\documentclass[10pt]{studiamnew} \usepackage{graphicx} \usepackage{amsmath} \usepackage{amssymb} \sloppy \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \theoremstyle{definition} \newtheorem{remark}[theorem]{Remark} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{problem}[theorem]{Problem} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{solution}[theorem]{Solution} \newtheorem{notation}[theorem]{Notation} \renewcommand{\theequation}{\thesection.\arabic{equation}} \numberwithin{equation}{section} \begin{document} % \setcounter{page}{1} \setcounter{firstpage}{1} \setcounter{lastpage}{4} \renewcommand{\currentvolume}{??} \renewcommand{\currentyear}{??} \renewcommand{\currentissue}{??} % \title[Coefficient estimates for a subclass of meromorphic]{Coefficient estimates for a subclass of meromorphic bi-univalent functions defined by subordination} \author{Ebrahim Analouei Adegani} \address{Faculty of Mathematical Sciences, \\ Shahrood University of Technology\\ P.O.Box 316-36155, Shahrood\\ Iran} \email{analoey.ebrahim@gmail.com} % \author{Ahmad Motamednezhad} \address{Faculty of Mathematical Sciences, \\ Shahrood University of Technology\\ P.O.Box 316-36155, Shahrood\\ Iran} \email[Corresponding author]{a.motamedne@gmail.com} % \author{Serap Bulut} \address{Kocaeli University, \\ Faculty of Aviation and Space Sciences\\ Arslanbey Campus,\\ 41285 Kartepe-Kocaeli,\\ Turkey} \email{serap.bulut@kocaeli.edu.tr} % \subjclass{30C45; 30C50} \keywords{Coefficient estimates, Faber polynomial expansion, meromorphic functions, subordinate} \begin{abstract} In this work, we use the Faber polynomial expansion by a new method to find upper bounds for $\left\vert b_{n}\right\vert $ coefficients for meromorphic bi-univalent functions class $\Sigma ^{\prime }$ which is defined by subordination. Further, we generalize and improve some of the previously published results. \end{abstract} \maketitle \section{Introduction and Preliminaries} Let $\mathcal{A}$ be a class of analytic functions in the open unit disk $% \mathbb{U}=\{z\in \mathbb{C}:\left\vert z\right\vert <1\mathbb{\}}$ of the form \begin{equation} f(z)=z+\sum\limits_{n=2}^{\infty }{a_{n}z^{n}}. \label{1.1} \end{equation}% Also denote by $\mathcal{S}$ the class of all functions in $\mathcal{A}$ which are univalent and normalized by the conditions $f(0)=0=f^{\prime }(0)-1 $. It is well known that every function $f\in \mathcal{S}$ has an inverse $f^{-1}$, which is defined by \begin{equation*} f^{-1}(f(z))=z\quad (z\in \mathbb{U})\quad \text{and}\quad f(f^{-1}(w))=w\quad \left( |w|\alpha \qquad (z\in \mathbb{U% }). \end{equation*}% This class is denoted by $\mathcal{ST}(\alpha )$. Note that $\mathcal{ST}(0)=% \mathcal{ST}$. \begin{definition} \label{dfn1} \cite{Du} For two functions $f$ and $g$ which are analytic in $% \mathbb{U},$ we say that the function $f$ is subordinate to $g$ in $\mathbb{U% },$ and write \begin{equation*} f\left( z\right) \prec g\left( z\right) \qquad \left( z\in \mathbb{U}\right) , \end{equation*}% if there exists a Schwarz function $\omega ,$ which is analytic in $\mathbb{U% }$ with \begin{equation*} \omega \left( 0\right) =0\qquad \text{and\qquad }\left\vert \omega \left( z\right) \right\vert <1\quad \left( z\in \mathbb{U}\right) , \end{equation*}% such that \begin{equation*} f\left( z\right) =g\left( \omega \left( z\right) \right) \quad \left( z\in \mathbb{U}\right) . \end{equation*}% In particular, if the function $g$ is univalent in $\mathbb{U},$ then $% f\prec g$ if and only if $f(0)=g(0)$ and $f(\mathbb{U})\subseteq g(\mathbb{U}% )$. \end{definition} Ma and Minda \cite{Ma} have given a unified treatment of various subclass consisting of starlike functions by replacing the superordinate function $% q(z)=\frac{1+z}{1-z}$ by a more general analytic function. For this purpose, they considered an analytic function $\varphi $ with positive real part on $% \mathbb{U}$, satisfying $\varphi (0)=1,\ \varphi ^{\prime }(0)>0$ and $% \varphi $ maps the unit disk $\mathbb{U}$ onto a region starlike with respect to $1$, symmetric with respect to the real axis. The class $\mathcal{% ST}(\varphi )$ of Ma-Minda starlike functions consists of functions $f\in \mathcal{S}$ satisfying \begin{equation*} \frac{zf^{\prime }(z)}{f(z)}\prec \varphi (z),\quad \text{for}~z\in \mathbb{U% }. \end{equation*}% It is clear that for special choices of $\varphi $, this class envelop several well-known subclasses of univalent function as special cases. The idea of subordination was used for defining many of classes of functions studied in the Geometric Function Theory, for example see \cite{Bulboaca, Miller-Mocanu}. Let $\Sigma ^{\prime }$ denote the class of meromorphic univalent functions $% g$ defined in $\Delta :=\{z\in \mathbb{C}:1<\left\vert z\right\vert <\infty \}$ of the form \begin{equation} g(z)=z+\sum\limits_{n=0}^{\infty }\frac{b_{n}}{z^{n}}. \label{1.3} \end{equation}% Since $g\in \Sigma ^{\prime }$ is univalent, it has an inverse $g^{-1}=G$ that satisfy \begin{equation*} g^{-1}(g(z))=z\quad (z\in \Delta )\quad \text{and}\quad g(g^{-1}(w))=w\quad (M<\left\vert w\right\vert <\infty ,\;M>0). \end{equation*}% Furthermore, the inverse function $g^{-1}=G$ has a series expansion of the form \begin{equation} G(w)=g^{-1}(w)=w+\sum\limits_{n=0}^{\infty }\frac{\tilde{b}_{n}}{w^{n}}% \qquad (M<\left\vert w\right\vert <\infty ). \label{1.x} \end{equation}% A simple calculation shows that the inverse function $g^{-1}=G$, is given by% \begin{equation} G(w)=g^{-1}(w)=w-b_{0}-\frac{b_{1}}{w}-\frac{b_{2}+b_{0}b_{1}}{w^{2}}+\cdots . \label{1.y} \end{equation} Let $\mathcal{(ST)}^{\prime }(\varphi )$ denote the class of functions $g\in \Sigma ^{\prime }$ which satisfy \begin{equation*} \frac{1}{z}\frac{g^{\prime }(1/z)}{g(1/z)}\prec \varphi (z),\quad \text{for}% \;z\in \mathbb{U}. \end{equation*}% The mapping $f(z)\mapsto g(z):=1/f(1/z)$ establishes a one-to-one correspondence between functions in the classes $\mathcal{S}$ and $\Sigma ^{\prime }$ and also between functions in the classes $\mathcal{ST}(\varphi ) $ and $\mathcal{(ST)}^{\prime }(\varphi )$ because (see for more details \cite{Fi}) \begin{equation*} \frac{zg^{\prime }(z)}{g(z)}=\frac{z(1/f(1/z))^{\prime }}{1/f(1/z)}=\frac{1}{% z}\frac{f^{\prime }(1/z)}{f(1/z)},\quad \text{for}\;\left\vert z\right\vert >1. \end{equation*}% Noth that if $g\in \mathcal{(ST)}^{\prime }(\varphi )$, then there exists a unique function $f\in \mathcal{ST}(\varphi )$ such that $g(z)=1/f(1/z)$. Also, it can be easily verified that $G(w)=1/F(1/w)$, where $F(w)$ is the inverse of $f(z)$. Analogous to the bi-univalent analytic functions, a function $g\in \Sigma ^{\prime }$ is said to be meromorphic bi-univalent if $g^{-1}\in \Sigma ^{\prime }$. Examples of the meromorphic bi-univalent functions are \begin{equation*} z+\frac{1}{z},\qquad z-1,\qquad -\frac{1}{\log \left( 1-\frac{1}{z}\right) }. \end{equation*} Determination of the sharp coefficient estimates of inverse functions in various subclasses of the class of analytic and univalent functions is an interesting problem in geometric function theory. Schiffer \cite{Sc} obtained the estimate $\left\vert b_{2}\right\vert \leq \frac{2}{3}$ for meromorphic univalent functions $g\in \Sigma ^{\prime }$ with $b_{0}=0$ and Duren \cite{Du} gave an elementary proof of the inequality $\left\vert b_{n}\right\vert \leq \frac{2}{n+1}$ on the coefficient of meromorphic univalent functions $g\in \Sigma ^{\prime }$ with $b_{k}=0$ for $1\leq k<% \frac{n}{2}$. But the interest on coefficient estimates of the meromorphic univalent functions keep on by many researchers, see for example, \cite% {KM,Ku,Sch,Sp}. Several authors by using Faber polynomial expansions obtained coefficient estimates $\left\vert a_{n}\right\vert $ for classes meromorphic bi-univalent functions and bi-univalent functions, see for example \cite% {Frasin-Aouf,Ham,SSJ,Ha2,HTJ, Jah,Ja,Zi1,Zi2}. First we recall some definitions and lemmas that used in this work. Faber \cite{Fa} introduced the Faber polynomials which play an important role in various areas of mathematical sciences, especially in geometric function theory. By using the Faber polynomial expansion of functions $g\in \Sigma ^{\prime }$ of the form \eqref{1.3}, the coefficients of its inverse map $g^{-1}=G$ defined in \eqref{1.y} may be expressed, (see for details \cite{Ai} and \cite{Air}), \begin{equation} G(w)=g^{-1}(w)=w-b_{0}-\sum_{n\geq 1}\frac{1}{n}K_{n+1}^{n}\frac{1}{w^{n}}, \label{1.4} \end{equation}% where \begin{eqnarray*} K_{n+1}^{n} &=&nb_{0}^{n-1}b_{1}+n(n-1)b_{0}^{n-2}b_{2}+\frac{n(n-1)(n-2)}{2}% b_{0}^{n-3}(b_{3}+b_{1}^{2}) \\ &&+\frac{n(n-1)(n-2)(n-3)}{3!}b_{0}^{n-3}(b_{4}+3b_{1}b_{2})+\sum_{j\geq 5}b_{0}^{n-j}V_{j}, \end{eqnarray*}% such that $V_{j}$ with $5\leq j\leq n$ is a homogeneous polynomial in the variables $b_{1},b_{2},\cdots ,b_{n}$, (see for details \cite{Air}). \begin{definition} \label{A}\cite{A} Let $\varphi $ is an analytic function with positive real part in the unit disk $\mathbb{U}$, satisfying $\varphi (0)=1,\ \varphi ^{\prime }(0)>0,$ $\varphi $\ maps the unit disk $\mathbb{U}$ onto a region starlike with respect to $1$, symmetric with respect to the real axis. Such a function has series expansion of the form% \begin{equation} \varphi (z)=1+B_{1}z+B_{2}z^{2}+\cdots \qquad (B_{1}>0). \label{2.1} \end{equation} \end{definition} \begin{lemma} \emph{\cite{Du}} Let $u(z)$ and $v(z)$ be two analytic functions in the unit disk $\mathbb{U}$ with% \begin{equation*} u(0)=v(0)=0\qquad \text{and\qquad }\max \left\{ \left\vert u(z)\right\vert ,\;\left\vert v(z)\right\vert \right\} <1. \end{equation*}% We suppose also that \begin{equation} u(z)=\sum_{n=1}^{\infty }p_{n}z^{n}\qquad \text{and}\qquad v(z)=\sum_{n=1}^{\infty }q_{n}z^{n}\quad (z\in \mathbb{U}). \label{a} \end{equation}% Then% \begin{equation} \left\vert p_{1}\right\vert \leq 1,\quad \left\vert p_{2}\right\vert \leq 1-\left\vert p_{1}\right\vert ^{2},\quad \left\vert q_{1}\right\vert \leq 1,\quad \left\vert q_{2}\right\vert \leq 1-\left\vert q_{1}\right\vert ^{2}. \label{b} \end{equation} \end{lemma} \begin{lemma} \emph{\cite{Ai1,Ai}}\label{K-p} Let the function $f\in \mathcal{A}$ be given by \eqref{1.1}. Then for any $p\in \mathbb{Z}$, there are the polynomials $% K_{n}^{p}$, such that \begin{equation*} (1+a_{2}z+a_{3}z^{2}+\cdots +a_{k}z^{k-1}+\cdots )^{p}=1+\sum_{n=1}^{\infty }K_{n}^{p}(a_{2},a_{3},\cdots ,a_{n+1})z^{n}, \end{equation*}% where \begin{equation*} K_{n}^{p}(a_{2},\cdots ,a_{n+1})=pa_{n+1}+\frac{p(p-1)}{2}D_{n}^{2}+\frac{p!% }{(p-3)!3!}D_{n}^{3}+\cdots +\frac{p!}{(p-n)!(n)!}D_{n}^{n}, \end{equation*}% and% \begin{equation*} D_{n}^{m}(a_{2},a_{3},\cdots ,a_{n})=\sum_{n=2}^{\infty} \frac{m!(a_{2})^{\mu _{1}}\cdots (a_{n})^{\mu _{n}}}{\mu _{1}!\cdots \mu _{n}!},~ \textnormal{for} ~ m\in \mathbb{N}=\{1,2,\ldots\}~ \textnormal{and}~ m\le n, \end{equation*}% the sum is taken over all nonnegative integers $\mu _{1},...,\mu _{n}$ satisfying \begin{equation*} \left\{ \begin{array}{l} \mu _{1}+\mu _{2}+\cdots +\mu _{n}=m, \\ \mu _{1}+2\mu _{2}+\cdots +n\mu _{n}=n. \\ \end{array}% \right. \end{equation*}% It is clear that $D_{n}^{n}(a_{2},a_{3},\cdots ,a_{n})=a_{2}^{n}$. In particular,% \begin{eqnarray*} K_{n}^{1} &=&a_{n+1},\qquad K_{1}^{2}=2a_{2},\qquad K_{2}^{2}=2a_{3}+a_{2}^{2}, \\ K_{3}^{2} &=&2a_{4}+2a_{2}a_{3},\qquad K_{4}^{2}=2a_{5}+2a_{2}a_{4}+a_{3}^{2}. \end{eqnarray*} \end{lemma} \begin{lemma} \emph{\cite{Ai,Air} and \cite[page 52]{Bo}}\label{Air} Let the function $% g\in \Sigma ^{\prime }$ be given by \eqref{1.3}. Then we have the following expansion \begin{equation*} \frac{zg^{\prime }(z)}{g(z)}=1+\sum_{n=0}^{\infty }F_{n+1}(b_{0},b_{1},\cdots b_{n})\frac{1}{z^{n+1}}, \end{equation*}% where \begin{equation*} F_{n+1}(b_{0},b_{1},\cdots b_{n})=\sum_{i_{1}+2i_{2}+\cdots +(n+1)i_{n+1}=n+1}A(i_{1},i_{2},\cdots ,i_{n+1})(b_{0}^{i_{1}}b_{1}^{i_{2}}\cdots b_{n}^{i_{n+1}}), \end{equation*}% and \begin{equation*} A(i_{1},i_{2},\cdots ,i_{n+1}):=(-1)^{(n+1)+2i_{1}+\cdots +(n+2)i_{n+1}}% \frac{(i_{1}+i_{2}+\cdots +i_{n+1}-1)!(n+1)}{i_{1}!i_{2}!\cdots i_{n+1}!}. \end{equation*}% The first four terms of the Faber polynomials $F_{n}$ are given by% \begin{eqnarray*} F_{1} &=&-b_{0},\qquad F_{2}=b_{0}^{2}-2b_{1},\qquad F_{3}=-b_{0}^{3}+3b_{1}b_{0}-3b_{2}, \\ F_{4} &=&b_{0}^{4}-4b_{0}^{2}b_{1}+4b_{0}b_{2}+2b_{1}^{2}-4b_{3}. \end{eqnarray*} \end{lemma} In this work, by using the Faber polynomial expansion we find upper bounds for $\left\vert b_{n}\right\vert $ coefficients by a new method for meromorphic bi-univalent functions class $\Sigma ^{\prime }$ which is defined by subordination. Further, we generalize and improve some of the previously published results. \section{Main Resuls} In this section, first we obtain estimates of coefficients $\left\vert b_{n}\right\vert $ of meromorphic bi-univalent functions in the class $% \mathcal{(ST)}^{\prime }(\varphi )$. Next we obtain an improvement of the bounds $\left\vert b_{0}\right\vert $ and $\left\vert b_{1}\right\vert $ for special choices of $\varphi $. \begin{theorem} \label{Main}Let the function $g$ given by \eqref{1.3} and its inverse map $% g^{-1}=G$ given by \eqref{1.x} be in the class $\mathcal{(ST)}^{\prime }(\varphi ),$ where $\varphi $ is given by Definition $\ref{A}$. If $b_{k}=0$ for $0\leq k\leq n-1$, then \begin{equation*} \left\vert b_{n}\right\vert \leq \frac{B_{1}}{n+1}. \end{equation*} \end{theorem} \begin{proof} From $g\in \mathcal{(ST)}^{\prime}(\varphi)$, we obtain \begin{eqnarray} \label{2.1} \frac{1}{z}\frac{g^{\prime}(1/z)}{g(1/z)}=\frac{1-b_1z^2-2b_2z^3-\cdots}{% 1+b_0z+b_1z^2+\cdots}=1-b_0z+(b_0^2-2b_1)z^2+\cdots. \end{eqnarray} Similar to Lemma \ref{Air}, for function $g\in \mathcal{(ST)}% ^{\prime}(\varphi)$ and for its inverse map $g^{-1}=G$, we have \begin{eqnarray} \label{2.1} \frac{1}{z}\frac{g^{\prime}(1/z)}{g(1/z)}=1+\sum_{n=0}^{\infty}F_{n+1}(b_0, b_1, \cdots b_{n}) z^{n+1}, \end{eqnarray} \begin{eqnarray} \label{2.2} \frac{1}{w}\frac{G^{\prime}(1/w)}{G(1/w)}=1+\sum_{n=0}^{\infty}F_{n+1}(% \tilde{b}_0, \tilde{b}_1, \cdots \tilde{b}_{n}) w^{n+1}, \end{eqnarray} respectively, where $\tilde{b}_0=-b_0,\ \tilde{b}_n=\frac{1}{n}K_{n+1}^{n}$. On the other hand, since $g,\ G \in \mathcal{(ST)}^{\prime}(\varphi)$, by the Definition \ref{dfn1}, there exist two Schwarz functions $u, v:\mathbb{U} \rightarrow \mathbb{U}$ where $u, \ v$ are given by \eqref{a}, so that \begin{eqnarray} \label{2.3} \frac{1}{z}\frac{g^{\prime}(1/z)}{g(1/z)}=\varphi(u(z))=1+\sum_{n=1}^{% \infty}\sum_{k=1}^{n}B_{k}D_{n}^{k}(p_1, p_{2}, \cdots,p_{n}) z^{n}, \end{eqnarray} and \begin{eqnarray} \label{2.4} \frac{1}{w}\frac{G^{\prime}(1/w)}{G(1/w)}=\varphi(v(w))=1+\sum_{n=1}^{% \infty}\sum_{k=1}^{n}B_{k}D_{n}^{k}(q_1, q_{2}, \cdots,q_{n}) w^{n}. \end{eqnarray} Comparing the corresponding coefficients of \eqref {2.1} and \eqref {2.3}, we get that \begin{eqnarray} \label{2.5} F_{n+1}(b_0, b_1, \cdots b_{n})= \sum_{k=1}^{n+1}B_{k}D_{n+1}^{k}(p_1, p_{2}, \cdots,p_{n+1}). \end{eqnarray} Similarly, by comparing the corresponding coefficients of \eqref{2.2} and % \eqref{2.4}, we get that \begin{eqnarray} \label{2.6} F_{n+1}(\tilde{b}_0, \tilde{b}_1, \cdots \tilde{b}_{n})=% \sum_{k=1}^{n+1}B_{k}D_{n+1}^{k}(q_1, q_{2}, \cdots,q_{n+1}). \end{eqnarray} Note that $b_k= 0$ for $0\le k\le n-1$, yields $\tilde{b}_n=-b_n$ and hence from \eqref{2.5} and \eqref{2.6}, respectively, we get \begin{eqnarray*} -(n+1)b_{n} =B_{1}p_{n+1}, \end{eqnarray*} and \begin{eqnarray*} -[-(n+1)]b_{n} =B_{1}q_{n+1}. \end{eqnarray*} By solving either of the above two equations for $b_n$ and applying $% \left\vert p_{n+1}\right\vert \le 1, \left\vert q_{n+1}\right\vert \le 1$, we obtain \begin{eqnarray*} \left\vert b_{n}\right\vert \le \frac{B_1}{n + 1}, \end{eqnarray*} this completes the proof. \end{proof} \begin{corollary} Let the function $g$ given by \eqref{1.3} and its inverse map $g^{-1}=G$ given by \eqref{1.x} be in the class $\mathcal{(ST)}^{\prime }\left( \left( \frac{1+z}{1-z}\right) ^{\alpha }\right) $. If $b_{k}=0$ for $0\leq k\leq n-1 $, then \begin{equation*} \left\vert b_{n}\right\vert \leq \frac{2\alpha }{n+1}\qquad \left( 0<\alpha \leq 1\right) . \end{equation*} \end{corollary} \begin{corollary} \cite{SSJ} \label{Co}Let the function $g$ given by \eqref{1.3} and its inverse map $g^{-1}=G$ given by \eqref{1.x} be in the class $\mathcal{(ST)}% ^{\prime }\left( \frac{1+(1-2\beta )z}{1-z}\right) $. If $b_{k}=0$ for $% 0\leq k\leq n-1$, then \begin{equation*} \left\vert b_{n}\right\vert \leq \frac{2(1-\beta )}{n+1}\qquad \left( 0\leq \beta <1\right) . \end{equation*} \end{corollary} \begin{corollary} \label{cor.1}Let the function $f$ given by \eqref{1.1} and its inverse map $% f^{-1}=F$ given by \eqref{1.2} be in the class $\mathcal{ST}(\varphi ).$ If $% a_{k}=0$ for $2\leq k\leq n-1$, then \begin{equation*} \left\vert a_{n}\right\vert \leq \frac{B_{1}}{n-1}. \end{equation*} \end{corollary} \begin{proof} Setting $f(1/z):=1/g(z)$ and $F(1/w)=1/G(w)$ in Theorem \ref{Main} we obtain the result and this completes the proof. \end{proof} \begin{corollary} \label{Main00}(\cite[Theorem 2.1]{Ha2}) Let the function $f$ given by % \eqref{1.1} and its inverse map $f^{-1}=F$ given by \eqref{1.2} be in the class $\mathcal{ST}\left( \frac{1+Az}{1+Bz}\right) ,$ where $A$ and $B$ are real numbers so that $-1\leq B