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\begin{document}

\title[Coefficient estimates and subordination properties $\cdots$]{%
Coefficient estimates and subordination properties for certain classes of
analytic functions of reciprocal order}
\author{Tariq Al-Hawary \and Basem Aref Frasin}
\email{tariq\_amh@yahoo.com}
\email{bafrasin@yahoo.com}
\subjclass{30C45, 30C80.}
\keywords{Analytic function, starlike function, convex function, $\beta$%
-spirallike of order $\alpha$, subordinating factor sequence.}

\begin{abstract}
In this work, we determine the coefficient bounds and subordination results
for functions in certain subclasses of analytic functions of reciprocal
order, which are introduced here by means of a Hadamard product of analytic
functions. The results presented in this paper improve or generalize the
recent works of other authors and also give rise to several new results.
\end{abstract}

\maketitle

\setcounter{page}{1} \setcounter{firstpage}{1} \setcounter{lastpage}{11} %
\renewcommand{\currentvolume}{56} \renewcommand{\currentyear}{2011} %
\renewcommand{\currentissue}{2}

\address{Ajloun College, Al-Balqa Applied University, \\ Department of Applied Science,\\
26816 Ajloun,\\
 Jordan}

\address{Al al-Bayt University, \\ Faculty of Science, Department of Mathematics,\\
25113 Mafraq, \\
Jordan}

\section{Introduction and preliminaries}

Let $\mathcal{A}$ denote the class of functions $f(z)$ defined by 
\begin{equation}
f(z)=z+\sum\limits_{n=2}^{\infty}a_{n}z^{n}   \label{tr}
\end{equation}
which are analytic and univalent in the open unit disk $\mathbb{U}%
=\{z:\,|z|<1\}.$ A function $f$ $\in\mathcal{A}$ is said to be starlike of
order $\alpha$ if it satisfies 
\begin{equation}
\mbox{Re}\left( \frac{zf^{\prime}(z)}{f(z)}\right) >\alpha\qquad(\text{ }%
0\leq\alpha<1,~z\in\mathbb{U)}.
\end{equation}
We denote by $\mathcal{S}^{\ast}(\alpha)$ the subclass of $\mathcal{A}$
consisting of functions which are starlike of order $\alpha$ in $\mathbb{U}.$
Also, a function $f$ $\in\mathcal{A}$ is said to be convex of order $\alpha$
if it satisfies%
\begin{equation}
\mbox{Re}\left( 1+\frac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\right)
>\alpha\qquad(\text{ }0\leq\alpha<1,~z\in\mathbb{U)}.
\end{equation}
We denote by $\mathcal{K}(\alpha)$ the subclass of $\mathcal{A}$ consisting
of functions which are convex of order $\alpha$ in $\mathbb{U}$. Clearly, we
have $\mathcal{S}^{\ast}(\alpha)\subseteq\mathcal{S}^{\ast}(0)=\mathcal{S}%
^{\ast},$ $\mathcal{K}(\alpha)\subseteq\mathcal{K}(0)=\mathcal{K}$ and $%
f(z)\in \mathcal{K}(\alpha)$ if and only if $zf^{\prime}(z)\in\mathcal{S}%
^{\ast }(\alpha)$ for $0\leq\alpha<1.$

For $\left\vert \beta\right\vert <\frac{\pi}{2}$ and $0\leq\alpha<1,$ a
function $f$ $\in\mathcal{A}$ is said to be $\beta$-spirallike of order $%
\alpha$ in $\mathbb{U}$ if it satisfies%
\begin{equation}
\mbox{Re}\left( e^{i\beta}\frac{zf^{\prime}(z)}{f(z)}\right) >\alpha
\cos\beta\qquad(z\in\mathbb{U)}.
\end{equation}
The class of all such functions is denoted by $\mathcal{S}_{\beta}(\alpha)$ 
\cite{lib}.

A function $f$ $\in\mathcal{A}$ is said to be starlike of reciprocal order $%
\alpha$ if 
\begin{equation}
\mbox{Re}\left\{ \frac{f(z)}{zf^{\prime}(z)}\right\} >\alpha\qquad\text{ }(%
\text{ }0\leq\alpha<1,~z\in\mathbb{U)}.
\end{equation}
We denote the class of such functions by $\mathcal{S}^{^{-1}\ast}(\alpha).$
Furthermore, a function $f$ $\in\mathcal{A}$ is said to be convex of
reciprocal order $\alpha$ if%
\begin{equation}
\mbox{Re}\left\{ \frac{1}{1+\frac{zf^{\prime\prime}(z)}{f^{\prime}(z)}}%
\right\} >\alpha\qquad(\text{ }0\leq\alpha<1,~z\in\mathbb{U)}.
\end{equation}
The class of all such convex functions of reciprocal order $\alpha$ is
denoted by $\mathcal{K}^{^{-1}}(\alpha).$

We note that $f(z)\in\mathcal{K}^{^{-1}}(\alpha)$ if and only if $zf^{\prime
}(z)\in\mathcal{S}^{^{-1}\ast}(\alpha)$.

In view of the fact that%
\begin{equation*}
\mbox{Re}p(z)>0\Rightarrow\mbox{Re}\frac{1}{p(z)}=\mbox{Re}\frac {p(z)}{%
\left\vert p(z)\right\vert ^{2}}>0 
\end{equation*}
it follows that $\mathcal{S}^{^{-1}\ast}(0)=\mathcal{S}^{\ast}$ and $%
\mathcal{K}^{^{-1}}(0)=\mathcal{K}$ . In particular, every starlike function
of reciprocal order $\alpha\geq0$ is starlike and hence univalent.

\begin{example}
The function $f(z)=$ $ze^{(1-\alpha)z}$ is a starlike function of reciprocal
order $1/(2-\alpha)$ \cite[Example 2]{nun}.
\end{example}

For functions $f\in\mathcal{A}$ given by (\ref{tr}) and $g\in\mathcal{A}$
given by $g(z)=z+\sum\limits_{n=2}^{\infty}b_{n}z^{n}$, we define the
Hadamard product (or Convolution ) of $f$ and $g$ by%
\begin{equation}
(f\ast g)(z)=\mathit{\ }z+\sum\limits_{n=2}^{\infty}a_{n}b_{n}z^{n}. 
\label{b2}
\end{equation}

Motivated and inspired by the work done by Owa et al. \cite{Kamali} and by
making use of the Hadamard product (\ref{b2}), we now introduce the
following subclass of $\mathcal{A}$.

\begin{definition}
\bigskip Let $\Phi(z)=z+\sum\limits_{n=2}^{\infty}\delta_{n}z^{n}$ and $%
\Psi(z)=z+\sum\limits_{n=2}^{\infty}\mu_{n}z^{n}$ \textit{be analytic in }$%
\mathbb{U}$\textit{, such that }$\delta_{n}\geq0,\mu_{n}\geq0$\textit{\ and }%
$\delta_{n}\geq\mu_{n}$\textit{\ for }$n\geq2,$ \textit{we say that \ }$%
f(z)\in\mathcal{A}$\textit{\ is in the class }$\mathcal{S}%
^{^{-1}}(\Phi,\Psi;\alpha,\beta)$\textit{\ if }$f(z)\ast\Phi(z)\neq0,$ 
\textit{\ }$f(z)\ast\Psi(z)\neq0$\textit{\ and} 
\begin{equation}
\left\vert \frac{1}{e^{i\beta}\left( \frac{f(z)\ast\Phi(z)}{f(z)\ast\Psi (z)}%
\right) }-\frac{1}{2\alpha}\right\vert <\frac{1}{2\alpha}\text{ \ \ }(\text{ 
}\beta\in\mathbb{R},~0<\alpha<1,~z\in\mathbb{U)}.   \label{d7}
\end{equation}
\end{definition}

Several known and new subclasses of analytic functions of reciprocal order $%
\alpha$ can be obtained from the class $\mathcal{S}^{^{-1}}(\Phi,\Psi
;\alpha,\beta)$, by suitably specializing the values of $\Phi$, $\Psi$ and $%
\beta$. We present below some of these subclasses of $\mathcal{S}%
^{^{-1}}(\Phi,\Psi;\alpha,\beta)$ consisting of functions of the form (\ref%
{tr}).

\begin{example}
\label{ex1}\bigskip If $\Phi(z)=z(1-z)^{-2}$ and $\Psi(z)=z(1-z)^{-1},$ then 
\begin{align*}
& \mathcal{S}^{^{-1}}\left( z(1-z)^{-2},z(1-z)^{-1};\alpha,\beta\right) \\
& \equiv\mathcal{S}_{\beta}(\alpha):=\left\{ f\in\mathcal{A}:\left\vert 
\frac{e^{-i\beta}f(z)}{zf^{\prime}(z)}-\frac{1}{2\alpha}\right\vert <\frac {1%
}{2\alpha}\text{ \ \ }(\text{ }\beta\in\mathbb{R},~0<\alpha<1,~z\in \mathbb{%
U)}\right\} .
\end{align*}
\end{example}

\begin{example}
\bigskip If $\Phi(z)=(z+z^{2})(1-z)^{-3}$ and $\Psi(z)=z(1-z)^{-2},$ then 
\begin{align*}
& \mathcal{S}^{^{-1}}\left( (z+z^{2})(1-z)^{-3},z(1-z)^{-2};\alpha
,\beta\right) \\
& \equiv\mathcal{K}_{\beta}(\alpha):=\left\{ f\in\mathcal{A}:\left\vert 
\frac{e^{-i\beta}f^{\prime}(z)}{f^{\prime}(z)+zf^{\prime\prime}(z)}-\frac {1%
}{2\alpha}\right\vert <\frac{1}{2\alpha}\text{ \ \ }(\text{ }\beta \in%
\mathbb{R},~0<\alpha<1,~z\in\mathbb{U)}\right\} .
\end{align*}
\end{example}

The class $\mathcal{S}_{\beta}(\alpha),$ the $\beta$-spirallike functions of
reciprocal order $\alpha$ and the class $\mathcal{K}_{\beta}(\alpha),$ the $%
\beta$-convexlike functions of reciprocal order $\alpha$ were studied by Owa
et al. \cite{Kamali}.

\begin{example}
\bigskip If $\Phi(z)=z(1-z)^{-2}$, $\Psi(z)=z(1-z)^{-1}$ and $\beta=0$, then 
\begin{align*}
& \mathcal{S}^{^{-1}}\left( z(1-z)^{-2},z(1-z)^{-1};\alpha,0\right) \\
& \equiv\mathcal{M}(\alpha):=\left\{ f\in\mathcal{A}:\left\vert \frac{%
f^{\prime}(z)}{zf(z)}-\frac{1}{2\alpha}\right\vert <\frac{1}{2\alpha }\text{
\ \ }(0<\alpha<1,~z\in\mathbb{U})\right\} .
\end{align*}
\end{example}

\begin{example}
\bigskip If $\Phi(z)=(z+z^{2})(1-z)^{-3}$,$\Psi(z)=z(1-z)^{-2}$ and $\beta=0$%
, then 
\begin{align*}
& \mathcal{S}^{^{-1}}\left( (z+z^{2})(1-z)^{-3},z(1-z)^{-2};\alpha,0\right)
\\
& \equiv\mathcal{N}(\alpha):=\left\{ f\in\mathcal{A}:\left\vert \frac{%
f^{\prime}(z)}{f^{\prime}(z)+zf^{\prime\prime}(z)}-\frac{1}{2\alpha }%
\right\vert <\frac{1}{2\alpha}\text{ \ \ }(0<\alpha<1,z\in\mathbb{U}%
)\right\} .
\end{align*}
\end{example}

The classes $\mathcal{M}(\alpha)$ and $\mathcal{N}(\alpha)$ were studied by
Owa et al. \cite{Ochi}.

Furthermore, we have the following new classes:

\begin{example}
\bigskip If $\Phi(z)=z(1-z)^{-2}$ and $\Psi(z)=z,$ then%
\begin{align*}
& \mathcal{S}^{^{-1}}\left( z(1-z)^{-2},z;\alpha,\beta\right) \\
& \equiv\mathcal{P}_{\beta}(\alpha):=\left\{ f\in\mathcal{A}:\left\vert 
\frac{1}{e^{i\beta}f^{\prime}(z)}-\frac{1}{2\alpha}\right\vert <\frac {1}{%
2\alpha}\text{ \ \ }(\text{ }\beta\in\mathbb{R},~0<\alpha<1,~z\in \mathbb{U)}%
\right\} .
\end{align*}
\end{example}

\begin{example}
\label{ex2}\bigskip If $\Phi(z)=(z+z^{2})(1-z)^{-3}$ and $\Psi(z)=z,$ then 
\begin{align*}
& \mathcal{S}^{^{-1}}(z+z^{2})(1-z)^{-3},z;\alpha,\beta) \\
& \equiv\mathcal{R}_{\beta}(\alpha):=\left\{ f\in\mathcal{A}:\left\vert 
\frac{1}{e^{i\beta}\left( (zf^{\prime}(z))^{\prime}\right) }-\frac {1}{%
2\alpha}\right\vert <\frac{1}{2\alpha}\text{ \ \ }(\text{ }\beta \in\mathbb{R%
},~0<\alpha<1,~z\in\mathbb{U)}\right\} .
\end{align*}
\end{example}

In fact many new subclasses of functions of reciprocal order can be defined
and studied by suitably choosing $\Phi(z),\Psi(z)$ and $\beta.$

The aim of the present paper is to investigate the coefficient estimates and
subordination properties for the class $\mathcal{S}^{^{-1}}(\Phi,\Psi
;\alpha,\beta)$. Some interesting consequences of the results are also
pointed out.

\section{Coefficient estimates}

The sufficient condition for $f(z)$ to be in the class $\mathcal{S}%
^{^{-1}}(\Phi,\Psi;\alpha,\beta)$ is given by using coefficient inequalities.

\begin{theorem}
\label{t} If $f$ $\in\mathcal{A}$ satisfies 
\begin{equation}
\sum\limits_{n=2}^{\infty}\left[ \delta_{n}+\left\vert \delta_{n}-2\alpha
e^{-i\beta}\mu_{n}\right\vert \right] \left\vert a_{n}\right\vert
\leq1-\left\vert 1-2\alpha e^{-i\beta}\right\vert   \label{dd}
\end{equation}
for some $\left\vert \beta\right\vert <\frac{\pi}{2}$ and $0<\alpha<\cos
\beta,$ then $f(z)\in\mathcal{S}^{^{-1}}(\Phi,\Psi;\alpha,\beta).$
\end{theorem}

\begin{proof}
It suffices to show that%
\[
\left\vert \frac{2\alpha\left(  f(z)\ast\Psi(z)\right)  -e^{i\beta}\left(
f(z)\ast\Phi(z)\right)  }{e^{i\beta}\left(  f(z)\ast\Phi(z)\right)
}\right\vert <1.
\]
We observe that%
\begin{align*}
\left\vert \frac{2\alpha\left(  f(z)\ast\Psi(z)\right)  -e^{i\beta}\left(
f(z)\ast\Phi(z)\right)  }{e^{i\beta}\left(  f(z)\ast\Phi(z)\right)
}\right\vert  &  =\left\vert \frac{\left(  1-2\alpha e^{-i\beta}\right)
+\sum\limits_{n=2}^{\infty}\left(  \delta_{n}-2\alpha e^{-i\beta}\mu
_{n}\right)  a_{n}z^{n-1}}{1+\sum\limits_{n=2}^{\infty}\delta_{n}a_{n}z^{n-1}%
}\right\vert \\
&  \leq\frac{\left\vert 1-2\alpha e^{i\beta}\right\vert +\sum\limits_{n=2}%
^{\infty}\left\vert \delta_{n}-2\alpha e^{-i\beta}\mu_{n}\right\vert
\left\vert a_{n}\right\vert \left\vert z\right\vert ^{n-1}}{1-\sum
\limits_{n=2}^{\infty}\delta_{n}\left\vert a_{n}\right\vert \left\vert
z\right\vert ^{n-1}}\\
&  <\frac{\left\vert 1-2\alpha e^{i\beta}\right\vert +\sum\limits_{n=2}%
^{\infty}\left\vert \delta_{n}-2\alpha e^{-i\beta}\mu_{n}\right\vert
\left\vert a_{n}\right\vert }{1-\sum\limits_{n=2}^{\infty}\delta_{n}\left\vert
a_{n}\right\vert }.
\end{align*}


It follows that the last term is bounded by $1$ if
\[
\sum\limits_{n=2}^{\infty}\left[  \delta_{n}+\left\vert \delta_{n}-2\alpha
e^{-i\beta}\mu_{n}\right\vert \right]  \left\vert a_{n}\right\vert
\leq1-\left\vert 1-2\alpha e^{-i\beta}\right\vert
\]
for some $\left\vert \beta\right\vert <\frac{\pi}{2}$ and $0<\alpha<\cos
\beta,$ which is equivalent to (\ref{dd}). Therefore, we have $f(z)\in
\mathcal{S}^{^{-1}}(\Phi,\Psi;\alpha,\beta)$ for some $\left\vert
\beta\right\vert <\frac{\pi}{2}$ and $0<\alpha<\cos\beta.$
\end{proof}

In the view of Examples \ref{ex1} to \ref{ex2}, we state the following
corollaries.

\begin{corollary}
(\cite{Kamali})If $f$ $\in\mathcal{A}$ satisfies%
\begin{equation}
\sum\limits_{n=2}^{\infty}\left[ n+\left\vert n-2\alpha e^{-i\beta
}\right\vert \right] \left\vert a_{n}\right\vert \leq1-\left\vert 1-2\alpha
e^{-i\beta}\right\vert   \label{a}
\end{equation}
for some $\left\vert \beta\right\vert <\frac{\pi}{2}$ and $0<\alpha<\cos
\beta,$ then $f(z)\in\mathcal{S}_{\beta}(\alpha).$
\end{corollary}

\begin{corollary}
(\cite{Kamali})If $f$ $\in\mathcal{A}$ satisfies%
\begin{equation}
\sum\limits_{n=2}^{\infty}n\left[ n+\left\vert n-2\alpha e^{-i\beta
}\right\vert \right] \left\vert a_{n}\right\vert \leq1-\left\vert 1-2\alpha
e^{-i\beta}\right\vert   \label{b}
\end{equation}
for some $\left\vert \beta\right\vert <\frac{\pi}{2}$ and $0<\alpha<\cos
\beta,$ then $f(z)\in\mathcal{K}_{\beta}(\alpha).$
\end{corollary}

\begin{corollary}
(\cite{Ochi})Let $\ 0<\alpha<1.$ If $f$ $\in\mathcal{A}$ satisfies 
\begin{equation}
\sum\limits_{n=2}^{\infty}\left( n-\alpha\right) \left\vert a_{n}\right\vert
\leq\frac{1}{2}\left( 1-\left\vert 1-2\alpha\right\vert \right) =\left\{ 
\begin{array}{c}
\alpha;\text{ \ \ \ if \ \ }\ 0<\alpha\leq\frac{1}{2} \\ 
1-\alpha;\text{ \ \ \ \ \ if \ }\ \frac{1}{2}\leq\alpha<1%
\end{array}
\right. ,   \label{cc}
\end{equation}
then $f(z)\in\mathcal{M}(\alpha).$
\end{corollary}

\begin{corollary}
(\cite{Ochi})Let $\ 0<\alpha<1.$ If $f$ $\in\mathcal{A}$ satisfies 
\begin{equation}
\sum\limits_{n=2}^{\infty}n\left( n-\alpha\right) \left\vert
a_{n}\right\vert \leq\frac{1}{2}\left( 1-\left\vert 1-2\alpha\right\vert
\right) =\left\{ 
\begin{array}{c}
\alpha;\text{ \ \ \ if \ \ }\ 0<\alpha\leq\frac{1}{2} \\ 
1-\alpha;\text{ \ \ \ \ \ if \ }\ \frac{1}{2}\leq\alpha<1%
\end{array}
\right. ,   \label{d}
\end{equation}
then $f(z)\in\mathcal{N}(\alpha).$
\end{corollary}

\begin{corollary}
If $f$ $\in\mathcal{A}$ satisfies%
\begin{equation}
\sum\limits_{n=2}^{\infty}2n\left\vert a_{n}\right\vert \leq1-\left\vert
1-2\alpha e^{-i\beta}\right\vert   \label{e}
\end{equation}
for some $\left\vert \beta\right\vert <\frac{\pi}{2}$ and $0<\alpha<\cos
\beta,$ then $f(z)\in\mathcal{P}_{\beta}(\alpha).$
\end{corollary}

\begin{corollary}
If $f$ $\in\mathcal{A}$ satisfies%
\begin{equation}
\sum\limits_{n=2}^{\infty}2n^{2}\left\vert a_{n}\right\vert \leq1-\left\vert
1-2\alpha e^{-i\beta}\right\vert   \label{f}
\end{equation}
for some $\left\vert \beta\right\vert <\frac{\pi}{2}$ and $0<\alpha<\cos
\beta,$ then $f(z)\in\mathcal{R}_{\beta}(\alpha).$
\end{corollary}

\section{Subordination results}

To proceed our main result in this section, let us first recall the
following definitions and lemma.

\begin{definition}
\textbf{(Subordination Principle).} For two functions $f$ and $g,$ analytic
in\ $\mathbb{U},$ we say that the function $f(z)$ is subordinate to $g(z)$
in $\mathbb{U}$, and write $f\prec g$ or $f(z)\prec g(z)$ $(z\in\mathbb{U)},$
if there exists a Schwarz function $w(z),$ analytic in $\mathbb{U}$ with $%
w(0)=0$ and $\left\vert w\right\vert <1$ $(z\in\mathbb{U)},$ such that $%
f(z)=g(w(z))$ $(z\in\mathbb{U)}.$ In particular, if the function $g$ is
univalent in $\mathbb{U},$ \ the above subordination is equivalent to $%
f(0)=g(0)$ and $f(\mathbb{U})\subset g(\mathbb{U}).$
\end{definition}

\begin{definition}
\label{def1} A sequence $\left\{ b_{n}\right\} _{n=1}^{\infty}$ of complex
numbers is said to be a subordinating factor sequence if, whenever $f(z)\ $%
of the form (\ref{tr}), $a_{1}=1$ is analytic, univalent and convex in $%
\mathbb{U}$, we have the subordination given by 
\begin{equation}
\sum\limits_{n=1}^{\infty}b_{n}a_{n}z^{n}\prec f(z),\ \ \ z\in\mathbb{U}. 
\label{qq}
\end{equation}
\end{definition}

\begin{lemma}
\label{lem1}\ ( \cite{a9}).\ The sequence $\left\{ b_{n}\right\}
_{n=1}^{\infty}$ is a subordinating factor sequence if and only if 
\begin{equation}
\mbox{Re}\left\{ 1+2\sum\limits_{n=1}^{\infty}b_{n}z^{n}\right\} >0\text{ }\
\ \text{ }(z\in\mathbb{U)}\text{.}   \label{oo}
\end{equation}
\end{lemma}

Let $\mathcal{S}^{\mathcal{\ast }^{-1}}(\Phi ,\Psi ;\alpha ,\beta )$ $%
\subseteq $ $\mathcal{S}^{^{-1}}(\Phi ,\Psi ;\alpha ,\beta )$ denote the
subclass of functions $f$ $\in \mathcal{A}$ whose coefficients $a_{n}$
satisfy the inequalities (\ref{dd}) .

Employing the techniques used by Attiya \cite{at}, Singh \cite{sin} and
Srivastava and Attiya \cite{sri} ( see also, \cite{auf}, \cite{aou}, \cite%
{fra1}, \cite{fra2}, \cite{fradar} and \cite{tar}), we state and prove the
following theorem.

\begin{theorem}
\label{th2} Let $f(z)\in$ $\mathcal{S}^{\mathcal{\ast}^{-1}}(\Phi,\Psi
;\alpha,\beta)$ and $\delta_{n}+\left\vert \delta_{n}-2\alpha
e^{-i\beta}\mu_{n}\right\vert $ \ is increasing function for $n\geq2,$ $%
\left\vert \beta\right\vert <\frac{\pi}{2}$ , $0<\alpha<\cos\beta$. Then%
\begin{equation}
\frac{\delta_{2}+\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert 
}{2\left( 1+\delta_{2}-\left\vert 1-2\alpha e^{-i\beta}\right\vert
+\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert \right) }(f\ast
g)(z)\prec g(z)   \label{ff}
\end{equation}
for every function $g(z)$ in the class $\mathcal{K}$

and%
\begin{equation}
\mbox{Re}f(z)>-\frac{1+\delta_{2}-\left\vert 1-2\alpha
e^{-i\beta}\right\vert +\left\vert \delta_{2}-2\alpha
e^{-i\beta}\mu_{2}\right\vert }{\delta _{2}+\left\vert \delta_{2}-2\alpha
e^{-i\beta}\mu_{2}\right\vert }   \label{yy}
\end{equation}
for \ $z\in\mathbb{U}$.

The constant $\frac{\delta_{2}+\left\vert \delta_{2}-2\alpha
e^{-i\beta}\mu_{2}\right\vert }{2\left( 1+\delta_{2}-\left\vert 1-2\alpha
e^{-i\beta }\right\vert +\left\vert \delta_{2}-2\alpha
e^{-i\beta}\mu_{2}\right\vert \right) }$ cannot be replace by any larger one.
\end{theorem}

\begin{proof}
Let\ $f(z)=z+\sum\limits_{n=2}^{\infty}a_{n}z^{n}\in\mathcal{S}^{\mathcal{\ast
}^{-1}}(\Phi,\Psi;\alpha,\beta),$ and let $g(z)=z+\sum\limits_{n=2}^{\infty
}c_{n}z^{n}\in\mathcal{K}$. Then
\begin{align*}
&  \frac{\delta_{2}+\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu
_{2}\right\vert }{2\left(  1+\delta_{2}-\left\vert 1-2\alpha e^{-i\beta
}\right\vert +\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert
\right)  }(f\ast g)(z)\\
&  =\frac{\delta_{2}+\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu
_{2}\right\vert }{2\left(  1+\delta_{2}-\left\vert 1-2\alpha e^{-i\beta
}\right\vert +\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert
\right)  }\left(  z+\sum\limits_{n=2}^{\infty}a_{n}c_{n}z^{n}\right)  .
\end{align*}
Thus, by Definition \ref{def1}, the assertion of our theorem will hold if the
sequence%
\[
\left\{  \frac{\delta_{2}+\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu
_{2}\right\vert }{2\left(  1+\delta_{2}-\left\vert 1-2\alpha e^{-i\beta
}\right\vert +\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert
\right)  }a_{n}\right\}  _{n=1}^{\infty}%
\]
is a subordinating factor sequence, with $a_{1}=1$. In view of Lemma
\ref{lem1}, this will be the case if and only if%
\begin{equation}
\mbox{Re}\left\{  1+\sum\limits_{n=1}^{\infty}\frac{\delta_{2}+\left\vert
\delta_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert }{1+\delta_{2}-\left\vert
1-2\alpha e^{-i\beta}\right\vert +\left\vert \delta_{2}-2\alpha e^{-i\beta}%
\mu_{2}\right\vert }a_{n}z^{n}\right\}  >0\text{ \ \ \ }(z\in\mathbb{U}).
\label{ss}%
\end{equation}
Since $\delta_{n}+\left\vert \delta_{n}-2\alpha e^{-i\beta}\mu_{n}\right\vert
$ increasing\ for all $n\geq2,$ $\left\vert \beta\right\vert <\frac{\pi}{2}$ ,
$0<\alpha<\cos\beta,$ we obtain
\begin{align*}
&  \mbox{Re}\left\{  1+\sum\limits_{n=1}^{\infty}\frac{\delta_{2}+\left\vert
\delta_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert }{1+\delta_{2}-\left\vert
1-2\alpha e^{-i\beta}\right\vert +\left\vert \delta_{2}-2\alpha e^{-i\beta}%
\mu_{2}\right\vert }a_{n}z^{n}\right\} \\
&  =\mbox{Re}\left\{  1+\frac{\delta_{2}+\left\vert \delta_{2}-2\alpha
e^{-i\lambda}\mu_{2}\right\vert }{1+\delta_{2}-\left\vert 1-2\alpha
e^{-i\lambda}\right\vert +\left\vert \delta_{2}-2\alpha e^{-i\lambda}\mu
_{2}\right\vert }z\right.  +\\
&  \left.  \frac{1}{1+\delta_{2}-\left\vert 1-2\alpha e^{-i\beta}\right\vert
+\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert }\sum
\limits_{n=2}^{\infty}\left(  \delta_{2}+\left\vert \delta_{2}-2\alpha
e^{-i\beta}\mu_{2}\right\vert \right)  a_{n}z^{n}\right\} \\
&  \geq1-\frac{\delta_{2}+\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu
_{2}\right\vert }{1+\delta_{2}-\left\vert 1-2\alpha e^{-i\beta}\right\vert
+\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert }r-\\
&  \frac{1}{1+\delta_{2}-\left\vert 1-2\alpha e^{-i\beta}\right\vert
+\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert }\sum
\limits_{n=2}^{\infty}\left(  \delta_{2}+\left\vert \delta_{2}-2\alpha
e^{-i\beta}\mu_{2}\right\vert \right)  \left\vert a_{n}\right\vert r^{n}\\
&  >1-\frac{\delta_{2}+\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu
_{2}\right\vert }{1+\delta_{2}-\left\vert 1-2\alpha e^{-i\beta}\right\vert
+\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert }r-\frac
{1-\left\vert 1-2\alpha e^{-i\beta}\right\vert }{1+\delta_{2}-\left\vert
1-2\alpha e^{-i\beta}\right\vert +\left\vert \delta_{2}-2\alpha e^{-i\beta}%
\mu_{2}\right\vert }r\\
&  >0,\qquad|z|=r<1.
\end{align*}


This evidently proves the inequality (\ref{ss}) and hence also the
subordination result (\ref{ff}). The inequality (\ref{yy}) follows from
(\ref{ff}) by taking $g(z)=\frac{z}{1-z}.$

To prove the sharpness of the constant $\frac{\delta_{2}+\left\vert \delta
_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert }{2\left(  1+\delta_{2}-\left\vert
1-2\alpha e^{-i\beta}\right\vert +\left\vert \delta_{2}-2\alpha e^{-i\beta}%
\mu_{2}\right\vert \right)  },$ we consider the function%
\[
f_{0}(z)=z-\frac{1-\left\vert 1-2\alpha e^{-i\beta}\right\vert }{\delta
_{2}+\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert }z^{2}\text{
\ \ \ }(\left\vert \beta\right\vert <\frac{\pi}{2},\text{ }0<\alpha<\cos
\beta),
\]
which is a member of the class $\mathcal{S}^{\mathcal{\ast}^{-1}}(\Phi
,\Psi;\alpha,\beta)$. Thus from the relation (\ref{ff}), we obtain%
\[
\frac{\delta_{2}+\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert
}{2\left(  1+\delta_{2}-\left\vert 1-2\alpha e^{-i\beta}\right\vert
+\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert \right)  }%
f_{0}(z)\prec\frac{z}{1-z}.
\]


It can be verified that%
\[
\underset{z\in\mathbb{U}}{\min}\left\{  \mbox{Re}\left(  \frac{\delta
_{2}+\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert }{2\left(
1+\delta_{2}-\left\vert 1-2\alpha e^{-i\beta}\right\vert +\left\vert
\delta_{2}-2\alpha e^{-i\beta}\mu_{2}\right\vert \right)  }f_{0}(z)\right)
\right\}  =\frac{-1}{2}.
\]
This shows that the constant $\frac{\delta_{2}+\left\vert \delta_{2}-2\alpha
e^{-i\beta}\mu_{2}\right\vert }{2\left(  1+\delta_{2}-\left\vert 1-2\alpha
e^{-i\beta}\right\vert +\left\vert \delta_{2}-2\alpha e^{-i\beta}\mu
_{2}\right\vert \right)  }$ is best possible.
\end{proof}

\bigskip By taking different choices of $\Phi$, $\Psi$ and $\beta$ in
Theorem \ref{th2} and in view of the Examples \ref{ex1} to \ref{ex2} in
Section 1, we state the following corollaries for the subclasses defined in
those examples.

\begin{corollary}
Let the function $f(z)$ defined by (\ref{tr}) be in the class $\mathcal{S}%
_{\beta}^{-1}(\alpha);$ $\left\vert \beta\right\vert <\frac{\pi}{2}$, $%
0<\alpha<\cos\beta$, where $\mathcal{S}_{\beta}^{-1}(\alpha)$ denote the
subclasses of functions whose coefficients $a_{n}$ satisfy the inequalities (%
\ref{a}) and suppose that $g(z)\in\mathcal{K}$. Then 
\begin{equation}
\frac{1+\left\vert 1-\alpha e^{-i\beta}\right\vert }{3-\left\vert 1-2\alpha
e^{-i\beta}\right\vert +2\left\vert 1-\alpha e^{-i\beta}\right\vert }(f\ast
g)(z)\prec g(z)   \label{c}
\end{equation}
and%
\begin{equation}
\mbox{Re}f(z)>-\frac{3-\left\vert 1-2\alpha e^{-i\beta}\right\vert
+2\left\vert 1-\alpha e^{-i\beta}\right\vert }{2(1+\left\vert 1-\alpha
e^{-i\beta}\right\vert )}.   \label{iii}
\end{equation}
The constant factor $\frac{1+\left\vert 1-\alpha e^{-i\beta}\right\vert }{%
3-\left\vert 1-2\alpha e^{-i\beta}\right\vert +2\left\vert 1-\alpha
e^{-i\beta}\right\vert }$ in the subordination result (\ref{c}) cannot be
replaced by a larger one.
\end{corollary}

\begin{corollary}
Let the function $f(z)$ defined by (\ref{tr}) be in the class $\mathcal{K}%
_{\beta}^{-1}(\alpha);$ $\left\vert \beta\right\vert <\frac{\pi}{2}$, $%
0<\alpha<\cos\beta$, where $\mathcal{K}_{\beta}^{-1}(\alpha)$ denote the
subclasses of functions whose coefficients $a_{n}$ satisfy the inequalities (%
\ref{b}) and suppose that $g(z)\in\mathcal{K}$. Then 
\begin{equation}
\frac{2(1+\left\vert 1-\alpha e^{-i\beta}\right\vert )}{5-\left\vert
1-2\alpha e^{-i\beta}\right\vert +4\left\vert 1-\alpha
e^{-i\beta}\right\vert }(f\ast g)(z)\prec g(z)   \label{h}
\end{equation}
and%
\begin{equation}
\mbox{Re}f(z)>-\frac{5-\left\vert 1-2\alpha e^{-i\beta}\right\vert
+4\left\vert 1-\alpha e^{-i\beta}\right\vert }{4(1+\left\vert 1-\alpha
e^{-i\beta}\right\vert )}.
\end{equation}
The constant factor $\frac{2(1+\left\vert 1-\alpha e^{-i\beta}\right\vert )}{%
5-\left\vert 1-2\alpha e^{-i\beta}\right\vert +4\left\vert 1-\alpha
e^{-i\beta}\right\vert }$ in the subordination result (\ref{h}) cannot be
replaced by a larger one.
\end{corollary}

\begin{corollary}
Let the function $f(z)$ defined by (\ref{tr}) be in the class $\mathcal{M}%
^{-1}(\alpha);0<\alpha<1$, where $\mathcal{M}^{-1}(\alpha)$ denote the
subclasses of functions whose coefficients $a_{n}$ satisfy the inequalities (%
\ref{cc}) and suppose that $g(z)\in\mathcal{K}$. Then 
\begin{equation}
\frac{2-\alpha}{5-2\alpha-\left\vert 1-2\alpha\right\vert }(f\ast g)(z)\prec
g(z)   \label{rr}
\end{equation}
and%
\begin{equation}
\mbox{Re}f(z)>-\frac{5-2\alpha-\left\vert 1-2\alpha\right\vert }{2(1-\alpha )%
}.   \label{k}
\end{equation}
The constant factor $\frac{2-\alpha}{5-2\alpha-\left\vert
1-2\alpha\right\vert }$ in the subordination result (\ref{rr}) cannot be
replaced by a larger one.
\end{corollary}

\begin{corollary}
Let the function $f(z)$ defined by (\ref{tr}) be in the class $\mathcal{N}%
^{-1}(\alpha);$ $0<\alpha<1$, where $\mathcal{N}^{-1}(\alpha)$ denote the
subclasses of functions whose coefficients $a_{n}$ satisfy the inequalities (%
\ref{d}) and suppose that $g(z)\in\mathcal{K}$. Then 
\begin{equation}
\frac{2(2-\alpha)}{9-4\alpha-\left\vert 1-2\alpha\right\vert }(f\ast
g)(z)\prec g(z)   \label{gh}
\end{equation}
and%
\begin{equation}
\mbox{Re}f(z)>-\frac{9-4\alpha-\left\vert 1-2\alpha\right\vert }{4(2-\alpha)}%
.
\end{equation}
The constant factor $\frac{2(2-\alpha)}{9-4\alpha-\left\vert 1-2\alpha
\right\vert }$ in the subordination result (\ref{gh}) cannot be replaced by
a larger one.
\end{corollary}

\begin{corollary}
Let the function $f(z)$ defined by (\ref{tr}) be in the class $\mathcal{P}%
_{\beta}^{-1}(\alpha);$ $\left\vert \beta\right\vert <\frac{\pi}{2}$, $%
0<\alpha<\cos\beta$, where $\mathcal{P}_{\beta}^{-1}(\alpha)$ denote the
subclasses of functions whose coefficients $a_{n}$ satisfy the inequalities (%
\ref{e}) and suppose that $g(z)\in\mathcal{K}$. Then 
\begin{equation}
\frac{2}{5-\left\vert 1-2\alpha e^{-i\beta}\right\vert }(f\ast g)(z)\prec
g(z)   \label{v}
\end{equation}
and%
\begin{equation}
\mbox{Re}f(z)>-\frac{5-\left\vert 1-2\alpha e^{-i\beta}\right\vert }{4}.
\end{equation}
The constant factor $\frac{2}{5-\left\vert 1-2\alpha e^{-i\beta}\right\vert }
$ in the subordination result (\ref{v}) cannot be replaced by a larger one.
\end{corollary}

\begin{corollary}
Let the function $f(z)$ defined by (\ref{tr}) be in the class $\mathcal{R}%
_{\beta}^{-1}(\alpha);$ $\left\vert \beta\right\vert <\frac{\pi}{2}$, $%
0<\alpha<\cos\beta$, where $\mathcal{R}_{\beta}^{-1}(\alpha)$ denote the
subclasses of functions whose coefficients $a_{n}$ satisfy the inequalities (%
\ref{f}) and suppose that $g(z)\in\mathcal{K}$. Then 
\begin{equation}
\frac{4}{9-\left\vert 1-2\alpha e^{-i\beta}\right\vert }(f\ast g)(z)\prec
g(z)   \label{gg}
\end{equation}
and%
\begin{equation}
\mbox{Re}f(z)>-\frac{9-\left\vert 1-2\alpha e^{-i\beta}\right\vert }{8}.
\end{equation}
The constant factor $\frac{4}{9-\left\vert 1-2\alpha e^{-i\beta}\right\vert }
$ in the subordination result (\ref{gg}) cannot be replaced by a larger one.
\end{corollary}

\textbf{Acknowledgements.} The authors would like to thank the referee for
his helpful comments and suggestions.

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\end{document}
