Second Hankel Determinant for certain class of bi-univalent functions defined by Chebyshev polynomials

Lemma 1. [34] If the function $p \in \mathcal{P}$ is given by the series (1.7), then the following sharp estimate holds: $$|c_k| \le 2, \qquad k = 1, 2, \cdots.$$ (1.8)

Lemma 2. [20] If the function $p \in \mathcal{P}$ is given by the series (1.7), then $$2c_2 = c_1^2 + x(4 - c_1^2)$$ $$4c_3 = c_1^3 + 2c_1(4 - c_1^2)x - c_1(4 - c_1^2)x^2 + 2(4 - c_1^2)(1 - |x|^2)z$$ for some x, z with $|x| \le 1$ and $|z| \le 1$ . In this present paper, we consider a subclass $\mathcal{N}_{\sigma}^{\mu}(\lambda, t)$ of analytic and bi-univalent functions using the Chebyshev polynomials expansions and find the second Hankel determinant estimates. Further we discuss its consequences.

Theorem 1. Let $f \in \sigma$ of the form (1.1) be in $\mathcal{N}^{\mu}_{\sigma}(\lambda; t)$ . Then $$|a_{2}a_{4} - a_{3}^{2}| \leq \begin{cases} K(2^{-}, t) & ; M_{1} \geq 0 \text{ and } M_{2} \geq 0 \\ \max\left\{\frac{4t^{2}}{(2\lambda + \mu)^{2}}, K(2^{-}, t)\right\} & ; M_{1} > 0 \text{ and } M_{2} < 0 \\ \frac{4t^{2}}{(2\lambda + \mu)^{2}} & ; M_{1} \leq 0 \text{ and } M_{2} \leq 0 \\ \max\left\{K(c_{0}, t), K(2^{-}, t)\right\} & ; M_{1} < 0 \text{ and } M_{2} > 0 \end{cases}$$ where $$K(2^{-}, t) = \frac{4t^{2}}{(2\lambda + \mu)^{2}} + \frac{M_{1} + 3M_{2}}{6(\lambda + \mu)^{4}(2\lambda + \mu)^{2}(3\lambda + \mu)},$$ $$K(c_{0}, t) = \frac{4t^{2}}{(2\lambda + \mu)^{2}} - \frac{3M_{2}^{2}}{8M_{1}(\lambda + \mu)^{4}(2\lambda + \mu)^{2}(3\lambda + \mu)}, \qquad c_{0} = \sqrt{\frac{-6M_{2}}{M_{1}}}$$ and $$\begin{split} M_1 &:= M_1(\lambda,\,\mu;\,t) &= 16t^2 \left| 3(2t^2-1)(\lambda+\mu)^3 - (\mu^2+3\mu+2)(3\lambda+\mu)t^2 \right| (2\lambda+\mu)^2 \\ &- 24t \left[ t^2(3\lambda+\mu) + (4t^2-1)(\lambda+\mu)(2\lambda+\mu) \right] (\lambda+\mu)^2 (2\lambda+\mu) \\ &- 24t^2\lambda^2(\lambda+\mu)^3, \\ M_2 &:= M_2(\lambda,\,\mu;\,t) &= 8t \left[ t^2(2\lambda+\mu)(3\lambda+\mu) + (4t^2-1)(\lambda+\mu)(2\lambda+\mu)^2 \right] \end{split}$$

Corollary 1. Let $f \in \sigma$ of the form (1.1) be in $\mathcal{B}_{\sigma}(\lambda, t)$ . Then $$|a_{2}a_{4} - a_{3}^{2}| \leq \begin{cases} K(2^{-}, t) & ; M_{3} \geq 0 \text{ and } M_{4} \geq 0 \\ \max\left\{\frac{4t^{2}}{(2\lambda+1)^{2}}, K(2^{-}, t)\right\} & ; M_{3} > 0 \text{ and } M_{4} < 0 \\ \frac{4t^{2}}{(2\lambda+1)^{2}} & ; M_{3} \leq 0 \text{ and } M_{4} \leq 0 \\ \max\left\{K(c_{0}, t), K(2^{-}, t)\right\} & ; M_{3} < 0 \text{ and } M_{4} > 0 \end{cases}$$ where $$K(2^{-}, t) = \frac{4t^{2}}{(2\lambda + 1)^{2}} + \frac{M_{3} + 3M_{4}}{6(\lambda + 1)^{4}(2\lambda + 1)^{2}(3\lambda + 1)},$$ $$K(c_{0}, t) = \frac{4t^{2}}{(2\lambda + 1)^{2}} - \frac{3M_{4}^{2}}{8M_{3}(\lambda + 1)^{4}(2\lambda + 1)^{2}(3\lambda + 1)}, \qquad c_{0} = \sqrt{\frac{-6M_{4}}{M_{3}}}$$ and $$M_{3} = 16t^{2} |3(2t^{2} - 1)(\lambda + 1)^{3} - 6(3\lambda + 1)t^{2}| (2\lambda + 1)^{2}$$ $$-24t [t^{2}(3\lambda + 1) + (4t^{2} - 1)(\lambda + 1)(2\lambda + 1)] (\lambda + 1)^{2}(2\lambda + 1)$$ $$-24t^{2}\lambda^{2}(\lambda + 1)^{3},$$ $$M_{4} = 8t [t^{2}(2\lambda + 1)(3\lambda + 1) + (4t^{2} - 1)(\lambda + 1)(2\lambda + 1)^{2}$$ $$+t(2\lambda + 1)^{2}(\lambda + 1) - 2t(\lambda + 1)^{2}(3\lambda + 1)] (\lambda + 1)^{2}.$$

Corollary 2. Let $f \in \sigma$ of the form (1.1) be in $\mathcal{B}^{\mu}_{\sigma}(t)$ . Then $$|a_{2}a_{4} - a_{3}^{2}| \leq \begin{cases} K(2^{-}, t) & ; M_{5} \geq 0 \text{ and } M_{6} \geq 0 \\ \max\left\{\frac{4t^{2}}{(2+\mu)^{2}}, K(2^{-}, t)\right\} & ; M_{5} > 0 \text{ and } M_{6} < 0 \\ \frac{4t^{2}}{(2+\mu)^{2}} & ; M_{5} \leq 0 \text{ and } M_{6} \leq 0 \\ \max\left\{K(c_{0}, t), K(2^{-}, t)\right\} & ; M_{5} < 0 \text{ and } M_{6} > 0 \end{cases}$$ where $$K(2^{-}, t) = \frac{4t^{2}}{(2+\mu)^{2}} + \frac{M_{5} + 3M_{6}}{6(1+\mu)^{4}(2+\mu)^{2}(3+\mu)},$$ $$K(c_{0}, t) = \frac{4t^{2}}{(2+\mu)^{2}} - \frac{3M_{6}^{2}}{8M_{5}(1+\mu)^{4}(2+\mu)^{2}(3+\mu)}, \qquad c_{0} = \sqrt{\frac{-6M_{6}}{M_{5}}}$$ and $$M_5 = 16t^2 |3(2t^2 - 1)(1 + \mu)^3 - (\mu^2 + 3\mu + 2)(3 + \mu)t^2 | (2 + \mu)^2$$ $$-24t [t^2(3 + \mu) + (4t^2 - 1)(1 + \mu)(2 + \mu)] (1 + \mu)^2 (2 + \mu)$$ $$-24t^2 (1 + \mu)^3,$$ $$M_6 = 8t [t^2(2 + \mu)(3 + \mu) + (4t^2 - 1)(1 + \mu)(2 + \mu)^2$$ $$+t(2 + \mu)^2 (1 + \mu) - 2t(1 + \mu)^2 (3 + \mu)] (1 + \mu)^2.$$

Corollary 3. Let $f \in \sigma$ of the form (1.1) be in $\mathcal{B}_{\sigma}(t)$ . Then $$|a_2 a_4 - a_3^2| \le \begin{cases} t^2 (1 - t^2), & \frac{1}{2} < t \le t_{0_1}; \\ \frac{t(260t^4 + 84t^3 - 139t^2 - 18t + 9)}{8(18t^3 + 42t^2 - 17t - 9)}, & t_{0_1} < t < 1, \end{cases}$$ where, the value of $t_{0_1}$ , which is approximately $t_{0_1} = 0.603615$ , is root of equation $M_1 = 0$ for $\lambda = \mu$ and $\frac{1}{2} < t < 1$ .

Corollary 4. Let $f \in \sigma$ of the form (1.1) be in $\mathcal{S}_{\sigma}^{*}(t)$ . Then $$|a_2 a_4 - a_3^2| \le \begin{cases} \frac{8t^2}{3}, & \frac{1}{2} < t \le \frac{7 + \sqrt{401}}{44}; \\ t^2 + \frac{t(2 + t - 11t^2)^2}{3(22t^2 - 7t - 4)}, & \frac{7 + \sqrt{401}}{44} < t < 1. \end{cases}$$ Remark 5. For specializing the parameters involving in Theorem 1, the results discussed are improve the results of Mustafa [28]. Acknowledgment: We record our sincere thanks to the referees for their insightful suggestions to improve the results as well as the present form of the article.

Definition 1. For $\lambda \geq 1$ , $\mu \geq 0$ and $t \in (1/2, 1]$ , a function $f \in \sigma$ given by (1.1) is said to be in the class $\mathcal{N}_{\sigma}^{\mu}(\lambda, t)$ if the following subordinations hold for all $z, w \in \mathbb{D}$ : $$(1 - \lambda) \left(\frac{f(z)}{z}\right)^{\mu} + \lambda f'(z) \left(\frac{f(z)}{z}\right)^{\mu - 1} \prec H(z, t) := \frac{1}{1 - 2tz + z^2}$$ (1.3) and $$(1 - \lambda) \left(\frac{g(w)}{w}\right)^{\mu} + \lambda g'(w) \left(\frac{g(w)}{w}\right)^{\mu - 1} \prec H(w, t) := \frac{1}{1 - 2tw + w^2},\tag{1.4}$$ where the function $g = f^{-1}$ is defined by (1.2). We note that if $t = \cos \alpha$ , where $\alpha \in (-\pi/3, \pi/3)$ , then $$H(z,t) = \frac{1}{1 - 2\cos\alpha z + z^2} = 1 + \sum_{n=1}^{\infty} \frac{\sin(n+1)\alpha}{\sin\alpha} z^n \quad (z \in \mathbb{D}).$$ Thus $$H(z,t) = 1 + 2\cos\alpha z + (3\cos^2\alpha - \sin^2\alpha)z^2 + \dots \quad (z \in \mathbb{D}).$$ It also can be write <span id="page-2-0"></span> $$H(z,t) = 1 + U_1(t)z + U_2(t)z^2 + \dots \quad (z \in \mathbb{D}, \quad t \in (-1,1))$$ (1.5) where $$U_{n-1} = \frac{\sin(n \operatorname{arc} \cos t)}{\sqrt{1 - t^2}} \quad (n \in \mathbb{N})$$ are the Chebyshev polynomials of the second kind and we have $$U_n(t) = 2tU_{n-1}(t) - U_{n-2}(t),$$ and $$U_1(t) = 2t$$ , $U_2(t) = 4t^2 - 1$ , $U_3(t) = 8t^3 - 4t$ , $U_4(t) = 16t^4 - 12t^2 + 1$ , .... (1.6) The generating function of the first kind of Chebyshev polynomial $T_n(t)$ , $t \in [-1, 1]$ is given by $$\sum_{n=0}^{\infty} T_n(t)z^n = \frac{1-tz}{1-2tz+z^2} \qquad (z \in \mathbb{D}).$$ The first kind of Chebyshev polynomial $T_n(t)$ and second kind of Chebyshev polynomial $U_n(t)$ are connected by: $$\frac{dT_n(t)}{dt} = nU_{n-1}(t); \quad T_n(t) = U_n(t) - tU_{n-1}(t); \quad 2T_n(t) = U_n(t) - U_{n-2}(t).$$ The class $\mathcal{N}^{\mu}_{\sigma}(\lambda, t)$ was introduced and studied by Bulut et al. [9]. Also, they discussed initial coefficient estimates and Fekete-Szegö bounds for the class $\mathcal{N}^{\mu}_{\sigma}(\lambda, t)$ and it's subclasses given in the following remark. Remark 1. For $\mu = 1$ , we get the class $\mathcal{N}_{\sigma}^{1}(\lambda, t) = \mathcal{B}_{\sigma}(\lambda, t)$ consists of functions $f \in \sigma$ satisfying the condition $$(1 - \lambda)\frac{f(z)}{z} + \lambda f'(z) \prec H(z, t) = \frac{1}{1 - 2tz + z^2} \qquad (z \in \mathbb{D})$$ and $$(1 - \lambda)\frac{g(w)}{w} + \lambda g'(w) \prec H(w, t) = \frac{1}{1 - 2tw + w^2} \qquad (w \in \mathbb{D})$$ where the function $g = f^{-1}$ is defined by (1.2). This class was introduced and studied by Bulut et al. [10] (see also [28]). Remark 2. For $\lambda = 1$ , we have a class $\mathcal{N}^{\mu}_{\sigma}(1,t) = \mathcal{B}^{\mu}_{\sigma}(t)$ consists of bi-Bazilevič functions: $$f'(z)\left(\frac{f(z)}{z}\right)^{\mu-1} \prec H(z,t) = \frac{1}{1 - 2tz + z^2} \qquad (z \in \mathbb{D})$$ and $$g'(w)\left(\frac{g(w)}{w}\right)^{\mu-1} \prec H(w,t) = \frac{1}{1 - 2tw + w^2} \qquad (w \in \mathbb{D})$$ where the function $g = f^{-1}$ is defined by (1.2). Remark 3. For $\lambda = 1$ and $\mu = 1$ , we have the class $\mathcal{N}_{\sigma}^{1}(1,t) = \mathcal{B}_{\sigma}(t)$ consists of functions f satisfying the condition $$f'(z) \prec H(z,t) = \frac{1}{1 - 2tz + z^2}$$ $(z \in \mathbb{D})$ and $$g'(w) \prec H(w,t) = \frac{1}{1 - 2tw + w^2}$$ $(w \in \mathbb{D})$ where the function $g = f^{-1}$ is defined by (1.2). Remark 4. For $\lambda = 1$ and $\mu = 0$ , we have the class $\mathcal{N}_{\sigma}^{0}(1,t) = \mathcal{S}_{\sigma}^{*}(t)$ consists of functions f satisfying the condition $$\frac{zf'(z)}{f(z)} \prec H(z,t) = \frac{1}{1 - 2tz + z^2} \qquad (z \in \mathbb{D})$$ and $$\frac{wg'(w)}{g(w)} \prec H(w,t) = \frac{1}{1 - 2tw + w^2} \qquad (w \in \mathbb{D})$$ where the function $g = f^{-1}$ is defined by (1.2). Next we state the following lemmas we shall use to establish the desired bounds in our study. Let $\mathcal{P}$ denote the class of functions p(z) of the form <span id="page-3-0"></span> $$p(z) = 1 + c_1 z + c_2 z^2 + c_3 z^3 + \cdots, (1.7)$$ which are analytic in the open unit disc $\mathbb{D}$ .