Bounds for Two New Subclasses of Bi-Univalent Functions Associated with Legendre Polynomials

Abstract

In this article, two new subclasses of the bi-univalent function class $\sigma$ related with Legendre polynomials are presented. Additionally, the first two Taylor–Maclaurin coefficients $\left|a_2\right|$ and $\left|a_3\right|$ for the functions belonging to these new subclasses are estimated.

1. Introduction

In 1782, Adrien-Marie Legendre discovered Legendre polynomials, which have numerous physical applications. The Legendre polynomials $P_n\left(x\right)$, sometimes called Legendre functions of the first kind, are the particular solutions to the Legendre differential equation

$$\left(1-x^2\right)y^{{}^{\prime \prime }}-2x{y}^{{}^{\prime }}+n\left(n+1\right)y=0,n\in {\mathbb{N}}_0,\left|x\right|<1.$$

Here and in the following, let $\mathbb{C}$ and $\mathbb{N}$ denote the sets of complex numbers and positive integers, respectively, and let ${\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\}$. The Legendre polynomials are defined by Rodrigues’ formula

$$P_n\left(x\right)=\frac{1}{2^{n}n!}\frac{d^n}{d{x}^n}{\left(x^2-1\right)}^n(n\in {\mathbb{N}}_0),$$

for arbitrary real or complex values of the variable x. The Legendre polynomials $P_n\left(x\right)$ are generated by the following function

$$\frac{1}{\sqrt{1-2xt+t^2}}=\sum _{n=0}^{\infty }{P}_n\left(x\right)t^n,$$

(1)

where the particular branch of ${\left(1-2xt+t^2\right)}^{-\frac{1}{2}}$ is taken to be 1 as $t\to 0$. The first few Legendre polynomials are

$$P_0\left(x\right)=1,P_1\left(x\right)=x,P_2\left(x\right)=\frac{1}{2}\left(3x^2-1\right),P_3\left(x\right)=\frac{1}{2}\left(5x^3-3x\right).$$

A general case of the Legendre polynomials and their applications can be found in [1,2]. Let $\mathcal{A}$ be the class of analytic functions in the open unit disc $U=\{z\in \mathbb{C}:\left|z\right|<1\}$ with the following Taylor–Maclaurin series expansion

$$f\left(z\right)=z+\sum _{n=2}^{\infty }{a}_n{z}^n,$$

(2)

and let $\mathcal{S}$ be the subclass of $\mathcal{A}$ consisting of univalent functions in U. An important member of the class $\mathcal{S}$ is the Koebe function

$$K\left(z\right)=z{(1-z)}^{-2}=\frac{1}{4}\left[{\left(\frac{1+z}{1-z}\right)}^2-1\right]=\sum _{n=1}^{\infty }n{z}^n,$$

for every $z\in U.$ This function maps U in a one-to-one manner onto the domain D that consists of the entire complex plane except for a slit along the negative real axis from $w=-\infty$ to $w=-\frac{1}{4}.$ The function

$$\varphi \left(z\right)=\frac{1-z}{\sqrt{1-2zcos\alpha +z^2}},$$

is in $\mathcal{P}$ for every real $\alpha$ (see [[3] Page 102]), where $\mathcal{P}$ is the Caratheodory class defined by

$$\mathcal{P}=\left\{p\left(z\right):\Re \left(p\left(z\right)\right)>0,z\in U\right\},$$

$p\left(z\right)=1+c_{1}z+c_2{z}^2+\dots$. By using (1), it is easy to check that

$$\begin{array}{cc}\hfill \varphi \left(z\right)& =1+{\displaystyle \sum _{n=1}^{\infty }}\left[P_n\left(cos\alpha \right)-P_{n-1}\left(cos\alpha \right)\right]z^n,\\ \hfill & =1+{\displaystyle \sum _{n=1}^{\infty }}{B}_n{z}^n,z\in U.\end{array}$$

(3)

If we consider

$$\begin{array}{cc}\hfill \frac{1}{{\left(\varphi \left(z\right)\right)}^2}& =\frac{1-2zcos\alpha +z^2}{{\left(1-z\right)}^2},\hfill \\ \hfill & =1+2\left(1-cos\alpha \right)\frac{z}{{\left(1-z\right)}^2}.\hfill \end{array}$$

From the geometric properties of the Koebe function, the function $\varphi$ maps the unit disc onto the right plane $\Re \left(w\right)>0$ minus the slit along the positive real axis from $\frac{1}{\left|cos\frac{\alpha }{2}\right|}$ to $\infty .$ $\varphi \left(U\right)$ is univalent, symmetric with respect to the real axis and starlike with respect to $\varphi \left(0\right)=1.$ It is well known, by using the Koebe one-quarter theorem [4], that every univalent function $f\in \mathcal{S}$ has an inverse function $f^{-1}$, which is defined by

$$f^{-1}\left(f\left(z\right)\right)=z(z\in U),$$

and

$$f\left(f^{-1}\left(w\right)\right)=w(w\in U^{*}=\{w\in \mathbb{C}:\left|w\right|<\frac{1}{4}\}),$$

where

$$g\left(w\right)=f^{-1}\left(w\right)=w-a_2{w}^2+(2a_2^2-a_3)w^3-(5a_2^3-5a_2{a}_3+a_4)w^4+\dots .$$

The function $f\in \mathcal{S}$ is said to be a bi-univalent function if its inverse $f^{-1}$ is also univalent in U. Let $\sigma$ be the class of all bi-univalent functions in U. Lewin [5] is the first author who introduced the class of analytic bi-univalent functions and estimated the second coefficient $\left|a_2\right|$. Many authors created several subclasses of analytic bi-univalent functions and found the bounds for the first two coefficients $\left|a_2\right|$ and $\left|a_3\right|$, see for example [6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23]. Let $\Omega$ be the class of all analytic functions $\omega$ in U which satisfy these conditions $\omega \left(0\right)=0$ and $\left|\omega \left(z\right)\right|<1$ for all $z\in U$. A function f is said to be subordinate to g, written as $f\left(z\right)\prec g\left(z\right)$ if there exists a Schwarz function $\omega \in \Omega$ such that $f\left(z\right)=g\left(\omega \left(z\right)\right).$ Furthermore, if the function g is univalent in $U,$ then f is subordinate to g is equivalent to $f\left(0\right)=g\left(0\right)$ and $f\left(U\right)\subset g\left(U\right).$

Definition 1.

A function $f\in \sigma$ belongs to the class $L_{\sigma }(\lambda ,\varphi )$ with $0\le \lambda \le 1$ if the following subordination conditions are satisfied

$$\lambda \left(1+\frac{z{f}^{{}^{\prime \prime }}\left(z\right)}{f^{{}^{\prime }}\left(z\right)}\right)+\left(1-\lambda \right)\left(\frac{z{f}^{{}^{\prime }}\left(z\right)}{f\left(z\right)}\right)\prec \varphi \left(z\right)\left(z\in U\right),$$

and

$$\lambda \left(1+\frac{w{g}^{{}^{\prime \prime }}\left(w\right)}{g^{{}^{\prime }}\left(w\right)}\right)+\left(1-\lambda \right)\left(\frac{w{g}^{{}^{\prime }}\left(w\right)}{g\left(w\right)}\right)\prec \varphi \left(w\right)\left(w\in U\right),$$

where $g\left(w\right)=f^{-1}\left(w\right)$.

Definition 2.

A function $f\in \sigma$ belongs to the class $L_{\sigma }\left(\gamma ,\rho ,\varphi \right)$ with $0\le \gamma ,\rho \le 1$ if the following subordination conditions are satisfied

$$\left(1-\gamma +2\rho \right)\frac{f\left(z\right)}{z}+\left(\gamma -2\rho \right)f^{{}^{\prime }}\left(z\right)+\rho z{f}^{{}^{\prime \prime }}\left(z\right)\prec \varphi \left(z\right)\left(z\in U\right),$$

and

$$\left(1-\gamma +2\rho \right)\frac{g\left(w\right)}{w}+\left(\gamma -2\rho \right)g^{{}^{\prime }}\left(w\right)+\rho w{g}^{{}^{\prime \prime }}\left(w\right)\prec \varphi \left(w\right)\left(w\in U\right),$$

where $g\left(w\right)=f^{-1}\left(w\right)$.

Remark 1.

In Definition 1, if $\lambda =1$ and $\alpha =\pi$, then the subclass in [15] will be obtained. If $\lambda =0$ and $\alpha =\pi$, then the subclass in [24] will be obtained. In addition, putting $\alpha =\pi$, this yields to the subclass in [25].

Remark 2.

In Definition 2, taking $\rho =0$ and $\alpha =\pi$, the subclass in [26] will be obtained. In addition, putting $\gamma =1,\rho =0$ and $\alpha =\pi$, this yields to the subclass in [27].

In this paper, the estimates for initial coefficients of functions in the two classes $L_{\sigma }(\lambda ,\varphi )$ and $L_{\sigma }\left(\gamma ,\rho ,\varphi \right)$ are found.

2. The Estimate of the Coefficients for the Classes $L_{\sigma }(\lambda ,\varphi )$ and $L_{\sigma }(\gamma ,\rho ,\varphi )$

Lemma 1 

([4]). Let $\omega \in \Omega$ with $\omega \left(z\right)={\displaystyle \sum _{n=1}^{\infty }}{\omega }_n{z}^n$$\left(z\in U\right)$. Then

$$\left|{\omega }_1\right|\le 1,\left|{\omega }_n\right|\le 1-{\left|{\omega }_1\right|}^2\left(n\in \mathbb{N}\setminus \left\{1\right\}\right).$$

Theorem 1.

Let the function $f\in L_{\sigma }(\lambda ,\varphi )$. Then

$$\left|a_2\right|\le \frac{\sqrt{2}\left(1-cos\alpha \right)}{\sqrt{\left(1+\lambda \right)[3\lambda (cos\alpha +1)+5+cos\alpha ]}},$$

and

$$\left|a_3\right|\le \left\{\begin{array}{c}\frac{1-cos\alpha }{2\left(1+2\lambda \right)}ifcos\alpha \ge 1-\frac{{\left(1+\lambda \right)}^2}{2\left(1+2\lambda \right)}\\ \\ \frac{\left(1-cos\alpha \right)[2\left(1+2\lambda \right)\left(1-cos\alpha \right)-{\left(1+\lambda \right)}^2]}{\left(1+2\lambda \right)(1+\lambda )[3\lambda (cos\alpha +1)+5+cos\alpha ]}+\frac{1-cos\alpha }{2\left(1+2\lambda \right)}ifcos\alpha <1-\frac{{\left(1+\lambda \right)}^2}{2\left(1+2\lambda \right)}\end{array}\right.$$

Proof .

Since $f\in L_{\sigma }(\lambda ,\varphi )$, from Definition 1, we have

$$\lambda \left(1+\frac{z{f}^{{}^{\prime \prime }}\left(z\right)}{f^{{}^{\prime }}\left(z\right)}\right)+\left(1-\lambda \right)\left(\frac{z{f}^{{}^{\prime }}\left(z\right)}{f\left(z\right)}\right)=\varphi \left(u\left(z\right)\right)\left(z\in U\right),$$

(4)

and

$$\lambda \left(1+\frac{w{g}^{{}^{\prime \prime }}\left(w\right)}{g^{{}^{\prime }}\left(w\right)}\right)+\left(1-\lambda \right)\left(\frac{w{g}^{{}^{\prime }}\left(w\right)}{g\left(w\right)}\right)=\varphi \left(v\left(w\right)\right)\left(w\in U\right),$$

(5)

for some $0\le \lambda \le 1$, where $g\left(w\right)=f^{-1}\left(w\right)$ and $u,v\in \Omega$ such that

$$u\left(z\right)=\sum _{n=1}^{\infty }{b}_n{z}^n,$$

and

$$v\left(w\right)=\sum _{n=1}^{\infty }{c}_n{w}^n.$$

Then

$$\varphi \left(u\left(z\right)\right)=1+B_1{b}_{1}z+\left(B_1{b}_2+B_2{b}_1^2\right)z^2+\left(B_1{b}_3+2b_1{b}_2{B}_2+B_3{b}_1^3\right)z^3+...,$$

(6)

and

$$\varphi \left(v\left(w\right)\right)=1+B_1{c}_{1}w+\left(B_1{c}_2+B_2{c}_1^2\right)w^2+\left(B_1{c}_3+2c_1{c}_2{B}_2+B_3{c}_1^3\right)w^3+...,$$

(7)

where

$$B_1=cos\alpha -1,B_2=\frac{1}{2}(cos\alpha -1)\left(1+3cos\alpha \right)\mathrm{and}{B}_3=\frac{1}{2}\left(5{cos}^3\alpha -3{cos}^2\alpha -3cos\alpha +1\right).$$

(8)

Then, Equations (4) and (5) become

$$\begin{array}{cc}\hfill & \lambda \left[1+2a_{2}z+\left(6a_3-4a_2^2\right)z^2+...\right]+\left(1-\lambda \right)\left[1+a_{2}z+\left(2a_3-a_2^2\right)z^2+...\right]\\ \hfill & =1+B_1{b}_{1}z+\left(B_1{b}_2+B_2{b}_1^2\right)z^2+\left(B_1{b}_3+2b_1{b}_2{B}_2+B_3{b}_1^3\right)z^3+...,\end{array}$$

(9)

and

$$\begin{array}{cc}\hfill & \lambda \left[1-2a_{2}w+\left(8a_2^2-6a_3\right)w^2+...\right]+\left(1-\lambda \right)\left[1-a_{2}w+\left(3a_2^2-2a_3\right)w^2+...\right]\\ \hfill & =1+B_1{c}_{1}w+\left(B_1{c}_2+B_2{c}_1^2\right)w^2+\left(B_1{c}_3+2c_1{c}_2{B}_2+B_3{c}_1^3\right)w^3+....\end{array}$$

(10)

Now, equating the corresponding coefficients in (9) and (10), we get

$$\left(1+\lambda \right)a_2=B_1{b}_1,$$

(11)

$$2\left(1+2\lambda \right)a_3-\left(1+3\lambda \right)a_2^2=B_1{b}_2+B_2{b}_1^2,$$

(12)

$$-\left(1+\lambda \right)a_2=B_1{c}_1,$$

(13)

$$\left(3+5\lambda \right)a_2^2-2\left(1+2\lambda \right)a_3=B_1{c}_2+B_2{c}_1^2.$$

(14)

(11) and (13) yield

$$b_1=-c_1,$$

(15)

and

$$b_1^2+c_1^2=\frac{2{\left(1+\lambda \right)}^2}{B_1^2}{a}_2^2.$$

(16)

From (12), (14) and (16), we have

$$a_2^2=\frac{B_1^3}{2\left(1+\lambda \right)\left[B_1^2-\left(1+\lambda \right)B_2\right]}\left(b_2+c_2\right).$$

By using Lemma 1, (11) and (15), we obtain

$$\begin{array}{cc}\hfill {\left|a_2\right|}^2& \le \frac{{\left|B_1\right|}^3\left(1-{\left|b_1\right|}^2\right)}{\left(1+\lambda \right)\left|B_1^2-\left(1+\lambda \right)B_2\right|},\hfill \\ \hfill & =\frac{{\left|B_1\right|}^3}{\left(1+\lambda \right)\left[\left|B_1^2-\left(1+\lambda \right)B_2\right|+\left(1+\lambda \right)\left|B_1\right|\right]}.\hfill \end{array}$$

Therefore,

$$\left|a_2\right|\le \frac{\left|B_1\right|\sqrt{\left|B_1\right|}}{\sqrt{\left(1+\lambda \right)\left[\left|B_1^2-\left(1+\lambda \right)B_2\right|+\left(1+\lambda \right)\left|B_1\right|\right]}}.$$

(17)

By noting that,

$$\begin{array}{cc}\hfill B_1^2-\left(1+\lambda \right)B_2& =\frac{1}{2}(1-cos\alpha )\left\{\left(1+3\lambda \right)cos\alpha +3+\lambda \right\},\hfill \\ \hfill & \ge (1-cos\alpha )\left(1-\lambda \right)\ge 0,\hfill \end{array}$$

now substituting the values of $B_1$ and $B_2$ from (8) in (17), we obtain

$$\left|a_2\right|\le \frac{\sqrt{2}\left(1-cos\alpha \right)}{\sqrt{\left(1+\lambda \right)[3\lambda (cos\alpha +1)+5+cos\alpha ]}},$$

which is the required estimation for $\left|a_2\right|.$

Next, in order to estimate $\left|a_3\right|$, subtracting (14) from (12), we obtain

$$a_3=a_2^2+\frac{B_1}{4\left(1+2\lambda \right)}\left(b_2-c_2\right).$$

By using Lemma 1 and (11), we find

$$\left|a_3\right|\le \left[1-\frac{{\left(1+\lambda \right)}^2}{2\left(1+2\lambda \right)\left|B_1\right|}\right]{\left|a_2\right|}^2+\frac{\left|B_1\right|}{2\left(1+2\lambda \right)}.$$

Case 1. If $1-\frac{{\left(1+\lambda \right)}^2}{2\left(1+2\lambda \right)\left|B_1\right|}\le 0$, then

$$\left|a_3\right|\le \frac{\left|B_1\right|}{2\left(1+2\lambda \right)}.$$

Case 2. If $1-\frac{{\left(1+\lambda \right)}^2}{2\left(1+2\lambda \right)\left|B_1\right|}>0$, then

$$\left|a_3\right|\le \left[1-\frac{{\left(1+\lambda \right)}^2}{2\left(1+2\lambda \right)\left|B_1\right|}\right]{\left|a_2\right|}^2+\frac{\left|B_1\right|}{2\left(1+2\lambda \right)}.$$

Therefore,

$$\left|a_3\right|\le \left\{\begin{array}{c}\frac{1-cos\alpha }{2\left(1+2\lambda \right)}ifcos\alpha \ge 1-\frac{{\left(1+\lambda \right)}^2}{2\left(1+2\lambda \right)}\\ \\ \frac{\left(1-cos\alpha \right)[2\left(1+2\lambda \right)\left(1-cos\alpha \right)-{\left(1+\lambda \right)}^2]}{\left(1+2\lambda \right)(1+\lambda )[3\lambda (cos\alpha +1)+5+cos\alpha ]}+\frac{1-cos\alpha }{2\left(1+2\lambda \right)}ifcos\alpha <1-\frac{{\left(1+\lambda \right)}^2}{2\left(1+2\lambda \right)}\end{array}\right.$$

which completes the proof. □

Theorem 2.

Let the function $f\in L_{\sigma }(\gamma ,\rho ,\varphi )$. Then

$$\left|a_2\right|\le \frac{\sqrt{2}(1-cos\alpha )}{\sqrt{2\left(1+2\gamma +2\rho \right)\left(1-cos\alpha \right)+3{\left(1+\gamma \right)}^2(1+cos\alpha )}},$$

and

$$\left|a_3\right|\le \left\{\begin{array}{c}\frac{1-cos\alpha }{\left(1+2\gamma +2\rho \right)}ifcos\alpha \ge 1-\frac{{\left(1+\gamma \right)}^2}{\left(1+2\gamma +2\rho \right)}\\ \\ \frac{2[\left(1+2\gamma +2\rho \right)\left(1-cos\alpha \right)-{\left(1+\gamma \right)}^2]\left(1-cos\alpha \right)}{\left(1+2\gamma +2\rho \right)[2\left(1+2\gamma +2\rho \right)\left(1-cos\alpha \right)+3{\left(1+\gamma \right)}^2(1+cos\alpha )]}+\frac{1-cos\alpha }{1+2\gamma +2\rho }ifcos\alpha <1-\frac{{\left(1+\gamma \right)}^2}{\left(1+2\gamma +2\rho \right)}\end{array}\right.$$

Proof. 

Since $f\in L_{\sigma }(\gamma ,\rho ,\varphi )$, from Definition 2, we have

$$\left(1-\gamma +2\rho \right)\frac{f\left(z\right)}{z}+\left(\gamma -2\rho \right)f^{{}^{\prime }}\left(z\right)+\rho z{f}^{{}^{\prime \prime }}\left(z\right)=\varphi \left(u\left(z\right)\right)\left(z\in U\right),$$

(18)

and

$$\left(1-\gamma +2\rho \right)\frac{g\left(w\right)}{w}+\left(\gamma -2\rho \right)g^{{}^{\prime }}\left(w\right)+\rho w{g}^{{}^{\prime \prime }}\left(w\right)=\varphi \left(v\left(w\right)\right)\left(w\in U\right),$$

(19)

for $0\le \gamma ,\rho \le 1$, where $g\left(w\right)=f^{-1}\left(w\right)$ and $u,v\in \Omega$ are defined as in Theorem 1. Then, rewriting (18) and (19) as

$$\begin{array}{cc}\hfill & \left(1-\gamma +2\rho \right)\left[1+a_{2}z+a_3{z}^2+...\right]+\left(\gamma -2\rho \right)\left[1+2a_{2}z+3a_3{z}^2+...\right]\hfill \\ \hfill & +\rho z\left[2a_2+6a_{3}z+...\right]=1+B_1{b}_{1}z+\left(B_1{b}_2+B_2{b}_1^2\right)z^2\hfill \\ \hfill & +\left(B_1{b}_3+2b_1{b}_2{B}_2+B_3{b}_1^3\right)z^3+...,\hfill \end{array}$$

(20)

and

$$\begin{array}{cc}\hfill & \left(1-\gamma +2\rho \right)\left[1-2a_{2}w+\left(2a_2^2-a_3\right)w^2+...\right]+\left(\gamma -2\rho \right)\left[1-2a_{2}w+\left(6a_2^2-3a_3\right)w^2+...\right]\hfill \\ \hfill & +\rho w\left[-2a_2+\left(12a_2^2-6a_3\right)w+...\right]=1+B_1{c}_{1}w+\left(B_1{c}_2+B_2{c}_1^2\right)w^2\hfill \\ \hfill & +\left(B_1{c}_3+2c_1{c}_2{B}_2+B_3{c}_1^3\right)w^3+...,\hfill \end{array}$$

(21)

where $B_1,B_2$ and $B_3$ are defined as in (8). Now, equating the coefficients in (20) and (21) yields

$$\left(1+\gamma \right)a_2=B_1{b}_1,$$

(22)

$$\left(1+2\gamma +2\rho \right)a_3=B_1{b}_2+B_2{b}_1^2,$$

(23)

$$-\left(1+\gamma \right)a_2=B_1{c}_1,$$

(24)

$$\left(1+2\gamma +2\rho \right)\left(2a_2^2-a_3\right)=B_1{c}_2+B_2{c}_1^2.$$

(25)

From (22) and (24), it is easy to see that

$$b_1=-c_1,$$

(26)

and

$$b_1^2+c_1^2=\frac{2{\left(1+\gamma \right)}^2}{B_1^2}{a}_2^2.$$

(27)

From (23), (25) and (27), we have

$$a_2^2=\frac{B_1^3}{2\left[\left(1+2\gamma +2\rho \right)B_1^2-{\left(1+\gamma \right)}^2{B}_2\right]}\left(b_2+c_2\right).$$

By using Lemma 1, (22) and (26), we obtain

$$\begin{array}{cc}\hfill {\left|a_2\right|}^2& \le \frac{{\left|B_1\right|}^3\left(1-{\left|b_1\right|}^2\right)}{\left|\left(1+2\gamma +2\rho \right)B_1^2-{\left(1+\gamma \right)}^2{B}_2\right|},\hfill \\ \hfill & =\frac{{\left|B_1\right|}^3}{\left|\left(1+2\gamma +2\rho \right)B_1^2-{\left(1+\gamma \right)}^2{B}_2\right|+{\left(1+\gamma \right)}^2\left|B_1\right|}.\hfill \end{array}$$

Therefore,

$$\left|a_2\right|\le \frac{\left|B_1\right|\sqrt{\left|B_1\right|}}{\sqrt{\left|\left(1+2\gamma +2\rho \right)B_1^2-{\left(1+\gamma \right)}^2{B}_2\right|+{\left(1+\gamma \right)}^2\left|B_1\right|}}.$$

(28)

By noting that,

$$\begin{array}{cc}\hfill & \left(1+2\gamma +2\rho \right)B_1^2-{\left(1+\gamma \right)}^2{B}_2\hfill \\ \hfill & =\frac{1}{2}(1-cos\alpha )\{2\left(1+2\gamma +2\rho \right)\left(1-cos\alpha \right)+{\left(1+\gamma \right)}^2(1+3cos\alpha )\},\hfill \\ \hfill & =\frac{1}{2}(1-cos\alpha )\{2{\left(1+\gamma \right)}^2(1+cos\alpha )+[1+\gamma (2-\gamma )+4\rho ]\left(1-cos\alpha \right)\}>0,\hfill \end{array}$$

now substituting the values of $B_1$ and $B_2$ from (8) in (28), we obtain

$$\left|a_2\right|\le \frac{\sqrt{2}(1-cos\alpha )}{\sqrt{2\left(1+2\gamma +2\rho \right)\left(1-cos\alpha \right)+3{\left(1+\gamma \right)}^2(1+cos\alpha )}},$$

which is the desired estimation for $\left|a_2\right|.$

Next, in order to estimate $\left|a_3\right|,$ subtracting (25) from (23), we obtain

$$a_3=a_2^2+\frac{B_1}{2\left(1+2\gamma +2\rho \right)}\left(b_2-c_2\right).$$

By using Lemma 1 and (22), we find

$$\left|a_3\right|\le \left[1-\frac{{\left(1+\gamma \right)}^2}{\left(1+2\gamma +2\rho \right)\left|B_1\right|}\right]{\left|a_2\right|}^2+\frac{\left|B_1\right|}{\left(1+2\gamma +2\rho \right)}.$$

Case 1. If $1-\frac{{\left(1+\gamma \right)}^2}{\left(1+2\gamma +2\rho \right)\left|B_1\right|}\le 0$, then

$$\left|a_3\right|\le \frac{\left|B_1\right|}{\left(1+2\gamma +2\rho \right)}.$$

Case 2. If $1-\frac{{\left(1+\gamma \right)}^2}{\left(1+2\gamma +2\rho \right)\left|B_1\right|}>0$, then

$$\left|a_3\right|\le \left[1-\frac{{\left(1+\gamma \right)}^2}{\left(1+2\gamma +2\rho \right)\left|B_1\right|}\right]{\left|a_2\right|}^2+\frac{\left|B_1\right|}{\left(1+2\gamma +2\rho \right)}.$$

Therefore,

$$\left|a_3\right|\le \left\{\begin{array}{c}\frac{1-cos\alpha }{\left(1+2\gamma +2\rho \right)}ifcos\alpha \ge 1-\frac{{\left(1+\gamma \right)}^2}{\left(1+2\gamma +2\rho \right)}\\ \\ \frac{2[\left(1+2\gamma +2\rho \right)\left(1-cos\alpha \right)-{\left(1+\gamma \right)}^2]\left(1-cos\alpha \right)}{\left(1+2\gamma +2\rho \right)[2\left(1+2\gamma +2\rho \right)\left(1-cos\alpha \right)+3{\left(1+\gamma \right)}^2(1+cos\alpha )]}+\frac{1-cos\alpha }{1+2\gamma +2\rho }ifcos\alpha <1-\frac{{\left(1+\gamma \right)}^2}{\left(1+2\gamma +2\rho \right)}\end{array}\right.$$

which completes the proof. □

3. Conclusions

In this paper, we have used the Legendre polynomials to define and study two new subclasses of the bi-univalent function class $\sigma$. Moreover, we have provided the estimations for the first two Taylor–Maclaurin coefficients $\left|a_2\right|$ and $\left|a_3\right|$ for the functions belonging to these new subclasses. Some special cases have been discussed as applications of our main results.