Initial Coefficient Bounds for Bi-Close-to-Convex Classes of n-Fold-Symmetric Bi-Univalent Functions

Abstract

In this article, the strong class of bi-close-to-convex functions of order $\alpha$ and $\beta$ in n-fold symmetric bi-univalent functions, which is the subclass of $\sigma$, is introduced. The upper bound value for $\left|a_{n+1}\right|$, $\left|a_{2n+1}\right|$ for functions in these classes are obtained. Moreover, the Fekete–Szegö relation for our classes of functions are established.

1. Introduction and Definitions

Let $\mathcal{A}$ denote the family of functions g analytic in the open unit disk $\mathbb{U}=\{z\in \mathbb{C}:|z|<1\}$ of the form

$$\begin{array}{c}\hfill g\left(z\right)=z+\sum _{j=2}^{\infty }{a}_j{z}^j,z\in \mathbb{U}.\end{array}$$

(1)

Let $\mathcal{S}$ denote the class of all functions in $\mathcal{A}$ that are univalent in $\mathbb{U}$. Refer to [1] for further explanations on univalent functions. Let ${\mathcal{S}}^{*}\left(\beta \right)$, $\mathcal{C}\left(\beta \right)$, $0\le \beta <1$ be the subclasses of $\mathcal{S}$ containing starlike and convex functions of order $\beta$. Their analytic representations are

$${\mathcal{S}}^{*}\left(\beta \right)=\left\{g:g\in \mathcal{S},\Re \left({\displaystyle \frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}}\right)>\beta (z\in \mathbb{U})\right\},$$

and

$$\mathcal{C}\left(\beta \right)=\left\{g:g\in \mathcal{S},\Re \left(1+{\displaystyle \frac{z{g}^{\prime \prime }\left(z\right)}{g^{\prime }\left(z\right)}}\right)>\beta (z\in \mathbb{U})\right\},$$

respectively. The class $\mathcal{C}\left(0\right)$ contains all convex univalent functions and is denoted by $\mathcal{C}$. Various classes, such as ${\mathcal{S}}^{*}\left(\beta \right)$ and $\mathcal{C}\left(\beta \right)$, of starlike and convex functions have been investigated by many authors. These functions are commonly described by the value $z{g}^{\prime }\left(z\right)/g\left(z\right)$ or $1+z{g}^{\prime \prime }\left(z\right)/g^{\prime }\left(z\right)$ lying in a starlike domain with respect to 1 in the right-half plane. Then, every function $g\in \mathcal{S}$ has an inverse $g^{-1}$ of the form

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

and

$$g\left(g^{-1}\left(w\right)\right)=w,\left|w\right|<r_0\left(g\right);r_0\left(g\right)\ge 1/4$$

where

$$\begin{array}{c}\hfill g^{-1}\left(w\right)=w-a_2{w}^2+(2a_2^2-a_3)w^3-(5a_2^3-5a_2{a}_3+a_4)w^4+\cdots .\end{array}$$

(2)

A function is said to be bi-univalent in $\mathbb{U}$ if both g and $g^{-1}$ are univalent in $\mathbb{U}$. Let $\sigma$ denote the class of bi-univalent function in $\mathbb{U}$ given by (2). The bound $|a_2| \le 1.51$ is found from Lewin [2]. Brannan and Taha [3] computed the values $|a_2|$ and $|a_3|$ of the functions in the classes ${\mathcal{S}}_{\sigma }^{*}\left(\beta \right)$ and ${\mathcal{C}}_{\sigma }\left(\beta \right)$, motivated by Lewin [2] (refer [3]). Brannan and Clunie [4] found $|a_2| \le \sqrt{2}$.

The coefficient computation problem is still unsolved: the bound for $|a_n|$$(n\in \mathbb{N},n\geqq 3)$ is unsolved ([5]).

Because of the work of Srivastava et al. [5], research on bi-univalent functions recently acquired more interest in this field. From this motivation, many researchers (see [5,6,7,8,9,10,11,12]) calculated various types of subclasses of the $\sigma$ and nonsharp values on the first two coefficients in the Taylor–Maclaurin series. Really, the calculation of $|a_n|$ for $n\ge 3$ is still an open problem. On the other hand, there have been a few intriguing publications on bi-univalent functions with Faber polynomials under a gap series condition (see Jahangiri and Hamidi [13,14]).

Let ${\mathcal{K}}_{\alpha }$, $0\le \alpha \le 1$ be the family of analytic functions $g\left(z\right)$ of the form (1). Then, there exists a convex function $\varphi$ satisfying

$$\left|arg\left({\displaystyle \frac{g^{\prime }\left(z\right)}{{\varphi }^{\prime }\left(z\right)}}\right)\right|<{\displaystyle \frac{\alpha \pi }{2}}.$$

Kaplan [15] and Reade [16] investigated these classes. Hence ${\mathcal{K}}_0=\mathcal{C}$ and ${\mathcal{K}}_1$ are the families of convex univalent functions and close-to-convex functions, respectively. Moreover, ${{\mathcal{K}}_{\alpha }}_1$ is a proper subclass of ${{\mathcal{K}}_{\alpha }}_2$ if ${\alpha }_1<{\alpha }_2$. The class of close-to-convex functions of order$\beta$, $0\le \beta \le 1$ from the work of Reade [16] is as follows:

$$\Re \left({\displaystyle \frac{g^{\prime }\left(z\right)}{{\varphi }^{\prime }\left(z\right)}}\right)>\beta .$$

Because of Brannan and Taha [3] and Reade [16], this is evident to the class of strongly bi-convex and bi-starlike functions of order $\alpha$ (see the work of Brannan and Taha [3]).

Let $\mathcal{P}$ be the categories of analytic functions $p\left(z\right)=1+{\displaystyle \sum _{k=1}^{\infty }}{p}_k{z}^k$ such that $\mathrm{Re}\left(p\right(z\left)\right)>0$ in $\mathbb{U}$. Srivastava et al. [17] have mentioned some illustrations of the class of n-fold symmetric bi-univalent functions and found the application on the coefficient estimates for $|a_{n+1}|$ and $|a_{2n+1}|$ to the new class of functions.

The function $h\left(z\right)=\sqrt[n]{g\left(z^n\right)}$ is univalent and maps the unit disk in $\mathbb{U}$ into a region with n-fold symmetry for each function in $\mathcal{S}$. Any function is called n-fold symmetric (see [18]) if it has the normalized form

$$\begin{array}{c}\hfill h\left(z\right)=z+\sum _{k=1}^{\infty }{a}_{nk+1}{z}^{nk+1}z\in \mathbb{U},\end{array}$$

(3)

and ${\mathcal{S}}_n$ represents the categories of n-fold symmetric univalent functions which are normalized by the above series expansion. In fact, the functions in class $\mathcal{S}$ are one-fold symmetric functions.

The n-fold symmetric bi-univalent functions are conceptualized similarly to n-fold symmetric univalent functions. For each integer n, each function in the class $h\left(z\right)$ in $\sigma$ produces an n-fold symmetric bi-univalent function. The normalization of $h\left(z\right)$ is mentioned in (3), and $h^{-1}$ is obtained from the following:

$$\begin{array}{ccc}\hfill f\left(w\right)=w& -& a_{k+1}{w}^{k+1}+\left[(k+1)a_{k+1}^2-a_{2k+1}\right]w^{2k+1}\hfill \\ & -& \left[{\displaystyle \frac{1}{2}}(k+1)(3k+2)a_{k+1}^3-(3k+2)a_{k+1}{a}_{2k+1}+a_{3k+1}\right]w^{3k+1}+\cdots .\hfill \end{array}$$

(4)

where $h^{-1}=f$. Here, ${\sigma }_n$ represents the the class of n-fold symmetric bi-univalent functions. The functions ${\displaystyle {\left(\frac{z^n}{1-z^n}\right)}^{\frac{1}{n}},{\left(\frac{1}{2}log\left(\frac{1+z^n}{1-z^n}\right)\right)}^{\frac{1}{n}}}$ and ${\displaystyle {\left(-log(1-z^n)\right)}^{\frac{1}{n}}}$ with the corresponding inverse functions ${\displaystyle {\left(\frac{w^n}{1+w^n}\right)}^{\frac{1}{n}},{\left(\frac{e^{2w^n}-1}{e^{2w^n}+1}\right)}^{\frac{1}{n}}}$, and ${\displaystyle {\left(\frac{e^{w^n}-1}{e^{w^n}}\right)}^{\frac{1}{n}}}$ are a few examples of n-fold symmetric bi-univalent functions.

From Pommerenke [18], the n-fold symmetric functions in $\mathcal{P}$ take the following form:

$$\begin{array}{c}\hfill p\left(z\right)=1+p_n{z}^n+p_{2n}{z}^{2n}+p_{3n}{z}^{3n}+\cdots .\end{array}$$

(5)

Finding the initial coefficient bound for bi-close-to-convex classes of n-fold-symmetric bi-univalent functions of order $\alpha$ and $\beta$ is the aim of this paper.

To obtain the main results, the following standard lemmas are used.

Lemma 1.

If $p\in \mathcal{P}$, for each $k\ge 1$, then $|p_k| \le 2$, and

$$\left|p_{2k}-{\displaystyle \frac{p_k^2}{2}}\right|\le 2-{\displaystyle \frac{|p_k{|}^2}{2}}.$$

for one-fold symmetric in Duren [1] and Ma and Minda [19].

Lemma 2.

Let $\varphi \in \mathcal{C}$ and $\lambda \in \mathbb{R}$. Then,

$$\left|c_{2k+1}-\lambda c_{k+1}^2\right|\le \left\{\begin{array}{ccc}1-\lambda & for& \lambda <{\displaystyle \frac{2}{3}}\\ 1& for& {\displaystyle \frac{2}{3}}\le \lambda \le {\displaystyle \frac{4}{3}}\\ \lambda -1& for& \lambda >{\displaystyle \frac{4}{3}}\end{array}\right.$$

for one-fold symmetry in Kanas [20].

2. The Bounds for Class ${\mathcal{K}}_{\mathit{\sigma },\mathit{n}}\left[\mathit{\alpha }\right]$

In this section, the bound of the first two coefficients of the strong class of bi-close-to-convex functions of order $\alpha$ is derived. The work starts from the following well-known definitions.

Definition 1.

Let ${\mathcal{A}}_{\sigma ,n}\left(R\right)$ be the class of functions of the form (3) in $\left|z\right|<R$ such that its inverse function has an analytic continuation to $\left|z\right|<R$ (see (4)). Hence, the functions in the class ${\mathcal{A}}_{\sigma ,n}\left(R\right)$ are called n-fold symmetric bi-analytic functions in $\left|z\right|<R$.

It will be easy to leave out the reference to the circular domain in Definition 1 when $R=1$ and $n=1$. Therefore, a bi-analytic function will lead to a bi-analytic on $\mathbb{U}$. It is shortened to ${\mathcal{A}}_{\sigma }\left(1\right)={\mathcal{A}}_{\sigma }$. It should be noted that the class ${\mathcal{A}}_{\sigma }$ of bi-analytic functions is a legitimate subclass of $\mathcal{A}$.

Definition 2.

Let $0\le \alpha \le 1$. A function $h\left(z\right)\in {\mathcal{A}}_{\sigma ,n}$, given by (3), with $h^{\prime }\left(z\right)\ne 0$, is called a strongly bi-close-to convex function of order α if there are bi-convex functions ϕ and ψ such that

$$\left|arg\left({\displaystyle \frac{g^{\prime }\left(z\right)}{{\varphi }^{\prime }\left(z\right)}}\right)\right|<{\displaystyle \frac{\alpha \pi }{2}}\left(z\in \mathbb{U}\right)$$

(6)

and

$$\left|arg\left({\displaystyle \frac{f^{\prime }\left(w\right)}{{\psi }^{\prime }\left(w\right)}}\right)\right|<{\displaystyle \frac{\alpha \pi }{2}}\left(w\in \mathbb{U}\right).$$

(7)

Here, $f=h^{-1}$. Here ${\mathcal{K}}_{\sigma ,n}\left[\alpha \right]$ represents the strong class of bi-close-to-convex n-fold symmetric functions.

The function $\psi$ is the inverse of $\varphi$ such that

$$\varphi \left(z\right)=z+c_{n+1}{z}^{n+1}+c_{2n+1}{z}^{2n+1}+\cdots .$$

(8)

and

$$\psi \left(w\right)=w-c_{n+1}{w}^{n+1}+\left[(n+1)c_{n+1}^2-c_{2n+1}\right]w^{2n+1}+\cdots .$$

(9)

Theorem 1.

Let $0\le \alpha \le 1$ and let $h\left(z\right)$ in ${\mathcal{K}}_{\sigma ,n}\left[\alpha \right]$ from (3). Then,

$$\begin{array}{c}\hfill \left|a_{n+1}\right|\le \sqrt{1+{\displaystyle \frac{4\left(n+2\right)\alpha }{\left(2n+1\right)\left(n+1\right)}}}\end{array}$$

(10)

and

$$\begin{array}{c}\hfill \left|a_{2n+1}\right|\le 1+{\displaystyle \frac{2\alpha \left(n+2\right)}{2n+1}}.\end{array}$$

(11)

Proof. 

The (6) and (7) can be written in the form

$$\begin{array}{c}\hfill {\displaystyle \frac{g^{\prime }\left(z\right)}{{\varphi }^{\prime }\left(z\right)}}={\left[p\left(z\right)\right]}^{\alpha },\end{array}$$

for some $p\in \mathcal{P}$. Thus,

$$g^{\prime }\left(z\right)={\left[p\left(z\right)\right]}^{\alpha }{\varphi }^{\prime }\left(z\right)$$

(12)

Similarly, there is a $q\in \mathcal{P}$ satisfying

$${\displaystyle \frac{f^{\prime }\left(w\right)}{{\psi }^{\prime }\left(w\right)}}={\left[q\left(w\right)\right]}^{\alpha }.$$

Therefore,

$$f^{\prime }\left(w\right)={\left[q\left(w\right)\right]}^{\alpha }{\psi }^{\prime }\left(w\right).$$

(13)

Now $p\left(z\right),q\left(w\right)$ are expressed in the following form:

$$p\left(z\right)=1+p_n{z}^n+p_{2n}{z}^{2n}+p_{3n}{z}^{3n}+\cdots$$

(14)

and

$$q\left(w\right)=1+q_n{w}^n+q_{2n}{w}^{2n}+q_{3n}{w}^{3n}+\cdots .$$

(15)

Then, from (12) and (13), one can obtain

$$\left(n+1\right)a_{n+1}=\left(n+1\right)c_{n+1}+\alpha p_n$$

(16)

$$\left(2n+1\right)a_{2n+1}=\left(2n+1\right)c_{2n+1}+\left(n+1\right)\alpha c_{n+1}{p}_n+\alpha p_{2n}+{\displaystyle \frac{\alpha (\alpha -1)}{2}}{p}_n^2$$

(17)

$$-\left(n+1\right)a_{n+1}=\alpha q_n-\left(n+1\right)c_{n+1}$$

(18)

and

$$\left(2n+1\right)\left(\left(n+1\right)a_{n+1}^2-a_{2n+1}\right)$$

$$=\left(2n+1\right)\left(\left(n+1\right)c_{n+1}^2-c_{2n+1}\right)-\left(n+1\right)\alpha c_{n+1}{q}_n+\alpha q_{2n}+{\displaystyle \frac{\alpha (\alpha -1)}{2}}{q}_n^2.$$

(19)

From (16) and (18), we obtain

$$p_n=-q_n.$$

(20)

By adding (17) and (19), we obtain

$$\left(n+1\right)\left(2n+1\right)a_{n+1}^2=\left(2n+1\right)\left(2n+1\right)c_{n+1}^2$$

$$\begin{array}{c}\hfill +\left(n+1\right)\alpha c_{n+1}(p_n-q_n)+\alpha \left(p_{2n}-{\displaystyle \frac{p_n^2}{2}}+q_{2n}-{\displaystyle \frac{q_n^2}{2}}\right)+{\displaystyle \frac{{\alpha }^2}{2}}(p_n^2+q_n^2).\end{array}$$

(21)

An application of Lemma 1 leads to

$$\left(n+1\right)\left(2n+1\right)\left|a_{n+1}^2\right|\le \left(2n+1\right)\left(2n+1\right)\left|c_{n+1}^2\right|$$

$$+\left(n+1\right)\alpha \left|c_{n+1}\right|\left|(p_n-q_n)\right|+\alpha \left(2-{\displaystyle \frac{1-\alpha }{2}}|p_n^2|+2-{\displaystyle \frac{1-\alpha }{2}}|q_n{|}^2\right).$$

Using $|c_k| \le 1$, $|p_k| \le 2$ and $|q_k| \le 2$ for $k=1,2,\dots$, it is noted

$$\left(n+1\left(2n+1\right)\right)\left|a_{n+1}^2\right|\le \left(n+1\right)\left(2n+1\right)+4\alpha \left(n+1\right)+4\alpha .$$

Therefore,

$$\left|a_{n+1}^2\right|\le 1+{\displaystyle \frac{4\left(n+2\right)\alpha }{\left(2n+1\right)\left(n+1\right)}}.$$

(22)

It produces the expected estimate of $\left|a_{n+1}\right|$.

For the bound (11), consider the relation (17).

$$\begin{array}{c}\hfill \left(2n+1\right)a_{2n+1}=\left(2n+1\right)c_{2n+1}+\alpha \left(n+1\right)c_{n+1}{p}_n+\alpha p_{2n}+{\displaystyle \frac{\alpha (\alpha -1)}{2}}{p}_n^2.\end{array}$$

The above equation reduces to

$$\begin{array}{c}\hfill \left(2n+1\right)a_{2n+1}=\left(2n+1\right)c_{2n+1}+\left(n+1\right)\alpha c_{n+1}{p}_n+\alpha \left(p_{2n}-{\displaystyle \frac{p_n^2}{2}}\right)+{\displaystyle \frac{{\alpha }^2}{2}}{p}_n^2.\end{array}$$

By applying Lemma 1 and $|c_k| \le 1$,

$$\begin{array}{c}\hfill \left(2n+1\right)\left|a_{2n+1}\right|\le 2\alpha \left(n+1\right)+\left(2n+1\right)+2\alpha .\end{array}$$

The bound for $\left|a_{2n+1}\right|$ is obtained as declared in (11). Hence, Theorem 1 is proved. □

${\mathcal{K}}_{\sigma }\left[\alpha \right]$ can be ${\mathcal{K}}_{\sigma }\left[1\right]={\mathcal{K}}_{\sigma }$, the class of bi-close-to-convex functions in the situation of one-fold symmetric functions for $\alpha =1$

${\mathcal{K}}_{\sigma }\left[\alpha \right]$ to ${\mathcal{K}}_{\sigma }\left[0\right]={\mathcal{C}}_{\sigma }$, the class of bi-convex in the situation of one-fold symmetric functions for $\alpha =0$ (see Brannan and Taha [3]).

The following corollaries are obtained from the theorem for one-fold symmetric functions when $\alpha =0$ and $0\le \alpha \le {\displaystyle \frac{1}{2}}$.

Corollary 1.

If $g\left(z\right)$ is in ${\mathcal{K}}_{\sigma }\left[0\right]={\mathcal{C}}_{\sigma }$, then $|a_2| \le 1.$

Corollary 2.

Let $g\left(z\right)$ represented in (1) be in ${\mathcal{K}}_{\sigma }\left[\alpha \right]$ whenever $0\le \alpha \le {\displaystyle \frac{1}{2}}$. Then, $|a_2| \le \sqrt{2}.$

Theorem 2.

If $h\left(z\right)$ is in ${\mathcal{K}}_{\sigma ,n}\left[\alpha \right]$, represented by (3) for $0\le \alpha \le 1$, then

$$\begin{array}{c}\hfill |a_{2n+1}-\lambda a_{n+1}^2| \le \left\{\begin{array}{ccc}(1-\lambda )+{\displaystyle \frac{2\alpha \left(\left(n+1\right)-2\lambda \right)}{\left(2n+1\right)}}+{\displaystyle \frac{\left(\left(n+1\right)-2\lambda \right)\alpha M}{\left(2n+1\right)\left(n+1\right)}}& for& \lambda <0,\\ (1-\lambda )+{\displaystyle \frac{2\alpha \left(n+1-2\lambda \right)}{\left(2n+1\right)}}+{\displaystyle \frac{\alpha M}{\left(2n+1\right)}}& for& 0\le \lambda <{\displaystyle \frac{2}{3}},\\ 1+{\displaystyle \frac{2\alpha \left(n+1-2\lambda \right)}{\left(2n+1\right)}}+{\displaystyle \frac{\alpha M}{\left(2n+1\right)}}& for& {\displaystyle \frac{2}{3}}\le \lambda <1,\\ 1+{\displaystyle \frac{2\alpha \left(2\lambda -\left(n+1\right)\right)}{\left(2n+1\right)}}+{\displaystyle \frac{\alpha M}{\left(2n+1\right)}}& for& 1\le \lambda \le {\displaystyle \frac{4}{3}},\\ (\lambda -1)+{\displaystyle \frac{2\alpha \left(2\lambda -\left(n+1\right)\right)}{\left(2n+1\right)}}+{\displaystyle \frac{\alpha M}{\left(2n+1\right)}}& for& {\displaystyle \frac{4}{3}}<\lambda <2,\\ (\lambda -1)+{\displaystyle \frac{2\alpha \left(2\lambda -\left(n+1\right)\right)}{\left(2n+1\right)}}+{\displaystyle \frac{\alpha M\left(2\lambda -\left(n+1\right)\right)}{\left(n+1\right)\left(2n+1\right)}}& for& \lambda \ge 2,\end{array}\right.\end{array}$$

(23)

where $M\le 2$.

Proof. 

From Equations (17) and (21), one can obtain

$$\begin{array}{c}\hfill a_{2n+1}-\lambda a_{n+1}^2=c_{2n+1}+{\displaystyle \frac{\left(n+1\right)\alpha c_{n+1}{p}_n}{2n+1}}+{\displaystyle \frac{\alpha p_{2n}}{2n+1}}+{\displaystyle \frac{\alpha (\alpha -1)}{2\left(2n+1\right)}}{p}_n^2\end{array}$$

$$\begin{array}{c}\hfill -\lambda \left[c_{n+1}^2+{\displaystyle \frac{\alpha c_{n+1}(p_n-q_n)}{2n+1}}\right.+\left.{\displaystyle \frac{\alpha (p_{2n}+q_{2n})}{\left(2n+1\right)\left(n+1\right)}}+{\displaystyle \frac{\alpha (\alpha -1)}{2\left(2n+1\right)\left(n+1\right)}}(p_n^2+q_n^2)\right].\end{array}$$

By using the relations $q_n=-p_n,\left|c_k\right| \le 1$, $|p_k| \le 2$, and $|q_k| \le 2$ for $k=1,2,\dots$, the above identity reduces to

$$\begin{array}{c}\hfill |a_{2n+1}-\lambda a_{n+1}^2| \le |c_{2n+1}-\lambda c_{n+1}^2|+{\displaystyle \frac{2\alpha |n+1-2\lambda |}{2n+1}}\end{array}$$

$$\begin{array}{c}\hfill +{\displaystyle \frac{\alpha |n+1-\lambda |}{\left(n+1\right)\left(2n+1\right)}}\left[\left|p_{2n}-{\displaystyle \frac{p_n^2}{2}}\right|+{\displaystyle \frac{\alpha |p_n^2|}{2}}\right]+{\displaystyle \frac{\alpha \left|\lambda \right|}{\left(n+1\right)\left(2n+1\right)}}\left[\left|q_{2n}-{\displaystyle \frac{q_n^2}{2}}\right|+{\displaystyle \frac{\alpha |q_n^2|}{2}}\right].\end{array}$$

The expressions $\left|p_{2n}-{\displaystyle \frac{p_n^2}{2}}\right|+{\displaystyle \frac{\alpha |p_n^2|}{2}}$ and $\left|q_{2n}-{\displaystyle \frac{q_n^2}{2}}\right|+{\displaystyle \frac{\alpha |q_n^2|}{2}}$ have the same bounds so that

$$\begin{array}{c}\hfill |a_{2n+1}-\lambda a_{n+1}^2| \le |c_{2n+1}-\lambda c_{n+1}^2|+{\displaystyle \frac{2\alpha |n+1-2\lambda |}{2n+1}}+{\displaystyle \frac{\alpha \left(|n+1-\lambda |+\left|\lambda \right|\right)}{\left(2n+1\right)\left(n+1\right)}}\left[\left|p_{2n}-{\displaystyle \frac{p_n^2}{2}}\right|+{\displaystyle \frac{\alpha |p_n^2|}{2}}\right].\end{array}$$

Use of Lemma 1 gives

$$\begin{array}{c}\hfill |a_{2n+1}-\lambda a_{n+1}^2| \le |c_{2n+1}-\lambda c_{n+1}^2|+{\displaystyle \frac{2\alpha |n+1-2\lambda |}{2n+1}}+{\displaystyle \frac{\alpha \left(|n+1-\lambda |+\left|\lambda \right|\right)}{\left(n+1\right)\left(2n+1\right)}}\left[\left|2-{\displaystyle \frac{p_n^2}{2}}\right|+{\displaystyle \frac{\alpha |p_n^2|}{2}}\right].\end{array}$$

That is,

$$\begin{array}{c}\hfill |a_{2n+1}-\lambda a_{n+1}^2| \le |c_{2n+1}-\lambda c_{n+1}^2| + {\displaystyle \frac{2\alpha |n+1-2\lambda |}{2n+1}} + {\displaystyle \frac{\alpha \left(|n+1-\lambda |+\left|\lambda \right|\right)}{\left(n+1\right)\left(2n+1\right)}}\left[2-{\displaystyle \frac{\left(1-\alpha \right)p_n^2}{2}}\right].\end{array}$$

Using $\left[2-{\displaystyle \frac{\left(1-\alpha \right)p_n^2}{2}}\right]<2$, the above equation reduces to

$$\begin{array}{c}\hfill |a_{2n+1}-\lambda a_{n+1}^2| \le |c_{2n+1}-\lambda c_{n+1}^2| + {\displaystyle \frac{2\alpha |n+1-2\lambda |}{2n+1}}+{\displaystyle \frac{2\alpha \left(|n+1-\lambda |+\left|\lambda \right|\right)}{\left(n+1\right)\left(2n+1\right)}}.\end{array}$$

(24)

For the case $\lambda <0$, by using Lemma 2, the above Equation (24) implies

$$\begin{array}{c}\hfill |a_{2n+1}-\lambda a_{n+1}^2| \le (1-\lambda )+{\displaystyle \frac{2\alpha \left(\left(n+1\right)-2\lambda \right)}{\left(2n+1\right)}}+{\displaystyle \frac{\left(\left(n+1\right)-\lambda -\lambda \right)2\alpha }{\left(n+1\right)\left(2n+1\right)}}\end{array}$$

which brings the bound declared in (23) for $\lambda <0.$

Again, for the case $0\le \lambda <{\displaystyle \frac{2}{3}},$ by using Lemma 2 in Equation (24), one can obtain

$$\begin{array}{c}\hfill |a_{2n+1}-\lambda a_{n+1}^2| \le (1-\lambda )+{\displaystyle \frac{2\alpha \left(\left(n+1\right)-2\lambda \right)}{\left(2n+1\right)}}+{\displaystyle \frac{\left(\left(n+1\right)-\lambda +\lambda \right)2\alpha }{\left(n+1\right)\left(2n+1\right)}}.\end{array}$$

This gives the bound as asserted in (23) for $0\le \lambda <{\displaystyle \frac{2}{3}}.$

Now, for the case ${\displaystyle \frac{2}{3}}\le \lambda <1,$ by using Lemma 2 in Equation (24), one can obtain

$$\begin{array}{c}\hfill |a_{2n+1}-\lambda a_{n+1}^2| \le 1 + {\displaystyle \frac{2\alpha \left(\left(n+1\right)-2\lambda \right)}{\left(2n+1\right)}}+{\displaystyle \frac{\left(\left(n+1\right)-\lambda +\lambda \right)2\alpha }{\left(n+1\right)\left(2n+1\right)}}.\end{array}$$

This gives the bound as asserted in (23) for ${\displaystyle \frac{2}{3}}\le \lambda <1.$

Now, for the case $1\le \lambda \le {\displaystyle \frac{4}{3}},$ by using Lemma 2 in Equation (24), one can obtain

$$\begin{array}{c}\hfill |a_{2n+1}-\lambda a_{n+1}^2| \le 1+{\displaystyle \frac{2\alpha \left(2\lambda -\left(n+1\right)\right)}{\left(2n+1\right)}}+{\displaystyle \frac{\left(\left(n+1\right)-\lambda +\lambda \right)2\alpha }{\left(n+1\right)\left(2n+1\right)}}.\end{array}$$

This gives the bound declared in (23) for $1\le \lambda \le {\displaystyle \frac{4}{3}}.$

Now, for the case ${\displaystyle \frac{4}{3}}<\lambda <2,$ by using Lemma 2 in Equation (24), one can obtain

$$\begin{array}{c}\hfill |a_{2n+1}-\lambda a_{n+1}^2|\le (\lambda -1)+{\displaystyle \frac{2\alpha \left(2\lambda -\left(n+1\right)\right)}{\left(2n+1\right)}}+{\displaystyle \frac{\left(\left(n+1\right)-\lambda +\lambda \right)2\alpha }{\left(n+1\right)\left(2n+1\right)}}.\end{array}$$

This gives the bound declared in (23) for ${\displaystyle \frac{4}{3}}<\lambda <2$.

Last, for the case $\lambda \ge 2,$ by using Lemma 2 in Equation (24), we obtain

$$\begin{array}{c}\hfill |a_{2n+1}-\lambda a_{n+1}^2|\le (\lambda -1)+{\displaystyle \frac{2\alpha \left(2\lambda -\left(n+1\right)\right)}{\left(2n+1\right)}}+{\displaystyle \frac{\left(\lambda -\left(n+1\right)+\lambda \right)2\alpha }{\left(n+1\right)\left(2n+1\right)}}\end{array}$$

This gives the bound declared in (23) for $\lambda \ge 2.$ □

Theorem 2 reduces Theorem 2.2 of Sivasubramanian et al. [21] to the case of one-fold symmetricity.

Corollary 3

([21]). If $g\left(z\right)$ is in the class ${\mathcal{K}}_{\sigma }\left[\alpha \right]$ given by (1) for $0\le \alpha \le 1$, then

$$\begin{array}{c}\hfill |a_3-\lambda a_2^2| \le \left\{\begin{array}{ccc}(1-\lambda )\left(1+{\displaystyle \frac{4}{3}}\alpha +{\displaystyle \frac{1}{3}}\alpha M\right)\hfill & for& \hfill \lambda <0,\\ (1-\lambda )\left(1+{\displaystyle \frac{4}{3}}\alpha \right)+{\displaystyle \frac{1}{3}}\alpha M\hfill & for& \hfill 0\le \lambda <{\displaystyle \frac{2}{3}},\\ 1+{\displaystyle \frac{4}{3}}\alpha (1-\lambda )+{\displaystyle \frac{1}{3}}\alpha M\hfill & for& \hfill {\displaystyle \frac{2}{3}}\le \lambda <1,\\ 1+{\displaystyle \frac{4}{3}}\alpha (\lambda -1)+{\displaystyle \frac{1}{3}}\alpha M\hfill & for& \hfill 1\le \lambda \le {\displaystyle \frac{4}{3}},\\ (\lambda -1)\left(1+{\displaystyle \frac{4}{3}}\alpha \right)+{\displaystyle \frac{1}{3}}\alpha M\hfill & for& \hfill {\displaystyle \frac{4}{3}}<\lambda <2,\\ (\lambda -1)\left(1+{\displaystyle \frac{4}{3}}\alpha +{\displaystyle \frac{1}{3}}\alpha M\right)\hfill & for& \hfill \lambda \ge 2\end{array}\right.\end{array}$$

where $M\le 2$.

3. The Bounds for Class ${\mathcal{K}}_{\mathit{\sigma },\mathit{n}}\left[\mathit{\beta }\right]$

Definition 3.

Let $0\le \beta <1$ and $h\left(z\right)\in {\mathcal{A}}_{\sigma ,n}$, given by (3) such that $h^{\prime }\left(z\right)\ne 0$ on $\mathbb{U}$. Then, $h\left(z\right)$ is said to be a bi-close-to convex function of order β, if there exist bi-convex functions $\varphi ,\psi \in {\mathcal{C}}_{\sigma }$ such that

$$\begin{array}{c}\hfill Re\left({\displaystyle \frac{g^{\prime }\left(z\right)}{{\varphi }^{\prime }\left(z\right)}}\right)>\beta ,z\in \mathbb{U}\end{array}$$

(25)

and

$$\begin{array}{c}\hfill Re\left({\displaystyle \frac{f^{\prime }\left(w\right)}{{\psi }^{\prime }\left(w\right)}}\right)>\beta ,w\in \mathbb{U}\end{array}$$

(26)

where $f=h^{-1}$. The term ${\mathcal{K}}_{\sigma ,n}\left[\beta \right]$ represents the class of bi-close-to-convex functions of order β.

Theorem 3.

If $h\left(z\right)$ is in ${\mathcal{K}}_{\sigma ,n}\left[\beta \right]$ given by (3) for $0\le \alpha \le 1$, then

$$\begin{array}{c}\hfill \left|a_{n+1}\right|\le \sqrt{1+{\displaystyle \frac{4(1-\beta )\left(n+2\right)}{\left(2n+1\right)\left(n+1\right)}}},\end{array}$$

(27)

and

$$\begin{array}{c}\hfill \left|a_{2n+1}\right|\le 1+{\displaystyle \frac{2(1-\beta )\left(n+2\right)}{\left(2n+1\right)}},\end{array}$$

(28)

and

$$\begin{array}{c}\hfill |a_{2n+1}-\lambda a_{n+1}^2| \le \left\{\begin{array}{ccc}(1-\lambda )+{\displaystyle \frac{2(1-\beta )(n+1-2\lambda )}{2n+1}}+{\displaystyle \frac{(n+1-2\lambda )N}{\left(n+1\right)\left(2n+1\right)}}& for& \lambda <0,\\ (1-\lambda )+{\displaystyle \frac{2(1-\beta )(n+1-2\lambda )}{2n+1}}+{\displaystyle \frac{N}{2n+1}}& for& 0\le \lambda <{\displaystyle \frac{2}{3}},\\ 1+{\displaystyle \frac{2(1-\beta )(n+1-2\lambda )}{2n+1}}+{\displaystyle \frac{N}{2n+1}}& for& {\displaystyle \frac{2}{3}}\le \lambda <1,\\ 1+{\displaystyle \frac{2(1-\beta )\left(2\lambda -\left(n+1\right)\right)}{2n+1}}+{\displaystyle \frac{N}{2n+1}}& for& 1\le \lambda \le {\displaystyle \frac{4}{3}},\\ (\lambda -1)+{\displaystyle \frac{2(1-\beta )\left(2\lambda -\left(n+1\right)\right)}{2n+1}}+{\displaystyle \frac{N}{2n+1}}& for& {\displaystyle \frac{4}{3}}<\lambda <2,\\ (\lambda -1)+{\displaystyle \frac{2(1-\beta )\left(2\lambda -\left(n+1\right)\right)}{2n+1}}+{\displaystyle \frac{\left(2\lambda -\left(n+1\right)\right)N}{\left(2n+1\right)\left(n+1\right)}}& for& \lambda \ge 2,\end{array}\right.\end{array}$$

(29)

where

$$\begin{array}{c}\hfill N\le 2(1-\beta ).\end{array}$$

Proof. 

The inequalities (25) and (26) imply

$${\displaystyle \frac{g^{\prime }\left(z\right)}{{\varphi }^{\prime }\left(z\right)}}=(1-\beta )p\left(z\right)+\beta$$

and

$${\displaystyle \frac{f^{\prime }\left(w\right)}{{\psi }^{\prime }\left(w\right)}}=(1-\beta )q\left(w\right)+\beta$$

for some p and q from the class $\mathcal{P}$ involved with the expansion (14) and (15). Therefore,

$$\begin{array}{c}\hfill g^{\prime }\left(z\right)={\varphi }^{\prime }\left(z\right)\left[(1-\beta )p\left(z\right)+\beta \right],f^{\prime }\left(w\right)={\psi }^{\prime }\left(w\right)\left[(1-\beta )q\left(w\right)+\beta \right].\end{array}$$

(30)

From the above two equations (30), it is easy to obtain

$$\begin{array}{c}\hfill \left(n+1\right)a_{n+1}=(1-\beta )p_n+\left(n+1\right)c_{n+1},\end{array}$$

(31)

$$\begin{array}{c}\hfill \left(2n+1\right)a_{2n+1}=\left(2n+1\right)c_{2n+1}+\left(n+1\right)(1-\beta )c_{n+1}{p}_n+(1-\beta )p_{2n},\end{array}$$

(32)

$$\begin{array}{c}\hfill -\left(n+1\right)a_{n+1}=-\left(n+1\right)c_{n+1}+(1-\beta )p_n,\end{array}$$

(33)

$$\begin{array}{c}\hfill \left(2n+1\right)\left(\left(n+1\right)a_{n+1}^2-a_{2n+1}\right)\end{array}$$

$$\begin{array}{c}\hfill =\left(2n+1\right)\left(\left(n+1\right)c_{n+1}^2-c_{2n+1}\right)-\left(n+1\right)(1-\beta )c_{n+1}{q}_n+(1-\beta )q_{2n}.\end{array}$$

(34)

Equations (31) and (33) yield that $q_n=-p_n$. The addition of (32) and (34) implies that

$$\begin{array}{c}\hfill \left(2n+1\right)\left(n+1\right)a_{n+1}^2=\left(2n+1\right)\left(n+1\right)c_{n+1}^2\end{array}$$

$$\begin{array}{c}\hfill +\left(n+1\right)(1-\beta )c_{n+1}(p_n-q_n)+(1-\beta )(p_{2n}+q_{2n}).\end{array}$$

(35)

By the relation $q_n=-p_n$, $|c_k| \le 1$ and Lemma 1, the equation reduces to

$$\begin{array}{c}\hfill \left(n+1\right)\left(2n+1\right)\left|a_{n+1}^2\right|\le \left(n+1\right)\left(2n+1\right)+4(1-\beta )\left(n+1\right)+4(1-\beta ).\end{array}$$

It brings out the desired bound for $\left|a_{n+1}\right|$, as declared in (27).

The same method as for $\left|a_{n+1}\right|$ is used in Equation (32) to obtain the bound (28). Now, by (32) and (35), one can obtain, for the real number $\lambda$,

$$\begin{array}{c}\hfill a_{2n+1}-\lambda a_{n+1}^2=c_{2n+1}+{\displaystyle \frac{(1-\beta )\left(n+1\right)c_{n+1}{p}_n}{\left(2n+1\right)}}+{\displaystyle \frac{(1-\beta )p_{2n}}{\left(2n+1\right)}}\end{array}$$

$$\begin{array}{c}\hfill -\lambda \left[c_{n+1}^2+{\displaystyle \frac{(1-\beta )c_{n+1}(p_n-q_n)}{\left(2n+1\right)}}+{\displaystyle \frac{(1-\beta )(p_{2n}+q_{2n})}{\left(2n+1\right)\left(n+1\right)}}\right].\end{array}$$

Hence,

$$\begin{array}{c}\hfill \left|a_{2n+1}-\lambda a_{n+1}^2\right|\le \left|c_{2n+1}-\lambda c_{n+1}^2\right|\end{array}$$

$$\begin{array}{c}\hfill +{\displaystyle \frac{2(1-\beta )\left|n+1-2\lambda \right|}{\left(2n+1\right)}}+{\displaystyle \frac{2(1-\beta )\left(\left|n+1-\lambda \right|+\left|\lambda \right|\right)}{\left(2n+1\right)\left(n+1\right)}}.\end{array}$$

(36)

When $\lambda <0$, by using Lemma 2 in the above Equation (36), one can obtain

$$\begin{array}{c}\hfill \left|a_{2n+1}-\lambda a_{n+1}^2\right|\le 1-\lambda +{\displaystyle \frac{2(1-\beta )\left(n+1-2\lambda \right)}{\left(2n+1\right)}}+{\displaystyle \frac{2(1-\beta )\left[\left(n+1-\lambda \right)-\lambda \right]}{\left(2n+1\right)\left(n+1\right)}}.\end{array}$$

This brings out the bound as declared in (29) for $\lambda <0.$

Also, for $0\le \lambda <{\displaystyle \frac{2}{3}},$ by using Lemma 2 in Equation (36), one can find

$$\begin{array}{c}\hfill \left|a_{2n+1}-\lambda a_{n+1}^2\right|\le 1-\lambda +{\displaystyle \frac{2(1-\beta )\left(n+1-2\lambda \right)}{\left(2n+1\right)}}+{\displaystyle \frac{2(1-\beta )\left[\left(n+1-\lambda \right)+\lambda \right]}{\left(2n+1\right)\left(n+1\right)}}.\end{array}$$

The bound is obtained as asserted in (29) for $0\le \lambda <{\displaystyle \frac{2}{3}}.$

Now, for the case ${\displaystyle \frac{2}{3}}\le \lambda <1,$ by using Lemma 2 in Equation (36), one can obtain

$$\begin{array}{c}\hfill \left|a_{2n+1}-\lambda a_{n+1}^2\right|\le 1+{\displaystyle \frac{2(1-\beta )\left(n+1-2\lambda \right)}{\left(2n+1\right)}}+{\displaystyle \frac{2(1-\beta )\left[\left(n+1-\lambda \right)+\lambda \right]}{\left(2n+1\right)\left(n+1\right)}}.\end{array}$$

The bound is obtained as asserted in (29) for ${\displaystyle \frac{2}{3}}\le \lambda <1.$

Now, for the case $1\le \lambda \le {\displaystyle \frac{4}{3}},$ by using Lemma 2 in Equation (36), one can obtain

$$\begin{array}{c}\hfill \left|a_{2n+1}-\lambda a_{n+1}^2\right|\le 1+{\displaystyle \frac{2(1-\beta )\left(2\lambda -\left(n+1\right)\right)}{\left(2n+1\right)}}+{\displaystyle \frac{2(1-\beta )\left[\left(n+1-\lambda \right)+\lambda \right]}{\left(2n+1\right)\left(n+1\right)}}.\end{array}$$

The bound is obtained as asserted in (29) for $1\le \lambda \le {\displaystyle \frac{4}{3}}.$

Now, for the case ${\displaystyle \frac{4}{3}}<\lambda <2,$ by using Lemma 2 in Equation (36), one can obtain

$$\begin{array}{c}\hfill \left|a_{2n+1}-\lambda a_{n+1}^2\right|\le \lambda -1+{\displaystyle \frac{2\left(2\lambda -\left(n+1\right)(1-\beta )\right)}{\left(2n+1\right)}}+{\displaystyle \frac{2\left[\left(n+1-\lambda \right)+\lambda \right](1-\beta )}{\left(n+1\right)\left(2n+1\right)}}.\end{array}$$

The bound is obtained as asserted in (29) for ${\displaystyle \frac{4}{3}}<\lambda <2.$

Last, for the case $\lambda \ge 2,$ by using Lemma 2 in Equation (36), one can obtain

$$\begin{array}{c}\hfill \left|a_{2n+1}-\lambda a_{n+1}^2\right|\le \lambda -1+{\displaystyle \frac{2(1-\beta )\left(2\lambda -\left(n+1\right)\right)}{\left(2n+1\right)}}+{\displaystyle \frac{2(1-\beta )\left(\lambda -\left(n+1\right)+\lambda \right)}{\left(n+1\right)\left(2n+1\right)}}.\end{array}$$

The bound is obtained as asserted in (29) for $\lambda \ge 2.$ □

Theorem 3 reduces Theorem 3.1 of Sivasubramanian et al. [21] for one-fold symmetric functions.

Corollary 4

([21]). If $g\left(z\right)$ of the form (1) is in ${\mathcal{K}}_{\sigma }\left(\beta \right)$ and $0\le \beta <1$, then

$$\begin{array}{c}\hfill |a_3-\lambda a_2^2| \le \left\{\begin{array}{ccc}(1-\lambda )\left(1+{\displaystyle \frac{4}{3}}(1-\beta )+{\displaystyle \frac{1}{3}}N\right)\hfill & for& \hfill \lambda <0,\\ (1-\lambda )\left(1+{\displaystyle \frac{4}{3}}(1-\beta )\right)+{\displaystyle \frac{1}{3}}N\hfill & for& \hfill 0\le \lambda <{\displaystyle \frac{2}{3}},\\ 1+{\displaystyle \frac{4}{3}}(1-\lambda )(1-\beta )+{\displaystyle \frac{1}{3}}N\hfill & for& \hfill {\displaystyle \frac{2}{3}}\le \lambda <1,\\ 1+{\displaystyle \frac{4}{3}}(\lambda -1)(1-\beta )+{\displaystyle \frac{1}{3}}N\hfill & for& \hfill 1\le \lambda \le {\displaystyle \frac{4}{3}},\\ (\lambda -1)\left(1+{\displaystyle \frac{4}{3}}(1-\beta )\right)+{\displaystyle \frac{1}{3}}N\hfill & for& \hfill {\displaystyle \frac{4}{3}}<\lambda <2,\\ (\lambda -1)\left(1+{\displaystyle \frac{4}{3}}(1-\beta )+{\displaystyle \frac{1}{3}}N\right)\hfill & for& \hfill \lambda \ge 2,\end{array}\right.\end{array}$$

For one-fold symmetric functions and ${\displaystyle \frac{1}{2}}\le \beta <1$, Theorem 3 reduces the following corollary:

Corollary 5.

If $g\left(z\right)$ given by (1) is in ${\mathcal{K}}_{\sigma }\left(\beta \right)$, then $|a_2| \le \sqrt{2}$ for ${\displaystyle \frac{1}{2}}\le \beta <1$.

4. Conclusions

This research article’s motivation is to introduce various classes and n-fold symmetric bi-univalent functions. The upper bounds for the second and third Taylor–Maclaurin coefficients are obtained for functions in each subclass. Furthermore, some of the results’ implications are discussed.