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\begin{document}
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\title{Starlike and convex properties for Poisson distribution series}
\author{N. Magesh}
\address{Post-Graduate and Research Department of Mathematics,\\
Government Arts College for Men,
Krishnagiri 635001, Tamilnadu, India}
\email{nmagi2000@gmail.com}
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\author{S. Porwal}
\address{Department of Mathematics, U.I.E.T., C.S.J.M. University \\ Kanpur-208024, (U.P.), India}
\email{saurabhjcb@rediffmail.com}
%
\author{C. Abirami}
\address{Faculty of Engineering and Technology \\
SRM University, Kattankulathur-603203, Tamilnadu, India}
\email{shreelekha07@yahoo.com}
%
\subjclass{30C45.}
\keywords{Starlike functions, convex functions, Poisson distribution series.}
\begin{abstract}
In this paper, we find the  necessary
and sufficient conditions,inclusion relations  for Poisson distribution series
belonging to the classes $\mathscr{S}^*(\alpha,\beta)$ and
$\mathscr{C}^*(\alpha,\beta).$ Further, we consider
an integral operator related to Poisson Distribution series.\end{abstract}
\maketitle

\section{Introduction}
Let $\mathscr{A}$ denote the class of functions of the form
\begin{equation}\label{e1.1}
 f(z) = z + \sum\limits_{n=2}^{\infty} a_nz^n
\end{equation}
which   are analytic and univalent in the open disc $\mathbb{U} = \{z :z
\in \mathbb{C}\,\, |z|<1 \}.$ Let $\mathscr{T}$ be a subclass of $\mathscr{A}$ consisting of functions
whose non-zero coefficients from second on is give by
\begin{equation}\label{NC2}
 f(z) = z - \sum\limits_{n=2}^{\infty} |a_n|z^n, \qquad z \in \mathbb{U}.
\end{equation}
%%======================================================================
In 2014, Porwal \cite{SP-JCA-14} introduced a power series whose coefficients are probabilities
of Poisson distribution
\[
\mathscr{K}(m,z):= z + \sum\limits_{n=2}^{\infty}
\frac{m^{n-1}}{(n-1)!}e^{-m}z^n,\quad z \in \mathbb{U},
\]
where $m > 0.$ By ratio test the radius of convergence of the above series is infinity.
Further, Porwal \cite{SP-JCA-14} defined a series
\begin{eqnarray*}
\mathscr{F}(m,z) & = & 2z - \mathscr{K}(m,z) = z - \sum\limits_{n=2}^{\infty}
\frac{m^{n-1}}{(n-1)!}e^{-m}z^n, \qquad z \in \mathbb{U}.
\end{eqnarray*}
Corresponding to the series $\mathscr{K}(m,z)$ using the Hadamard product for $f \in \mathscr{A}$,
Porwal and Kumar \cite{SP-MK-16} introduced a new linear operator
$\mathscr{I}(m): \mathscr{A} \rightarrow \mathscr{A}$ defined by
\begin{eqnarray*}
\mathscr{I}(m)f(z) :& = & \mathscr{K}(m,z) * f(z) \nonumber\\
          & = & z + \sum\limits_{n=2}^{\infty}
\frac{m^{n-1}}{(n-1)!}e^{-m}a_n z^n, \qquad z \in \mathbb{U},
\end{eqnarray*}
where $*$ denotes the convolution (or Hadamard product) of
two series $f(z)=\sum\limits_{n=0}^{\infty}a_nz^n$ and
$g(z)=\sum\limits_{n=0}^{\infty}b_nz^n$ is defined by
$(f*g)(z)=\sum\limits_{n=0}^{\infty}a_nb_nz^n.$

Let $\mathscr{S}^*(\alpha, \beta)$ be the subclass of $\mathscr{T}$ consisting of functions which satisfy the condition:
\[
\left |
\frac{\frac{zf'(z)}{f(z)} - 1}
{\frac{zf'(z)}{f(z)} + 1 - 2\alpha}
\right| < \beta, \qquad z \in \mathbb{U},
\]
where $0 \leq \alpha < 1$ and $0 < \beta \leq 1.$

Also, let $\mathscr{C}^*(\alpha, \beta)$ be the subclass of $\mathscr{T}$ consisting of functions which satisfy the condition:
\[
\left |
\frac{\frac{zf''(z)}{f'(z)}}
{\frac{zf''(z)}{f'(z)} + 2(1 - \alpha)}
\right| < \beta, \qquad z \in \mathbb{U},
\]
where $0 \leq \alpha < 1$ and $0 < \beta \leq 1.$

The classes $\mathscr{S}^*(\alpha, \beta)$ and $\mathscr{C}^*(\alpha, \beta),$
were introduced and studied by Gupta and Jain \cite{Gupta-Jain} (see \cite{Mostafa-2010}).
Also, we note that for $\beta = 1$ the classes $\mathscr{S}^*(\alpha, \beta)$ and
$\mathscr{C}^*(\alpha, \beta)$  reduce to the class of starlike and convex functions
of order $\alpha (0 \leq \alpha < 1)$ (see \cite{Silverman-1975}).
\par A function $f \in \mathscr{A}$ is said to be in the class $\mathscr{R}^{\tau}(A, B),$
$(\tau \in \mathbb{C}\setminus\{0\}, -1 \leq B < A\leq 1),$ if it satisfies the inequality
\[
\left |\frac{f'(z) - 1}{(A-B)\tau - B[f'(z) - 1]} \right| < 1, \qquad z \in \mathbb{U}.
\]
This class was introduced by Dixit and Pal \cite{Dixit-Pal}.
\begin{lemma}\label{tb1.1a}\cite{Gupta-Jain}
A function $f(z)$ of the form (\ref{NC2}) is in $\mathscr{S}^*(
\alpha, \beta)$ if and only if
\begin{equation}\label{e1.13a}
\sum\limits_{n=2}^{\infty}  [n(1+\beta) - 1 + \beta(1 - 2\alpha)]~|a_n| \leq 2\beta(1-\alpha).
\end{equation}
\end{lemma}
\begin{lemma}\label{tb1.2a}\cite{Gupta-Jain}
A function $f(z)$ of the form (\ref{NC2}) is in $\mathscr{C}^*(
\alpha, \beta)$ if and only if
\begin{equation}\label{e1.3a}
\sum\limits_{n=2}^{\infty}  n[n(1+\beta) - 1 + \beta(1 - 2\alpha)]~|a_n| \leq 2\beta(1-\alpha).
\end{equation}
\end{lemma}
To obtain our main results, we need the following lemmas:
\begin{lemma}\label{tb1.6}\cite{Dixit-Pal}
If $f \in \mathscr{R}^{\tau}(A,B)$ is of the form (\ref{e1.1}), then
\begin{equation}\label{e1.16}
|a_n| \leq (A-B) \frac{|\tau|}{n}, \qquad n \in \mathbb{N}\setminus\{1\}.
\end{equation}
\end{lemma}
In the present investigation, inspired by the works of Porwal
\cite{SP-JCA-14} and Porwal and Kumar \cite{SP-MK-16},
we find the necessary and sufficient conditions for $\mathscr{F}(m,z)$ belonging to the classes
$\mathscr{S}^*(\alpha, \beta)$ and $\mathscr{C}^*(\alpha, \beta).$
Also, we obtain inclusion relations for aforecited classes
with $\mathscr{R}^{\tau}(A,B).$
%%================================================================================
\section{Necessary and Sufficient Conditions}
\begin{theorem}\label{Gt1a} If $m > 0,$ $0 \leq \alpha < 1$ and $0 < \beta \leq 1,$
then $\mathscr{F}(m,z) \in \mathscr{S}^*(\alpha, \beta)$ if and only if
\begin{eqnarray}\label{Ge1a}
&& e^{m}m (1+\beta) \leq 2\beta(1- \alpha).
\end{eqnarray}
\end{theorem}
\begin{proof}
Since
\begin{eqnarray*}
\mathscr{F}(m,z) = z - \sum\limits_{n=2}^{\infty}\frac{m^{n-1}}{(n-1)!}e^{-m}z^n,
\end{eqnarray*}
in view of Lemma \ref{tb1.1a}, it is enough to show that
\begin{eqnarray*}
\sum\limits_{n=2}^{\infty}  [n(1+\beta) - 1 + \beta(1 - 2\alpha)]~\frac{m^{n-1}}{(n-1)!}e^{-m} \leq 2\beta(1-\alpha).
\end{eqnarray*}
Let
\[
T_1 = \sum\limits_{n=2}^{\infty}  [n(1+\beta) - 1
        + \beta(1 - 2\alpha)]~\frac{m^{n-1}}{(n-1)!}e^{-m}.
\]
Now,
\begin{eqnarray*}
T_1 & = & \sum\limits_{n=2}^{\infty}  [n(1+\beta) - 1 + \beta(1 - 2\alpha)]~\frac{m^{n-1}}{(n-1)!}e^{-m}  \\
    & = &  e^{-m}
            \sum\limits_{n=2}^{\infty}  \left[(n-1)(1+\beta)+2\beta(1-\alpha)\right]
            ~\frac{m^{n-1}}{(n-1)!}  \\
    & = & e^{-m}
            \left [ (1+\beta)
            \sum\limits_{n=2}^{\infty}  ~
            \frac{m^{n-1}}{(n-2)!}
            + 2\beta(1-\alpha)
            \sum\limits_{n=2}^{\infty} \frac{m^{n-1}}{(n-1)!}
            \right ]\\
  & = & e^{-m}
            \left [
            (1+\beta)me^{m} +
            2\beta(1-\alpha)(e^{m} - 1)
            \right ]\\
  & = & (1+\beta)m +
            2\beta(1-\alpha)(1 - e^{-m}).
\end{eqnarray*}
But this last expression is bounded by $2\beta(1-\alpha)$, if and only if (\ref{Ge1a}) holds.
This completes the proof of Theorem \ref{Gt1a}.
\end{proof}
\begin{theorem}\label{Gt2a}
If $m > 0,$ $0 \leq \alpha < 1$ and $0< \beta \leq 1,$  then $\mathscr{F}(m,z) \in \mathscr{C}^*(\alpha, \beta)$
if and only if
\begin{eqnarray}\label{Ge2a}
e^{m}\left[(1+\beta)m^2+2(1+\beta(2-\alpha))m\right] \leq 2\beta(1- \alpha).
\end{eqnarray}
\end{theorem}
\begin{proof}
Since
\begin{eqnarray*}
\mathscr{F}(m,z) = z - \sum\limits_{n=2}^{\infty}\frac{m^{n-1}}{(n-1)!}e^{-m}z^n,
\end{eqnarray*}
in view of Lemma \ref{tb1.2a}, it is enough to show that
\begin{eqnarray*}
\sum\limits_{n=2}^{\infty}  n[n(1+\beta) - 1 + \beta(1 - 2\alpha)]~\frac{m^{n-1}}{(n-1)!}e^{-m} \leq 2\beta(1-\alpha).
\end{eqnarray*}
Let
\begin{eqnarray*}
T_2 & = & \sum\limits_{n=2}^{\infty}  n[n(1+\beta) - 1 + \beta(1 - 2\alpha)]~\frac{m^{n-1}}{(n-1)!}e^{-m}.
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
T_2 & = & e^{-m} \left [
                \sum\limits_{n=2}^{\infty}  (n-1)(n-2)(1+\beta)~\frac{m^{n-1}}{(n-1)!}
                \right. \\ && \left. \qquad
                + \sum\limits_{n=2}^{\infty} (n-1)[3(1+\beta) - 1+\beta(1-2\alpha)]~\frac{m^{n-1}}{(n-1)!}
                + \sum\limits_{n=2}^{\infty} 2\beta(1-\alpha)~\frac{m^{n-1}}{(n-1)!}
                \right ]\\
    & = & e^{-m} \left [
                (1+\beta)\sum\limits_{n=3}^{\infty} \frac{m^{n-1}}{(n-3)!}
                \right. \\ && \left. \qquad
                + 2[1+\beta(2-\alpha)]\sum\limits_{n=2}^{\infty} \frac{m^{n-1}}{(n-2)!}
                + 2\beta(1-\alpha)\sum\limits_{n=2}^{\infty} \frac{m^{n-1}}{(n-1)!}
                \right ]\\
    & = & e^{-m} \left [
                (1+\beta)m^2e^m
                + 2(1+\beta(2-\alpha))me^m
                + 2\beta(1-\alpha)(e^m - 1)
                \right ]\\
    & = & (1+\beta)m^2
                + 2(1+\beta(2-\alpha))m
                + 2\beta(1-\alpha)(1 - e^{-m}).
\end{eqnarray*}
But this last expression is bounded by $2\beta(1-\alpha)$, if and only if  (\ref{Ge2a}) holds.
This completes the proof of Theorem \ref{Gt2a}.
\end{proof}
%%================================================================================
\section{Inclusion Results}
\begin{theorem}\label{Gt1} Let $m > 0,$ $0 \leq \alpha < 1$ and $0 < \beta \leq 1.$
If $f \in \mathscr{R}^{\tau}(A,B),$  then $\mathscr{I}(m)f \in \mathscr{S}^*(\alpha, \beta)$
if and only if
\begin{eqnarray}\label{Ge1}
&& (A-B)|\tau|\left[(1+\beta)(1 - e^{-m})+\frac{(\beta(1-2\alpha)-1)}{m}
     (1-e^{-m}-me^{-m})\right] \leq 2\beta(1- \alpha).
\end{eqnarray}
\end{theorem}
\begin{proof}
In view of Lemma \ref{tb1.1a}, it suffices to show that
\begin{eqnarray*}
P_1 = \sum\limits_{n=2}^{\infty}  [n(1+\beta) - 1 + \beta(1 - 2\alpha)]~\frac{m^{n-1}}{(n-1)!}e^{-m}|a_n| \leq 2\beta(1-\alpha).
\end{eqnarray*}
\noindent Since $f \in \mathscr{R}^{\tau}(A,B),$ then by Lemma \ref{tb1.6}, we have
\[
|a_n| \leq  \frac{(A-B)|\tau|}{n}.
\]
Therefore,
\begin{eqnarray*}
P_1 &\leq& \sum\limits_{n=2}^{\infty}  [n(1+\beta) - 1 + \beta(1 - 2\alpha)]
            ~\frac{m^{n-1}}{(n-1)!}e^{-m} \frac{(A-B)|\tau|}{n} \\
    & = &  (A-B)|\tau|e^{-m}
            \sum\limits_{n=2}^{\infty}  [n(1+\beta) - 1 + \beta(1 - 2\alpha)]
            ~\frac{m^{n-1}}
            {n!}  \\
    & = & (A-B)|\tau|e^{-m}
            \left [ (1+\beta)
            \sum\limits_{n=2}^{\infty}  ~
            \frac{m^{n-1}}{(n-1)!}
            + \frac{(\beta(1-2\alpha)-1)}{m}
            \sum\limits_{n=2}^{\infty} \frac{m^{n}}{n!}
            \right ]\\
  & = & (A-B)|\tau|e^{-m}
            \left [(1+\beta)
            (e^{m} - 1)
            + \frac{(\beta(1-2\alpha)-1)}{m}
            (e^{m} - 1 - m)
            \right ]\\
  & = & (A-B)|\tau|
            \left [(1+\beta)
            [1 - e^{-m}]
            + \frac{(\beta(1-2\alpha)-1)}{m}
            (1 - e^{-m} - me^{-m})
            \right ].
\end{eqnarray*}
But this last expression is bounded by $2\beta(1-\alpha)$, if (\ref{Ge1}) holds.
This completes the proof of Theorem \ref{Gt1}.
\end{proof}
%%================Theorem==========================================================
\begin{theorem}\label{Gt2} Let $m > 0,$ $0 \leq \alpha < 1$ and $0 < \beta \leq 1.$
If $f \in \mathscr{R}^{\tau}(A,B),$ then $\mathscr{I}(m)f \in \mathscr{C}^*(\alpha, \beta)$ if and only if
\begin{eqnarray}\label{Ge2}
&& (A-B)|\tau|\left[m(1+\beta)+2\beta(1-\alpha)
                     (1-e^{-m})\right] \leq 2\beta(1- \alpha).
\end{eqnarray}
\end{theorem}
\begin{proof}
In view of Lemma \ref{tb1.2a}, it suffices to show that
\begin{eqnarray*}
P_2 = \sum\limits_{n=2}^{\infty}  n[n(1+\beta) - 1 + \beta(1 - 2\alpha)]
        ~\frac{m^{n-1}}{(n-1)!}e^{-m}|a_n| \leq 2\beta(1 - \alpha).
\end{eqnarray*}
\noindent Since $f \in \mathscr{R}^{\tau}(A,B),$ then by Lemma \ref{tb1.6}, we have
\[
|a_n| \leq  \frac{(A-B)|\tau|}{n}.
\]
Therefore,
\begin{eqnarray*}
P_2 &\leq& \sum\limits_{n=2}^{\infty}  n[n(1+\beta) - 1 + \beta(1 - 2\alpha)]
            ~\frac{m^{n-1}}{(n-1)!}e^{-m} \frac{(A-B)|\tau|}{n} \\
    & = &  (A-B)|\tau|e^{-m}
            \sum\limits_{n=2}^{\infty}  [n(1+\beta) - 1 + \beta(1 - 2\alpha)]
            ~\frac{m^{n-1}}{(n-1)!}  \\
    & = &  (A-B)|\tau|e^{-m}
            \sum\limits_{n=2}^{\infty}  [(n-1)(1+\beta)+2\beta(1-\alpha)]
            ~\frac{m^{n-1}}{(n-1)!}  \\
    & = &   (A-B)|\tau|e^{-m}
            \left [
            \sum\limits_{n=2}^{\infty}  (1+\beta)~
            \frac{m^{n-1}}{(n-2)!}
            + 2\beta(1-\alpha)
            \sum\limits_{n=2}^{\infty} \frac{m^{n-1}}{(n-1)!}
            \right ] \\
    & = &   (A-B)|\tau|e^{-m}
            \left [ (1+\beta)
            \sum\limits_{n=2}^{\infty}  ~
            \frac{m^{n-1}}{(n-2)!}
            + 2\beta(1-\alpha)
            \sum\limits_{n=2}^{\infty} \frac{m^{n-1}}{(n-1)!}
            \right ]\\
    & = &   (A-B)|\tau|e^{-m}
            \left [ me^{m}(1+\beta)
              + 2\beta(1-\alpha)(e^{m} - 1)
            \right ].
\end{eqnarray*}
But this last expression is bounded by $2\beta(1-\alpha)$, if (\ref{Ge2}) holds. This completes the proof of Theorem \ref{Gt2}.
\end{proof}
\section{An Integral Operator}
\begin{theorem}\label{Gt1b}
If $m > 0,$ $0 \leq \alpha < 1$ and $0< \beta \leq 1,$ then $\mathscr{G}(m,z)=\int\limits_{0}^{z}\frac{\mathscr{F}(m,t)}{t}dt$
is in $\mathscr{C}^*(\alpha, \beta)$ if and only if inequality (\ref{Ge1a}) is satisfied.
\end{theorem}
\begin{proof}
Since
\begin{eqnarray*}
\mathscr{G}(m,z) &=& z - \sum\limits_{n=2}^{\infty}
            \frac{e^{-m}m^{n-1}}{(n-1)!} \frac{z^n}{n}
        =   z - \sum\limits_{n=2}^{\infty}
            \frac{e^{-m}m^{n-1}}{n!}z^n
\end{eqnarray*}
by Lemma \ref{tb1.2a}, we need only to show that
\begin{eqnarray*}
\sum\limits_{n=2}^{\infty}  n[n(1+\beta) - 1 + \beta(1 - 2\alpha)]
                            ~\frac{m^{n-1}}{n!}e^{-m} \leq 2\beta(1 - \alpha).
\end{eqnarray*}
Let
\begin{eqnarray*}
Q_1 = \sum\limits_{n=2}^{\infty}  n[n(1+\beta) - 1 + \beta(1 - 2\alpha)]
        ~\frac{m^{n-1}}{n!}e^{-m}.
\end{eqnarray*}
Now,
\begin{eqnarray*}
Q_1 &=& \sum\limits_{n=2}^{\infty}  [n(1+\beta) - 1 + \beta(1 - 2\alpha)]
            ~\frac{m^{n-1}}{(n-1)!}e^{-m}\\
    &=& e^{-m}\sum\limits_{n=2}^{\infty}  [(n-1)
    (1+\beta) + 2\beta(1 - \alpha)]~\frac{m^{n-1}}{(n-1)!}\\
    &=& e^{-m}
        \left[
            \sum\limits_{n=2}^{\infty}  (n-1)(1+\beta)~\frac{m^{n-1}}{(n-1)!}
            + \sum\limits_{n=2}^{\infty} 2\beta(1 - \alpha)~\frac{m^{n-1}}{(n-1)!}
        \right] \\
    &=& e^{-m}
        \left[
            (1+\beta)
            \sum\limits_{n=2}^{\infty} ~\frac{m^{n-1}}{(n-2)!}
            + 2\beta(1 - \alpha)\sum\limits_{n=2}^{\infty} ~\frac{m^{n-1}}{(n-1)!}
        \right ] \\
    &=& e^{-m}
        \left [
            (1+\beta)me^m
            + 2\beta(1 - \alpha)(e^m - 1)
        \right ] \\
    &=& (1+\beta)m + 2\beta(1 - \alpha)(1 - e^{-m}).
\end{eqnarray*}
But this last expression is bounded by $2\beta(1-\alpha)$, if and only if (\ref{Ge1a}) holds. This completes the proof of Theorem \ref{Gt1b}.
\end{proof}
\begin{theorem}\label{Gt2b}
If $m > 0,$ $0 \leq \alpha < 1$ and $0< \beta \leq 1,$ then $\mathscr{G}(m,z)=\int\limits_{0}^{z}\frac{\mathscr{F}(m,t)}{t}dt$
is in $\mathscr{S}^*(\alpha, \beta)$ if and only if
\begin{equation*}\label{Ge2b}
(1+\beta)(1-e^{-m}) + \frac{(\beta(1-2\alpha)-1)}{m}(1-e^{-m}-me^{-m}) \leq 2\beta(1- \alpha).
\end{equation*}
\end{theorem}
The proof of Theorem \ref{Gt2b} is lines similar to the proof of Theorem \ref{Gt1b}, so we omitted the proof of Theorem \ref{Gt2b}.

\section*{Acknowledgement}
The authors are thankful to the referee for his/her valuable comments and observations which helped in improving the paper.


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\end{thebibliography}

\end{document}