha)}{\mu+\lambda+2\xi\delta}, & \frac{\mu+2\lambda-\lambda^2}{(\mu+2\lambda+6\xi\delta)(\mu+1)} \le \alpha \le 1 \end{cases}$$ (3.13) $$|a_3| \le \begin{cases} \min\left\{\frac{4(1-\alpha)^2}{(\mu+\lambda+2\xi\delta)^2} + \frac{2(1-\alpha)}{\mu+2\lambda+6\xi\delta}, \frac{4(1-\alpha)}{(\mu+2\lambda+6\xi\delta)(\mu+1)}\right\}, & 0 \le \mu < 1\\ \frac{2(1-\alpha)}{\mu+2\lambda+2\xi\delta}, & \mu \ge 1 \end{cases}$$ (3.14) and $$\left| a_3 - \frac{\mu + 3}{2} a_2^2 \right| \le \frac{2(1 - \alpha)}{\mu + 2\lambda + 6\xi\delta}.$$

Corollary 3.4. For $\lambda \geq 1$ , $\delta \geq 0$ and $0 \leq \alpha < 1$ , let the function $f \in \mathfrak{B}_{\Sigma}(\alpha, \lambda, \delta)$ be given by (1.1). Then one has the following $$|a_2| \le \begin{cases} \sqrt{\frac{2(1-\alpha)}{(1+2\lambda+6\xi\delta)}}, & 0 \le \alpha \le \frac{1+2\lambda-\lambda^2}{2(1+2\lambda+6\xi\delta)} \\ \frac{2(1-\alpha)}{1+\lambda+2\xi\delta}, & \frac{1+2\lambda-\lambda^2}{2(1+2\lambda+6\xi\delta)} \le \alpha \le 1 \end{cases}$$ $$|a_3| \le \frac{2(1-\alpha)}{1+2\lambda+6\xi\delta},$$ and $$\left| a_3 - 2a_2^2 \right| \le \frac{2(1-\alpha)}{1+2\lambda+6\xi\delta}$$ By setting $\lambda = 1$ in Theorem 3.3, we obtain the following consequence.

Corollary 3.5. For $\mu \geq 0$ , $\delta \geq 0$ and $0 \leq \alpha < 1$ , let the function $f \in \mathfrak{B}^{\mu}_{\Sigma}(\alpha, \delta)$ be given by (1.1). Then one has the following $$\begin{split} |a_2| & \leq \left\{ \begin{array}{l} \sqrt{\frac{4(1-\alpha)}{(\mu+6\xi\delta+2)(\mu+1)}}, & 0 \leq \alpha \leq \frac{1}{\mu+6\xi\delta+2} \\ \frac{2(1-\alpha)}{\mu+2\xi\delta+1}, & \frac{1}{\mu+6\xi\delta+2} \leq \alpha \leq 1 \end{array} \right. \\ |a_3| & \leq \left\{ \begin{array}{l} \min\left\{ \frac{4(1-\alpha)^2}{(\mu+2\xi\delta+1)^2} + \frac{2(1-\alpha)}{\mu+6\xi\delta+2}, \frac{4(1-\alpha)}{(\mu+6\xi\delta+2)(\mu+1)} \right\}, & 0 \leq \mu < 1 \\ \frac{2(1-\alpha)}{\mu+6\xi\delta+2}, & \mu \geq 1 \end{array} \right. \end{split}$$ Remark 3.6. As a final remark, for $\delta = 0$ in - (i) Theorem 3.1 we obtain Theorem 1 in [7]. - (ii) Theorem 3.3 we obtain Theorem 2 in [7]. - (iii) Corollary 3.2 we obtain Theorem 1 in [16]. - (iv) Corollary 3.4 we obtain Theorem 2 in [16]. - (v) Corollary 3.5 we obtain Corollary 3 in [7].

Definition 2.1. (See [25]) For $\lambda \geq 1$ , $\mu \geq 0$ , $\delta \geq 0$ and $0 \leq \alpha < 1$ , a function $f \in \Sigma$ given by (1.1) is said to be in the class $\mathfrak{B}^{\mu}_{\Sigma}(\alpha, \lambda, \delta)$ if the following conditions hold for all $z, w \in \mathbb{U}$ : $$\operatorname{Re}\left((1-\lambda)\left(\frac{f(z)}{z}\right)^{\mu} + \lambda f'(z)\left(\frac{f(z)}{z}\right)^{\mu-1} + \xi \delta z f''(z)\right) > \alpha \tag{2.1}$$ and $$\operatorname{Re}\left((1-\lambda)\left(\frac{g(w)}{w}\right)^{\mu} + \lambda g'(w)\left(\frac{g(w)}{w}\right)^{\mu-1} + \xi \delta w g''(w)\right) > \alpha,\tag{2.2}$$ where the function $g(w) = f^{-1}(w)$ is defined by (1.4) and $\xi = \frac{2\lambda + \mu}{2\lambda + 1}$ . - Remark 2.2. In the following special cases of Definition 2.1; we show how the class of analytic bi-univalent functions $\mathfrak{B}^{\mu}_{\Sigma}(\alpha,\lambda,\delta)$ for suitable choices of $\lambda$ , $\mu$ and $\delta$ lead to certain new as well as known classes of analytic bi-univalent functions studied earlier in the literature. - (i) For $\delta = 0$ , we obtain the bi-univalent function class $\mathfrak{B}^{\mu}_{\Sigma}(\alpha, \lambda, 0) := \mathfrak{B}^{\mu}_{\Sigma}(\alpha, \lambda)$ introduced by Cağlar et al. [8]. - (ii) For $\delta = 0$ and $\mu = 1$ , we obtain the bi-univalent function class $\mathfrak{B}^1_{\Sigma}(\alpha, \lambda, 0) := \mathfrak{B}_{\Sigma}(\alpha, \lambda)$ introduced by Frasin and Aouf [11]. - (iii) For $\delta = 0$ , $\mu = 1$ , and $\lambda = 1$ , we obtain the bi-univalent function class $\mathfrak{B}^1_{\Sigma}(\alpha, 1, 0) := \mathfrak{B}_{\Sigma}(\alpha)$ introduced by Srivastava et al. [19]. - (iv) For $\delta = 0$ , $\mu = 0$ , and $\lambda = 1$ , we obtain the well-known class $\mathfrak{B}^{0}_{\Sigma}(\alpha, 1, 0) := \mathcal{S}^{*}_{\Sigma}(\alpha)$ of bi-starlike functions of order $\alpha$ . - (iv) For $\mu = 1$ , we obtain the well-known class $\mathfrak{B}^1_{\Sigma}(\alpha, \lambda, \delta) := \mathfrak{B}_{\Sigma}(\alpha, \lambda, \delta)$ of bi-univalent functions.