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\begin{document}
\title[]{CERTAIN \ CLASS OF ANALYTIC FUNCTIONS DEFINED BY $q-$ANALOGUE OF
RUSCHEWEYH DIFFERENTIAL OPERATOR }
\author{M. K. Aouf, A. O. Mostafa}
\author{F. Y. Al-Quhali}
\address{Department of Mathematics, Faculty of Science, Mansoura University,
Mansoura 35516, Egypt.}
\email{mkaouf127@yahoo.com\\
adelaeg254@yahoo.com\\
fyalquhali89@gmail.com}
\keywords{Analytic functions, Coefficient estimates, Distortion, $q-$%
Ruscheweyh type differential operator, Neighborhoods, Partial sums.\\
\textbf{2010 Mathematical Subject Classification:} 30C45.}

\begin{abstract}
In this paper, we obtain coefficient estimates, distortion theorems, radii
of close-to-convexity, starlikeness and convexity for functions belonging to
the class $\QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta )$ of analytic starlike
and convex functions defined by $q-$analogue of Ruscheweyh differential
operator. Also we find closure theorems, $N_{k,q,\delta }(e,g)$ neighborhood
and partial sums for functions in this class.
\end{abstract}

\maketitle

\section{Introduction}

Let $\mathcal{S}$ be the class of analytic and univalent functions of the
form:%
\begin{equation}
f(z)=z+\sum_{k=2}^{\infty }a_{k}z^{k}\ ,z\in \mathbb{U}=\left\{ z:z\in 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
:\left\vert z\right\vert <1\right\} .  \tag{1.1}
\end{equation}

Also let $\mathcal{S}^{\ast }(\alpha )$ \ and \ $C(\alpha )$ denote the
subclasses of $\mathcal{S}$ which are, respectively, starlike and convex
functions of order $\alpha (0\leq \alpha <1),$ satisfying (see Robertson
[30]) 
\begin{equation}
\mathcal{S}^{\ast }(\alpha )=\left\{ f:f\in \mathcal{S}\text{ \ and }\func{Re%
}\left( \frac{zf^{^{\prime }}(z)}{f(z)}\right) >\alpha \right\} ,  \tag{1.2}
\end{equation}%
and 
\begin{equation}
C(\alpha )=\left\{ f:f\in \mathcal{S}\text{ \ and }\func{Re}\left( 1+\frac{%
zf^{^{^{\prime \prime }}}(z)}{f^{^{\prime }}(z)}\right) >\alpha \right\} . 
\tag{1.3}
\end{equation}

It readily follows from (1.2) and (1.3) that 
\begin{equation*}
f(z)\in C(\alpha )\Leftrightarrow zf^{^{\prime }}(z)\in S^{\ast }(\alpha ).
\end{equation*}

For $0<q<1$ the Jackson's $q-$derivative of a function $f(z)\in \mathcal{S}$
\ is given by [22] (see also [2, 3, 8, 13, 17, 20, 24, 34, 35, 39])

\begin{equation}
D_{q}f(z)=\left\{ 
\begin{array}{c}
\frac{f(z)-f(qz)}{(1-q)z}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ \ \
z\neq 0, \\ 
f^{^{\prime }}(0)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for\ \
\ z=0,\ \ 
\end{array}%
\right. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \   \tag{1.4}
\end{equation}%
For $f(z)$ of the form (1.1), we have%
\begin{equation}
D_{q}f(z)=1+\sum_{k=2}^{\infty }\left[ k\right] _{q}a_{k}z^{k-1},  \tag{1.5}
\end{equation}%
where$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ $ 
\begin{equation}
\left[ n\right] _{q}=\frac{1-q^{n}}{1-q}\text{ \ \ }\medskip (0<q<1;\text{ }%
n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
=\left\{ 1,2,...\right\} ).  \tag{1.6}
\end{equation}%
Kanas and Raducanu [23] ( see also Aldweby and Darus [1]) defined the $q-$%
analogue of Ruscheweyh operator by%
\begin{equation}
R_{q}^{\lambda }f(z)=z+\sum_{k=2}^{\infty }\tfrac{\left[ k+\lambda -1\right]
_{q}!}{\left[ \lambda \right] _{q}!\left[ k-1\right] _{q}!}a_{k}z^{k}\text{
\ \ }(0<q<1;\lambda \geq 0),  \tag{1.7}
\end{equation}%
where%
\begin{equation}
\left[ n\right] _{q}!=\left\{ 
\begin{array}{c}
\left[ n\right] _{q}\left[ n-1\right] _{q}...\left[ 1\right] _{q},\text{ \ \
\ \ \ \ }n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
; \\ 
1,\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }n=0%
\end{array}%
\right. ,  \tag{1.8}
\end{equation}%
From (1.7) we obtain that 
\begin{equation*}
R_{q}^{0}f(z)=f(z)\text{ \ \ \ and }R_{q}^{1}f(z)=zD_{q}f(z),
\end{equation*}%
and 
\begin{equation}
\underset{q\longrightarrow 1^{-}}{\lim }R_{q}^{\lambda
}f(z)=z+\sum_{k=2}^{\infty }\frac{\left( k+\lambda -1\right) !}{\lambda
!\left( k-1\right) !}a_{k}z^{k}=R^{\lambda }f(z),  \tag{1.9}
\end{equation}%
where $R^{\lambda }$ is the Ruscheweyh differential operator ( see [32] and
[4, 7, 10, 14, 18]).

\begin{definition}
For $\ 0<q<1,0\leq \alpha <1,\beta \geq 0$ and $\lambda \geq 0,$ let \textsl{%
B}$_{q}^{\lambda }(\alpha ,\beta )$ be the class of functions $f\in \mathcal{%
S}$ satisfying%
\begin{equation}
\func{Re}\left\{ \frac{zD_{q}(R_{q}^{\lambda }f(z))}{R_{q}^{\lambda }f(z)}%
-\alpha \right\} >\beta \left\vert \frac{zD_{q}(R_{q}^{\lambda }f(z))}{%
R_{q}^{\lambda }f(z)}-1\right\vert .  \tag{1.10}
\end{equation}
\end{definition}

Let $\mathcal{T}\subset \mathcal{S}$ such that: 
\begin{equation}
\mathcal{T}=\left\{ f\in \mathcal{S}:f(z)=z-\sum_{k=2}^{\infty }a_{k}z^{k}\
,a_{k}\geq 0\right\} ,  \tag{1.11}
\end{equation}%
and 
\begin{equation}
\QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta )=\QTR{sl}{B}_{q}^{\lambda
}(\alpha ,\beta )\cap \mathcal{T}.  \tag{1.12}
\end{equation}%
Note that

$(i)$ \ \textsl{TB}$_{q}^{0}(\alpha ,\beta )=$\textsl{S}$_{p}^{q}(\alpha
,\beta )=\left\{ f\in T:\func{Re}\left\{ \frac{zD_{q}f(z)}{f(z)}-\alpha
\right\} >\beta \left\vert \frac{zD_{q}f(z)}{f(z)}-1\right\vert ,z\in 
\mathbb{U}\right\} ;$

$(ii)$ \ \textsl{TB}$_{q}^{0}(\alpha ,0)=$\textsl{TB}$_{q}(\alpha )=\left\{
f\in T:\func{Re}\left\{ \frac{zD_{q}f(z)}{f(z)}\right\} >\alpha \right\} ;$
\ 

$(iii)\underset{q\rightarrow 1^{-}}{\lim }$\textsl{TB}$_{q}^{0}(\alpha
,\beta )=\QTR{sl}{S}_{p}(\alpha ,\beta )=\left\{ f\in T:\func{Re}\left\{ 
\frac{zf^{^{\prime }}(z)}{f(z)}-\alpha \right\} >\beta \left\vert \frac{%
zf^{^{\prime }}(z)}{f(z)}-1\right\vert ,z\in \mathbb{U}\right\} $

(see [29] and [36]);

$(iv)$\textsl{TB}$_{q}^{1}(\alpha ,\beta )=\QTR{sl}{UCS}_{p}^{q}(\alpha
,\beta )=\left\{ f\in T:\func{Re}\left\{ \frac{D_{q}(zD_{q}f(z))}{D_{q}f(z)}%
-\alpha \right\} >\beta \left\vert \frac{D_{q}(zD_{q}f(z))}{D_{q}f(z)}%
-1\right\vert ,z\in \mathbb{U}\right\} ;$

$(v)$\textsl{TB}$_{q}^{1}(\alpha ,0)=C_{q}(\alpha )=\left\{ f\in T:\func{Re}%
\left\{ \frac{D_{q}(zD_{q}f(z))}{D_{q}f(z)}\right\} >\alpha ,z\in \mathbb{U}%
\right\} ;$

$(vi)\underset{q\rightarrow 1^{-}}{\lim }$\textsl{TB}$_{q}^{1}(\alpha ,\beta
)=\QTR{sl}{UCS}_{p}(\alpha ,\beta )=\left\{ f\in T:\func{Re}\left\{ 1+\frac{%
zf^{^{^{\prime \prime }}}(z)}{f^{^{\prime }}(z)}-\alpha \right\} >\beta
\left\vert \frac{zf^{^{^{\prime \prime }}}(z)}{f^{^{\prime }}(z)}\right\vert
,z\in \mathbb{U}\right\} $

(see [29]);

$(vii)$ \ $\underset{q\rightarrow 1^{-}}{\lim }$\textsl{TB}$_{q}^{\lambda
}(\alpha ,\beta )=\QTR{sl}{S}_{p}^{\lambda }(\alpha ,\beta )$ (see Rosy et
al. [31]).

\section{COEFFICIENT ESTIMATES}

Unless indicated, we assume that $0\leq \alpha <1,$ $\beta \geq 0,\lambda
\geq 0,$ $0<q<1$ and $f(z)\in \mathcal{T}.$

\begin{theorem}
A function $f(z)\in $\textsl{TB}$_{q}^{\lambda }(\alpha ,\beta )$ if and
only if%
\begin{equation}
\sum_{k=2}^{\infty }\left[ \left[ k\right] _{q}(1+\beta )-(\alpha +\beta )%
\right] \tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda \right] _{q}!%
\left[ k-1\right] _{q}!}a_{k}\leq 1-\alpha .  \tag{2.1}
\end{equation}
\end{theorem}

\begin{proof}
Assume that (2.1) holds. Then it is suffices to show that 
\begin{equation*}
\beta \left\vert \frac{zD_{q}(R_{q}^{\lambda }f(z))}{R_{q}^{\lambda }f(z)}%
-1\right\vert -\func{Re}\left\{ \frac{zD_{q}(R_{q}^{\lambda }f(z))}{%
R_{q}^{\lambda }f(z)}-1\right\} \leq 1-\alpha .
\end{equation*}%
We have%
\begin{eqnarray*}
&&\beta \left\vert \frac{zD_{q}(R_{q}^{\lambda }f(z))}{R_{q}^{\lambda }f(z)}%
-1\right\vert -\func{Re}\left\{ \frac{zD_{q}(R_{q}^{\lambda }f(z))}{%
R_{q}^{\lambda }f(z)}-1\right\} \\
&\leq &(1+\beta )\left\vert \frac{zD_{q}(R_{q}^{\lambda }f(z))}{%
R_{q}^{\lambda }f(z)}-1\right\vert \\
&\leq &\tfrac{(1+\beta )\underset{k=2}{\overset{\infty }{\sum }}\tfrac{\left[
k+\lambda -1\right] _{q}!}{\left[ \lambda \right] _{q}!\left[ k-1\right]
_{q}!}\left( \left[ k\right] _{q}-1\right) a_{k}}{1-\overset{\infty }{%
\underset{k=2}{\sum }}\tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[
\lambda \right] _{q}!\left[ k-1\right] _{q}!}a_{k}}.
\end{eqnarray*}%
This last expression is bounded above by $(1-\alpha )$ since (2.1) holds.

Conversely if $f(z)\in $\textsl{TB}$_{q}^{\lambda }(\alpha ,\beta )$ and $z$
is real, then 
\begin{equation*}
\func{Re}\left\{ \tfrac{1-\underset{k=2}{\overset{\infty }{\sum }}\tfrac{%
\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda \right] _{q}!\left[ k-1%
\right] _{q}!}\left[ k\right] _{q}a_{k}z^{k-1}}{1-\underset{k=2}{\overset{%
\infty }{\sum }}\tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda %
\right] _{q}!\left[ k-1\right] _{q}!}a_{k}z^{k-1}}-\alpha \right\} \geq
\beta \left\vert \tfrac{\underset{k=2}{\overset{\infty }{\sum }}\tfrac{\left[
k+\lambda -1\right] _{q}!}{\left[ \lambda \right] _{q}!\left[ k-1\right]
_{q}!}(\left[ k\right] _{q}-1)a_{k}z^{k-1}}{1-\underset{k=2}{\overset{\infty 
}{\sum }}\tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda \right]
_{q}!\left[ k-1\right] _{q}!}a_{k}z^{k-1}}\right\vert .
\end{equation*}%
Letting $z\rightarrow 1^{-}$ along the real axis, we obtain (2.1). Hence the
proof is completed.
\end{proof}

\begin{corollary}
For $f(z)\in \QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta ),$ 
\begin{equation}
a_{k}\leq \frac{1-\alpha }{\left[ \left[ k\right] _{q}(1+\beta )-(\alpha
+\beta )\right] \tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda %
\right] _{q}!\left[ k-1\right] _{q}!}}\ (k\geq 2)  \tag{2.2}
\end{equation}%
and 
\begin{equation}
f(z)=z-\frac{1-\alpha }{\left[ \left[ k\right] _{q}(1+\beta )-(\alpha +\beta
)\right] \tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda \right]
_{q}!\left[ k-1\right] _{q}!}}z^{k}\ (k\geq 2),  \tag{2.3}
\end{equation}%
gives the sharpness.
\end{corollary}

\textbf{Remark 1.} Letting $q\rightarrow 1^{-}$ in the results of Section 2,
we get the results of Section 2 for the class $\QTR{sl}{S}_{p}^{\lambda
}(\alpha ,\beta )$ studied by Rosy et al. [31].

\section{GROWTH AND DISTORTION THEOREMS}

\begin{theorem}
For $f(z)\in \QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta )$ and $\left\vert
z\right\vert =r<1,$ we have 
\begin{equation}
\left\vert f(z)\right\vert \geq r-\frac{1-\alpha }{\left[ \left[ 2\right]
_{q}(1+\beta )-(\alpha +\beta )\right] \left[ 1+\lambda \right] _{q}}r^{2}, 
\tag{3.1}
\end{equation}%
and%
\begin{equation}
\left\vert f(z)\right\vert \leq r+\frac{1-\alpha }{\left[ \left[ 2\right]
_{q}(1+\beta )-(\alpha +\beta )\right] \left[ 1+\lambda \right] _{q}}r^{2}. 
\tag{3.2}
\end{equation}%
Equalities hold for 
\begin{equation}
f(z)=z-\frac{1-\alpha }{\left[ \left[ 2\right] _{q}(1+\beta )-(\alpha +\beta
)\right] \left[ 1+\lambda \right] _{q}}z^{2},  \tag{3.3}
\end{equation}%
at $z=r$ and $z=re^{i(2k+1)\pi }$ $(k\geq 2).$
\end{theorem}

\begin{proof}
Since for $k\geq 2,$

\begin{eqnarray}
&&\left[ \left[ 2\right] _{q}(1+\beta )-(\alpha +\beta )\right] \left[
1+\lambda \right] _{q}\sum_{k=2}^{\infty }a_{k}  \notag \\
&\leq &\sum_{k=2}^{\infty }\left[ \left[ k\right] _{q}(1+\beta )-(\alpha
+\beta )\right] \tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda %
\right] _{q}!\left[ k-1\right] _{q}!}a_{k}\leq 1-\alpha ,  \TCItag{3.4}
\end{eqnarray}%
then 
\begin{equation}
\sum_{k=2}^{\infty }a_{k}\leq \frac{1-\alpha }{\left[ \left[ 2\right]
_{q}(1+\beta )-(\alpha +\beta )\right] \left[ 1+\lambda \right] _{q}}. 
\tag{3.5}
\end{equation}%
From (1.12) and (3.5), we have 
\begin{equation}
\left\vert f(z)\right\vert \geq r-r^{2}\sum_{k=2}^{\infty }a_{k}\geq r-\frac{%
1-\alpha }{\left[ \left[ 2\right] _{q}(1+\beta )-(\alpha +\beta )\right] %
\left[ 1+\lambda \right] _{q}}r^{2}  \tag{3.6}
\end{equation}%
and 
\begin{equation}
\left\vert f(z)\right\vert \leq r+r^{2}\sum_{k=2}^{\infty }a_{k}\leq r+\frac{%
1-\alpha }{\left[ \left[ 2\right] _{q}(1+\beta )-(\alpha +\beta )\right] %
\left[ 1+\lambda \right] _{q}}r^{2}.  \tag{3.7}
\end{equation}%
This completes the proof.
\end{proof}

Letting $q\rightarrow 1^{-}$ in Theorem 3.1, we have

\begin{corollary}
For $f(z)\in \QTR{sl}{S}_{p}^{\lambda }(\alpha ,\beta ),$ then 
\begin{equation}
\left\vert f(z)\right\vert \geq r-\frac{1-\alpha }{\left( 2+\beta -\alpha
\right) (1+\lambda )}r^{2},  \tag{3.8}
\end{equation}%
and%
\begin{equation}
\left\vert f(z)\right\vert \leq r+\frac{1-\alpha }{\left( 2+\beta -\alpha
\right) \left( 1+\lambda \right) }r^{2}.  \tag{3.9}
\end{equation}%
Equalities hold for 
\begin{equation}
f(z)=z-\frac{1-\alpha }{\left( 2+\beta -\alpha \right) \left( 1+\lambda
\right) }z^{2},  \tag{3.10}
\end{equation}%
at $z=r$ and $z=re^{i(2k+1)\pi }$ $(k\geq 2).$
\end{corollary}

\begin{proof}
Letting $q\rightarrow 1^{-}$ in Theorem 3.1, we can show (3.8) and (3.9).
\end{proof}

\begin{theorem}
Let \ $f(z)\in \QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta ).$ Then for $%
\left\vert z\right\vert =r<1,$ 
\begin{equation}
\left\vert f^{^{\prime }}(z)\right\vert \geq 1-\tfrac{2\left( 1-\alpha
\right) }{\left[ \left[ 2\right] _{q}(1+\beta )-(\alpha +\beta )\right] %
\left[ 1+\lambda \right] _{q}}r,  \tag{3.11}
\end{equation}%
and 
\begin{equation}
\left\vert f^{^{\prime }}(z)\right\vert \leq 1+\tfrac{2\left( 1-\alpha
\right) }{\left[ \left[ 2\right] _{q}(1+\beta )-(\alpha +\beta )\right] %
\left[ 1+\lambda \right] _{q}}r.  \tag{3.12}
\end{equation}%
\ The sharpness are attained for $f(z)$ given by (3.3).
\end{theorem}

\begin{proof}
For $k\geq 2,$ we have%
\begin{equation*}
\left\vert f^{^{\prime }}(z)\right\vert \leq 1-r\sum_{k=2}^{\infty }ka_{k}.
\end{equation*}%
We find from (2.1) and (3.5) that 
\begin{eqnarray*}
\left[ 2\right] _{q}(1+\beta )\left[ \lambda +1\right] _{q}\sum_{k=2}^{%
\infty }ka_{k} &\leq &2\left( 1-\alpha \right) +2(\alpha +\beta )\left[
\lambda +1\right] _{q}\underset{k=2}{\overset{\infty }{\sum }}a_{k} \\
&\leq &2\left( 1-\alpha \right) +\frac{2(\alpha +\beta )(1-\alpha )}{\left[ %
\left[ 2\right] _{q}(1+\beta )-(\alpha +\beta )\right] } \\
&\leq &\frac{2\left[ 2\right] _{q}(1+\beta )(1-\alpha )}{\left[ \left[ 2%
\right] _{q}(1+\beta )-(\alpha +\beta )\right] },
\end{eqnarray*}%
that is, that%
\begin{equation}
\sum_{k=2}^{\infty }ka_{k}\leq \frac{2(1-\alpha )}{\left[ \left[ 2\right]
_{q}(1+\beta )-(\alpha +\beta )\right] \left[ \lambda +1\right] _{q}}. 
\tag{3.13}
\end{equation}%
From (3.11) and (3.12) that 
\begin{equation}
\left\vert f^{^{\prime }}(z)\right\vert \geq 1-r\sum_{k=2}^{\infty
}ka_{k}\geq 1-\tfrac{2\left( 1-\alpha \right) }{\left[ \left[ 2\right]
_{q}(1+\beta )-(\alpha +\beta )\right] \left[ 1+\lambda \right] _{q}}r 
\tag{3.14}
\end{equation}%
and 
\begin{equation}
\left\vert f^{^{\prime }}(z)\right\vert \leq 1+r\sum_{k=2}^{\infty
}ka_{k}\leq 1+\tfrac{2\left( 1-\alpha \right) }{\left[ \left[ 2\right]
_{q}(1+\beta )-(\alpha +\beta )\right] \left[ 1+\lambda \right] _{q}}r. 
\tag{3.15}
\end{equation}%
This completes the proof.
\end{proof}

\begin{theorem}
For \ $f(z)\in \QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta )$ and $\left\vert
z\right\vert =r<1,$ 
\begin{equation}
\left\vert D_{q}f(z)\right\vert \geq 1-\tfrac{\left[ 2\right] _{q}\left(
1-\alpha \right) }{\left[ \left[ 2\right] _{q}(1+\beta )-(\alpha +\beta )%
\right] \left[ 1+\lambda \right] _{q}}r,  \tag{3.16}
\end{equation}%
and 
\begin{equation}
\left\vert D_{q}f(z)\right\vert \leq 1+\tfrac{\left[ 2\right] _{q}\left(
1-\alpha \right) }{\left[ \left[ 2\right] _{q}(1+\beta )-(\alpha +\beta )%
\right] \left[ 1+\lambda \right] _{q}}r.  \tag{3.17}
\end{equation}%
\ The sharpness are attained for $f(z)$ given by (3.3).
\end{theorem}

\begin{proof}
For $k\geq 2,$ we have%
\begin{equation*}
\left\vert D_{q}f(z)\right\vert \leq 1-r\sum_{k=2}^{\infty }\left[ k\right]
_{q}a_{k}.
\end{equation*}%
We find from (2.1) and (3.5) that 
\begin{eqnarray*}
(1+\beta )\left[ \lambda +1\right] _{q}\sum_{k=2}^{\infty }\left[ k\right]
_{q}a_{k} &\leq &\left( 1-\alpha \right) +(\alpha +\beta )\left[ \lambda +1%
\right] _{q}\underset{k=2}{\overset{\infty }{\sum }}a_{k} \\
&\leq &\left( 1-\alpha \right) +\frac{\left[ 2\right] _{q}(\alpha +\beta
)(1-\alpha )}{\left[ \left[ 2\right] _{q}(1+\beta )-(\alpha +\beta )\right] }
\\
&\leq &\frac{\left[ 2\right] _{q}(1+\beta )(1-\alpha )}{\left[ \left[ 2%
\right] _{q}(1+\beta )-(\alpha +\beta )\right] },
\end{eqnarray*}%
that is, that%
\begin{equation}
\sum_{k=2}^{\infty }\left[ k\right] _{q}a_{k}\leq \frac{\left[ 2\right]
_{q}(1-\alpha )}{\left[ \left[ 2\right] _{q}(1+\beta )-(\alpha +\beta )%
\right] \left[ \lambda +1\right] _{q}},  \tag{3.18}
\end{equation}%
From (3.16) and (3.17) that 
\begin{equation}
\left\vert D_{q}f(z)\right\vert \geq 1-r\sum_{k=2}^{\infty }\left[ k\right]
_{q}a_{k}\geq 1-\tfrac{\left[ 2\right] _{q}\left( 1-\alpha \right) }{\left[ %
\left[ 2\right] _{q}(1+\beta )-(\alpha +\beta )\right] \left[ 1+\lambda %
\right] _{q}}r  \tag{3.19}
\end{equation}%
and 
\begin{equation}
\left\vert D_{q}f(z)\right\vert \leq 1+r\sum_{k=2}^{\infty }\left[ k\right]
_{q}a_{k}\leq 1+\tfrac{\left[ 2\right] _{q}\left( 1-\alpha \right) }{\left[ %
\left[ 2\right] _{q}(1+\beta )-(\alpha +\beta )\right] \left[ 1+\lambda %
\right] _{q}}r.  \tag{3.20}
\end{equation}%
This completes the proof.
\end{proof}

Letting $q\rightarrow 1^{-}$ in Theorem 3.4, we have

\begin{corollary}
For $f(z)\in \QTR{sl}{S}_{p}^{\lambda }(\alpha ,\beta ),$ then 
\begin{equation}
\left\vert f^{^{\prime }}(z)\right\vert \geq 1-\tfrac{2\left( 1-\alpha
\right) }{\left( 2+\beta -\alpha \right) \left( 1+\lambda \right) }r, 
\tag{3.21}
\end{equation}%
and 
\begin{equation}
\left\vert f^{^{\prime }}(z)\right\vert \leq 1+\tfrac{2\left( 1-\alpha
\right) }{\left( 2+\beta -\alpha \right) \left( 1+\lambda \right) }r. 
\tag{3.22}
\end{equation}%
\ The sharpness are attained for $f(z)$ given by (3.10).
\end{corollary}

\begin{proof}
Letting $q\rightarrow 1^{-}$ in Theorem 3.4, we can show (3.21) and (3.22).
Then Corollary 3.5 corresponds to Theorem 3.3 when $q\rightarrow 1^{-}.$
\end{proof}

\section{CLOSURE THEOREMS}

Let $f_{j}(z)$ be defined, for $j=1,2,...,m,$ by%
\begin{equation}
f_{j}(z)=z-\sum_{k=2}^{\infty }a_{k,j}z^{k}\text{ \ \ }(a_{k,j}\geq 0,\text{ 
}z\in \mathbb{U)}.  \tag{4.1}
\end{equation}

\begin{theorem}
Let $f_{j}(z)\in \QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta )$ for $%
j=1,2,...,m.$ Then 
\begin{equation}
g(z)=\sum_{j=1}^{m}c_{j}f_{j}(z),  \tag{4.2}
\end{equation}%
is also in the same class, where $c_{j}\geq 0,$ $\overset{m}{\underset{j=1}{%
\sum }}c_{j}=1.$
\end{theorem}

\begin{proof}
According to (4.2), we can write 
\begin{equation}
g(z)=z-\sum_{k=2}^{\infty }\left( \sum_{j=1}^{m}c_{j}a_{k,j}\right) z^{k}. 
\tag{4.3}
\end{equation}%
Further, since $f_{j}(z)\in \QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta ),$ we
get%
\begin{equation}
\sum_{k=2}^{\infty }\left[ \left[ k\right] _{q}(1+\beta )-(\alpha +\beta )%
\right] \tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda \right] _{q}!%
\left[ k-1\right] _{q}!}a_{k,j}\leq 1-\alpha .  \tag{4.4}
\end{equation}%
Hence 
\begin{eqnarray}
&&\sum_{k=2}^{\infty }\left[ \left[ k\right] _{q}(1+\beta )-(\alpha +\beta )%
\right] \tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda \right] _{q}!%
\left[ k-1\right] _{q}!}\left( \sum_{j=1}^{m}c_{j}a_{k,j}\right)  \notag \\
&=&\sum_{j=1}^{m}c_{j}\left[ \sum_{k=2}^{\infty }\left[ \left[ k\right]
_{q}(1+\beta )-(\alpha +\beta )\right] \tfrac{\left[ k+\lambda -1\right]
_{q}!}{\left[ \lambda \right] _{q}!\left[ k-1\right] _{q}!}a_{k,j}\right] 
\notag \\
&\leq &\left( \sum_{j=1}^{m}c_{j}\right) \left( 1-\alpha \right) =1-\alpha ,
\TCItag{4.5}
\end{eqnarray}%
which implies that $g(z)\in \QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta ).$
Thus we have the theorem.
\end{proof}

\begin{corollary}
The class $\QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta )$ is closed under
convex linear combination.
\end{corollary}

\begin{proof}
Let $f_{j}(z)\in \QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta )$ $(j=1,2)$ and 
\begin{equation}
g(z)=\mu f_{1}(z)+(1-\mu )f_{2}(z)\ \ \ (0\leq \mu \leq 1),  \tag{4.6}
\end{equation}%
Then by, taking $m=2,$ $c_{1}=\mu $ and $c_{2}=1-\mu $ in Theorem 5, we have 
$g(z)\in \QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta ).$
\end{proof}

\begin{theorem}
Let $f_{1}(z)=z$ and 
\begin{equation}
f_{k}(z)=z-\tfrac{1-\alpha }{\left[ \left[ k\right] _{q}(1+\beta )-(\alpha
+\beta )\right] \tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda %
\right] _{q}!\left[ k-1\right] _{q}!}}z^{k}\text{ \ \ }(k\geq 2).  \tag{4.7}
\end{equation}%
Then $f(z)\in \QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta )$ if and only if 
\begin{equation}
f(z)=\sum_{k=1}^{\infty }\mu _{k}f_{k}(z),  \tag{4.8}
\end{equation}%
where $\mu _{k}\geq 0$ $(k\geq 1)$ and $\overset{\infty }{\underset{k=1}{%
\sum }}\mu _{k}=1.$
\end{theorem}

\begin{proof}
Suppose that 
\begin{equation}
f(z)=\sum_{k=1}^{\infty }\mu _{k}f_{k}(z)=z-\sum_{k=2}^{\infty }\tfrac{%
1-\alpha }{\left[ \left[ k\right] _{q}(1+\beta )-(\alpha +\beta )\right] 
\tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda \right] _{q}!\left[
k-1\right] _{q}!}}\mu _{k}z^{k}.  \tag{4.9}
\end{equation}%
Then it follows that 
\begin{eqnarray}
&&\sum_{k=2}^{\infty }\tfrac{\left[ \left[ k\right] _{q}(1+\beta )-(\alpha
+\beta )\right] \tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda %
\right] _{q}!\left[ k-1\right] _{q}!}}{1-\alpha }\cdot \tfrac{1-\alpha }{%
\left[ \left[ k\right] _{q}(1+\beta )-(\alpha +\beta )\right] \tfrac{\left[
k+\lambda -1\right] _{q}!}{\left[ \lambda \right] _{q}!\left[ k-1\right]
_{q}!}}\mu _{k}  \notag \\
&=&\sum_{k=2}^{\infty }\mu _{k}=1-\mu _{1}\leq 1.  \TCItag{4.10}
\end{eqnarray}%
So by Theorem 2.1, $f(z)\in \QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta ).$

Conversely, assume that $f(z)\in \QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta
). $ Then 
\begin{equation}
a_{k}\leq \tfrac{1-\alpha }{\left[ \left[ k\right] _{q}(1+\beta )-(\alpha
+\beta )\right] \tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda %
\right] _{q}!\left[ k-1\right] _{q}!}}\ \ (k\geq 2).  \tag{4.11}
\end{equation}%
Setting 
\begin{equation}
\mu _{k}=\tfrac{\left[ \left[ k\right] _{q}(1+\beta )-(\alpha +\beta )\right]
\tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda \right] _{q}!\left[
k-1\right] _{q}!}}{1-\alpha }a_{k}\text{ \ \ }(k\geq 2),  \tag{4.12}
\end{equation}%
and 
\begin{equation}
\mu _{1}=1-\sum_{k=2}^{\infty }\mu _{k},  \tag{4.13}
\end{equation}%
we see that $f(z)$ can be expressed in the form (4.8). This completes the
proof.
\end{proof}

\begin{corollary}
The extreme points of $\QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta )$ are $%
f_{k}(z)$ $(k\geq 1)$ given by Theorem 4.3.
\end{corollary}

\section{SOME RADII OF THE CLASS $\QTR{sl}{TB}_{q}^{\protect\lambda }(%
\protect\alpha ,\protect\beta )$}

\begin{theorem}
Let $f(z)\in \QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta ).$ Then for $0\leq
\rho <1,k\geq 2,f(z)$ is

\begin{itemize}
\item[(i)] close -to- convex of order $\rho $ in $\left\vert z\right\vert
<r_{1},$ where%
\begin{equation}
r_{1}=r_{1}(q,\alpha ,\beta ,\lambda ,\rho ):=\underset{k}{\inf }\left[ 
\tfrac{(1-\rho )\left[ \left[ k\right] _{q}(1+\beta )-(\alpha +\beta )\right]
\tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda \right] _{q}!\left[
k-1\right] _{q}!}}{k(1-\alpha )}\right] ^{\frac{1}{(k-1)}}.  \tag{5.1}
\end{equation}%
\ 

\item[(ii)] starlike of order $\rho $ in $\left\vert z\right\vert <r_{2},$
where%
\begin{equation}
r_{2}=r_{2}(q,\alpha ,\beta ,\lambda ,\rho ):=\underset{k}{\inf }\left[ 
\tfrac{(1-\rho )\left[ \left[ k\right] _{q}(1+\beta )-(\alpha +\beta )\right]
\tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda \right] _{q}!\left[
k-1\right] _{q}!}}{(k-\rho )(1-\alpha )}\right] ^{\frac{1}{(k-1)}}. 
\tag{5.2}
\end{equation}

\item[(iii)] convex of order $\rho $ in $\left\vert z\right\vert <r_{3},$
where 
\begin{equation}
r_{3}=r_{3}(q,\alpha ,\beta ,\lambda ,\rho ):=\underset{k}{\inf }\left[ 
\tfrac{(1-\rho )\left[ \left[ k\right] _{q}(1+\beta )-(\alpha +\beta )\right]
\tfrac{\left[ k+\lambda -1\right] _{q}!}{\left[ \lambda \right] _{q}!\left[
k-1\right] _{q}!}}{k(k-\rho )(1-\alpha )}\right] ^{\frac{1}{(k-1)}}. 
\tag{5.3}
\end{equation}%
The result is sharp for $f(z)$ is given by (2.3).
\end{itemize}
\end{theorem}

\begin{proof}
To prove (i) we must show that$\ \ \ \ $%
\begin{equation*}
\left\vert f^{^{\prime }}(z)-1\right\vert \leq 1-\rho \ \ for\ \left\vert
z\right\vert <r_{1}(q,\alpha ,\beta ,\rho ).
\end{equation*}%
From (1.12), we have 
\begin{equation*}
\left\vert f^{^{\prime }}(z)-1\right\vert \leq \sum_{k=2}^{\infty
}ka_{k}\left\vert z\right\vert ^{k-1}.
\end{equation*}%
Thus 
\begin{equation*}
\left\vert f^{^{\prime }}(z)-1\right\vert \leq 1-\rho ,
\end{equation*}%
if 
\begin{equation}
\sum_{k=2}^{\infty }\left( \frac{k}{1-\rho }\right) a_{k}\left\vert
z\right\vert ^{k-1}\leq 1.  \tag{5.4}
\end{equation}%
But, by Theorem 2.1, (5.4) will be true if 
\begin{equation*}
\left( \frac{k}{1-\rho }\right) \left\vert z\right\vert ^{k-1}\leq \tfrac{%
\left[ \left[ k\right] _{q}(1+\beta )-(\alpha +\beta )\right] \tfrac{\left[
k+\lambda -1\right] _{q}!}{\left[ \lambda \right] _{q}!\left[ k-1\right]
_{q}!}}{1-\alpha },
\end{equation*}%
that is, if 
\begin{equation}
\left\vert z\right\vert \leq \left[ \tfrac{(1-\rho )\left[ \left[ k\right]
_{q}(1+\beta )-(\alpha +\beta )\right] \tfrac{\left[ k+\lambda -1\right]
_{q}!}{\left[ \lambda \right] _{q}!\left[ k-1\right] _{q}!}}{k(1-\alpha )}%
\right] ^{\frac{1}{(k-1)}}\text{ \ \ }(k\geq 2),  \tag{5.5}
\end{equation}%
which gives (5.1).

To prove (ii) and (iii) it is suffices to show 
\begin{equation}
\left\vert \frac{zf^{^{\prime }}(z)}{f(z)}-1\right\vert \leq 1-\rho \text{ \
for }\left\vert z\right\vert <r_{2},  \tag{5.6}
\end{equation}%
\begin{equation}
\left\vert \frac{zf^{^{^{\prime \prime }}}(z)}{f^{^{\prime }}(z)}\right\vert
\leq 1-\rho \text{ \ for }\left\vert z\right\vert <r_{3},  \tag{5.7}
\end{equation}%
respectively, by using arguments as in proving (i), we have the results.
\end{proof}

\section{INCLUSION RELATIONS INVOLVING $N_{k,q,\protect\delta }(e)$}

In this section following the works of Goodman [21] and Ruscheweyh [33] \ (
see also [5], [6], [9], [16], [26] and [28]) defined the $k,\delta $
neighborhood of function $f(z)\in T$ \ by 
\begin{equation}
N_{k,\delta }(f;g)=\left\{ g\in T:g(z)=z-\underset{k=2}{\overset{\infty }{%
\sum }}b_{k}z^{k}\text{ \ \ and \ }\underset{k=2}{\overset{\infty }{\sum }}%
k\left\vert a_{k}-b_{k}\right\vert \leq \delta \text{ }\right\} .  \tag{6.1}
\end{equation}%
In particular, for the identity function $e(z)=z,$ we have 
\begin{equation}
N_{k,\delta }(e;g)=\left\{ g\in T:g(z)=z-\underset{k=2}{\overset{\infty }{%
\sum }}b_{k}z^{k}\text{ \ \ and \ }\underset{k=2}{\overset{\infty }{\sum }}%
k\left\vert b_{k}\right\vert \leq \delta \text{ }\right\} .  \tag{6.2}
\end{equation}%
Aouf et al. [12] defined the $k,q,\delta $ neighborhood of function $f(z)\in
T$ \ by%
\begin{equation}
N_{k,q,\delta }(f;g)=\left\{ g\in T:g(z)=z-\underset{k=2}{\overset{\infty }{%
\sum }}b_{k}z^{k}\text{ \ \ and }\underset{k=2}{\overset{\infty }{\sum }}%
\left[ k\right] _{q}\left\vert a_{k}-b_{k}\right\vert \leq \delta _{q}\text{ 
}\right\} .  \tag{6.3}
\end{equation}%
In particular, for the identity function $e(z)=z,$ we have

\begin{equation}
N_{k,q,\delta }(e;g)=\left\{ g\in T:g(z)=z-\underset{k=2}{\overset{\infty }{%
\sum }}b_{k}z^{k}\text{ \ \ and \ }\underset{k=2}{\overset{\infty }{\sum }}%
\left[ k\right] _{q}\left\vert b_{k}\right\vert \leq \delta _{q}\text{ }%
\right\} .  \tag{6.4}
\end{equation}

\begin{theorem}
Let 
\begin{equation}
\delta _{q}=\tfrac{\left( 1-\alpha \right) }{\left[ \left[ 2\right]
_{q}(1+\beta )-(\alpha +\beta )\right] \left[ \lambda +1\right] _{q}}. 
\tag{6.5}
\end{equation}%
Then \textsl{TB}$_{q}^{\lambda }(\alpha ,\beta )\subset N_{k,q,\delta }(e).$
\end{theorem}

\begin{proof}
For $f\in \QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta ),$ Theorem 2.1, (3.5)
and (3.18), and in view of the (6.4), Theorem 6.1 follows.
\end{proof}

A function $f\in T$ is in the class \textsl{TB}$_{q}^{\lambda }(\alpha
,\beta ,\xi )$ if there exists a function $g\in $\textsl{TB}$_{q}^{\lambda
}(\alpha ,\beta )$ such that 
\begin{equation}
\left\vert \frac{f(z)}{g(z)}-1\right\vert <1-\xi _{q}\text{ \ }(z\in \mathbb{%
U},\text{ }0\leq \xi _{q}<1).  \tag{6.6}
\end{equation}%
Now we determine the neighborhood for the class \textsl{TB}$_{q}^{\lambda
}(\alpha ,\beta ,\xi )$.

\begin{theorem}
If $g\in $\textsl{TB}$_{q}^{\lambda }(\alpha ,\beta )$ and%
\begin{equation}
\xi _{q}=1-\tfrac{\delta _{q}\left[ \left[ 2\right] _{q}(1+\beta )-(\alpha
+\beta )\right] \left[ \lambda +1\right] _{q}}{2\left\{ \left[ \left[ 2%
\right] _{q}(1+\beta )-(\alpha +\beta )\right] \left[ \lambda +1\right]
_{q}-(1-\alpha )\right\} },  \tag{6.7}
\end{equation}%
where $\delta _{q}\leq \frac{2\left\{ \left[ \left[ 2\right] _{q}(1+\beta
)-(\alpha +\beta )\right] \left[ \lambda +1\right] _{q}-(1-\alpha )\right\} 
}{\left[ \left[ 2\right] _{q}(1+\beta )-(\alpha +\beta )\right] \left[
\lambda +1\right] _{q}}.$Then $N_{k,q,\delta }(g)\subset \QTR{sl}{TB}%
_{q}^{\lambda }(\alpha ,\beta ,\xi ).$
\end{theorem}

\begin{proof}
Suppose that $f\in N_{k,q,\delta }(g)$ then 
\begin{equation*}
\underset{k=2}{\overset{\infty }{\sum }}\left[ k\right] _{q}\left\vert
a_{k}-b_{k}\right\vert \leq \delta _{q},
\end{equation*}%
where $\delta _{q}$ is given by (6.5), which implies that the coefficient
inequality 
\begin{equation*}
\underset{k=2}{\overset{\infty }{\sum }}\left\vert a_{k}-b_{k}\right\vert
\leq \frac{\delta _{q}}{\left[ 2\right] _{q}}.
\end{equation*}%
Next, since $g\in \QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta ),$ we have 
\begin{equation*}
\underset{k=2}{\overset{\infty }{\sum }}b_{k}\leq \tfrac{1-\alpha }{\left[ %
\left[ 2\right] _{q}(1+\beta )-(\alpha +\beta )\right] \left[ \lambda +1%
\right] _{q}},
\end{equation*}%
so that 
\begin{equation*}
\left\vert \tfrac{f(z)}{g(z)}-1\right\vert <\tfrac{\underset{k=2}{\overset{%
\infty }{\sum }}\left\vert a_{k}-b_{k}\right\vert }{1-\underset{k=2}{\overset%
{\infty }{\sum }}b_{k}}\leq \frac{\delta _{q}}{\left[ 2\right] _{q}}\times 
\tfrac{\left[ \left[ 2\right] _{q}(1+\beta )-(\alpha +\beta )\right] \left[
\lambda +1\right] _{q}}{\left[ \left[ 2\right] _{q}(1+\beta )-(\alpha +\beta
)\right] \left[ \lambda +1\right] _{q}-\left( 1-\alpha \right) }\leq 1-\xi
_{q}.
\end{equation*}%
Provided that $\xi _{q}$ is given precisely by (6.7). Thus, by definition, $%
f\in \QTR{sl}{TB}_{q}^{\lambda }(\alpha ,\beta ,\xi ),$ which completes the
proof.
\end{proof}

\section{PARTIAL SUMS}

For $f(z)$ of the form (1.1), the sequence of partial sums is given by 
\begin{equation*}
f_{m}(z)=z+\sum_{k=2}^{m}a_{k}z^{k}\text{ \ }(m\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
\backslash \left\{ 1\right\} ).
\end{equation*}

Now following the work of [38] and also the works cited in [11, 15, 19, 25,
27, 31, 37] on partial sums of analytic functions, to obtain our results.
Let 
\begin{equation}
\Phi _{q,k}^{\lambda }=\Phi _{q}^{\lambda }(k,\alpha ,\beta )=\left[ \left[ k%
\right] _{q}(1+\beta )-(\alpha +\beta )\right] \tfrac{\left[ k+\lambda -1%
\right] _{q}!}{\left[ \lambda \right] _{q}!\left[ k-1\right] _{q}!}. 
\tag{7.1}
\end{equation}

\begin{theorem}
If $f\in \mathcal{S}$, satisfies the condition (2.1), then 
\begin{equation}
\func{Re}\left( \frac{f(z)}{f_{m}(z)}\right) \geq \frac{\Phi
_{q,m+1}^{\lambda }-1+\alpha }{\Phi _{q,m+1}^{\lambda }},  \tag{7.2}
\end{equation}%
\newline
where 
\begin{equation}
\Phi _{q,k}^{\lambda }\geq \left\{ 
\begin{array}{c}
1-\alpha ,\text{ \ \ \ \ \ \ \ \ \ \ \ \ }if\text{ \ }k=2,3,...,m\text{\ \ \
\ \ \ \ \ } \\ 
\Phi _{q,m+1}^{\lambda },\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }if\text{ \ }%
k=m+1,m+2,...\text{ }.%
\end{array}%
\right.  \tag{7.3}
\end{equation}%
The result (7.2) is sharp for 
\begin{equation}
f(z)=z+\frac{1-\alpha }{\Phi _{q,m+1}^{\lambda }}z^{m+1}.  \tag{7.4}
\end{equation}
\end{theorem}

\begin{proof}
Define $g(z)$ by 
\begin{equation}
\tfrac{1+g(z)}{1-g(z)}=\tfrac{\Phi _{q,m+1}^{\lambda }}{1-\alpha }\left[ 
\tfrac{f(z)}{f_{m}(z)}-\tfrac{\Phi _{q,m+1}^{\lambda }-1+\alpha }{\Phi
_{q,m+1}^{\lambda }}\right] =\tfrac{1+\underset{k=2}{\overset{m}{\sum }}%
a_{k}z^{k-1}+\left( \tfrac{\Phi _{q,m+1}^{\lambda }}{1-\alpha }\right) 
\underset{k=m+1}{\overset{\infty }{\sum }}a_{k}z^{k-1}}{1+\underset{k=2}{%
\overset{m}{\sum }}a_{k}z^{k-1}}.  \tag{7.5}
\end{equation}

It suffices to show that $\left\vert g(z)\right\vert \leq 1.$ Now from (7.5)
we have 
\begin{equation*}
g(z)=\tfrac{\left( \frac{\Phi _{q,m+1}^{\lambda }}{1-\alpha }\right) 
\underset{k=m+1}{\overset{\infty }{\sum }}a_{k}z^{k-1}}{2+\underset{k=2}{%
\overset{m}{2\sum }}a_{k}z^{k-1}+\left( \frac{\Phi _{q,m+1}^{\lambda }}{%
1-\alpha }\right) \underset{k=m+1}{\overset{\infty }{\sum }}a_{k}z^{k-1}}.
\end{equation*}

Hence we obtain 
\begin{equation*}
\left\vert g(z)\right\vert \leq \tfrac{\left( \frac{\Phi _{q,m+1}^{\lambda }%
}{1-\alpha }\right) \underset{k=m+1}{\overset{\infty }{\sum }}\left\vert
a_{k}\right\vert }{2-\underset{k=2}{\overset{m}{2\sum }}\left\vert
a_{k}\right\vert -\left( \frac{\Phi _{q,m+1}^{\lambda }}{1-\alpha }\right) 
\underset{k=m+1}{\overset{\infty }{\sum }}\left\vert a_{k}\right\vert }.
\end{equation*}

Now $\left\vert g(z)\right\vert \leq 1$ if and only if 
\begin{equation*}
2\left( \frac{\Phi _{q,m+1}^{\lambda }}{1-\alpha }\right) \underset{k=m+1}{%
\overset{\infty }{\sum }}\left\vert a_{k}\right\vert \leq 2-\underset{k=2}{%
\overset{m}{2\sum }}\left\vert a_{k}\right\vert ,
\end{equation*}%
or, equivalently, 
\begin{equation*}
\underset{k=2}{\overset{m}{\sum }}\left\vert a_{k}\right\vert +\underset{%
k=m+1}{\overset{\infty }{\sum }}\frac{\Phi _{q,m+1}^{\lambda }}{1-\alpha }%
\left\vert a_{k}\right\vert \leq 1.
\end{equation*}%
From (2.1), it is sufficient to show that 
\begin{equation*}
\underset{k=2}{\overset{m}{\sum }}\left\vert a_{k}\right\vert +\underset{%
k=m+1}{\overset{\infty }{\sum }}\frac{\Phi _{q,m+1}^{\lambda }}{1-\alpha }%
\left\vert a_{k}\right\vert \leq \underset{k=2}{\overset{\infty }{\sum }}%
\frac{\Phi _{q,k}^{\lambda }}{1-\alpha }\left\vert a_{k}\right\vert ,
\end{equation*}%
which is equivalent to 
\begin{equation}
\underset{k=2}{\overset{m}{\sum }}\left( \tfrac{\Phi _{q,k}^{\lambda
}-1+\alpha }{1-\alpha }\right) \left\vert a_{k}\right\vert +\underset{k=m+1}{%
\overset{\infty }{\sum }}\left( \tfrac{\Phi _{q,k}^{\lambda }-\Phi
_{q,m+1}^{\lambda }}{1-\alpha }\right) \left\vert a_{k}\right\vert \geq 0. 
\tag{7.6}
\end{equation}%
For $z=re^{i\pi \diagup m}$ we have

\begin{equation*}
\tfrac{f(z)}{f_{m}(z)}=1+\tfrac{1-\alpha }{\Phi _{q,m+1}^{\lambda }}%
z^{k}\rightarrow 1-\tfrac{1-\alpha }{\Phi _{q,m+1}^{\lambda }}=\tfrac{\Phi
_{q,m+1}^{\lambda }-1+\alpha }{\Phi _{q,m+1}^{\lambda }}\text{ \ where }%
r\rightarrow 1^{-},
\end{equation*}%
which shows that $f(z)$ is given by (7.4) gives the sharpness.
\end{proof}

\bigskip \textbf{Remark 2. }$(i)$ \ Putting $\lambda =0$ and $(ii)$ $\lambda
=1$ in Theorem 7.1, we obtain the following results, respectively.

\begin{corollary}
If $f\in \mathcal{S}$, satisfies the condition (2.1) and $\frac{f(z)}{z}\neq
0(0<\left\vert z\right\vert <1),$ then%
\begin{equation}
\func{Re}\left( \frac{f(z)}{f_{m}(z)}\right) \geq \tfrac{\left[ \left[ m+1%
\right] _{q}(1+\beta )-(\alpha +\beta )\right] -1+\alpha }{\left[ \left[ m+1%
\right] _{q}(1+\beta )-(\alpha +\beta )\right] }.  \tag{7.7}
\end{equation}%
The result is sharp for 
\begin{equation}
f(z)=z+\tfrac{1-\alpha }{\left[ \left[ m+1\right] _{q}(1+\beta )-(\alpha
+\beta )\right] }z^{m+1}.  \tag{7.8}
\end{equation}
\end{corollary}

\begin{corollary}
If $f\in \mathcal{S}$, satisfies the condition (2.1) and $\frac{f(z)}{z}\neq
0(0<\left\vert z\right\vert <1),$ then%
\begin{equation}
\func{Re}\left( \frac{f(z)}{f_{m}(z)}\right) \geq 1-\tfrac{1-\alpha }{\left[
m+1\right] _{q}\left[ \left[ m+1\right] _{q}(1+\beta )-(\alpha +\beta )%
\right] }.  \tag{7.9}
\end{equation}%
The result is sharp for 
\begin{equation}
f(z)=z+\tfrac{1-\alpha }{\left[ m+1\right] _{q}\left[ \left[ m+1\right]
_{q}(1+\beta )-(\alpha +\beta )\right] }z^{m+1}.  \tag{7.10}
\end{equation}
\end{corollary}

\begin{theorem}
If $f\in \mathcal{S}$, satisfies the condition (2.1), then 
\begin{equation}
\func{Re}\left( \frac{f_{m}(z)}{f(z)}\right) \geq \tfrac{\Phi
_{q,m+1}^{\lambda }}{\Phi _{q,m+1}^{\lambda }+1-\alpha },  \tag{7.11}
\end{equation}%
\newline
where $\Phi _{q,m+1}^{\lambda }$ is defined by (7.1) and satisfies (7.3) and 
$f(z)$ given by (7.4) gives the sharpness.
\end{theorem}

\begin{proof}
The proof follows by defining 
\begin{equation*}
\frac{1+g(z)}{1-g(z)}=\tfrac{\Phi _{q,m+1}^{\lambda }+1-\alpha }{1-\alpha }%
\left[ \frac{f_{m}(z)}{f(z)}-\tfrac{\Phi _{q,m+1}^{\lambda }}{\Phi
_{q,m+1}^{\lambda }+1-\alpha }\right]
\end{equation*}%
and much akin are to similar arguments in Theorem 7.1. So, we omit it.
\end{proof}

\bigskip \textbf{Remark 3. }$(i)$ \ Putting $\lambda =0$ and $(ii)$ $\lambda
=1$ in Theorem 7.4, we obtain the following sharp results, respectively.

\begin{corollary}
If $f\in \mathcal{S}$, satisfies the condition (2.1) and $\frac{f(z)}{z}\neq
0(0<\left\vert z\right\vert <1),$ then%
\begin{equation}
\func{Re}\left( \frac{f_{m}(z)}{f(z)}\right) \geq \tfrac{\left[ m+1\right]
_{q}(1+\beta )-(\alpha +\beta )}{\left[ m+1\right] _{q}(1+\beta )-(\alpha
+\beta )+1-\alpha }.  \tag{7.12}
\end{equation}%
\newline
\end{corollary}

\begin{corollary}
If $f\in \mathcal{S}$, satisfies the condition (2.1) and $\frac{f(z)}{z}\neq
0(0<\left\vert z\right\vert <1),$ then%
\begin{equation}
\func{Re}\left( \frac{f_{m}(z)}{f(z)}\right) \geq \tfrac{\left[ m+1\right]
_{q}\left[ \left[ m+1\right] _{q}(1+\beta )-(\alpha +\beta )\right] }{\left[
m+1\right] _{q}\left[ \left[ m+1\right] _{q}(1+\beta )-(\alpha +\beta )%
\right] +1-\alpha }.  \tag{7.13}
\end{equation}%
\newline
\end{corollary}

\begin{theorem}
If $f\in \mathcal{S}$, satisfies the condition (2.1), then 
\begin{equation}
\func{Re}\left( \frac{f^{^{\prime }}(z)}{f_{m}^{^{\prime }}(z)}\right) \geq 
\tfrac{\Phi _{q,m+1}^{\lambda }-\left( m+1\right) \left( 1-\alpha \right) }{%
\Phi _{q,m+1}^{\lambda }},  \tag{7.14}
\end{equation}%
\newline
and 
\begin{equation}
\func{Re}\left( \frac{f_{m}^{^{\prime }}(z)}{f^{^{\prime }}(z)}\right) \geq 
\tfrac{\Phi _{q,m+1}^{\lambda }}{\Phi _{q,m+1}^{\lambda }+\left( m+1\right)
\left( 1-\alpha \right) },  \tag{7.15}
\end{equation}%
where $\Phi _{q,m+1}^{\lambda }\geq \left( m+1\right) \left( 1-\alpha
\right) $ and 
\begin{equation}
\Phi _{q,k}^{\lambda }\geq \left\{ 
\begin{array}{c}
k\left( 1-\alpha \right) ,\text{ \ \ \ \ \ \ \ \ \ \ \ \ }if\text{ \ }%
k=2,3,...,m\text{\ \ \ \ \ \ \ \ } \\ 
k\left( \frac{\Phi _{q,m+1}^{\lambda }}{\left( m+1\right) }\right) ,\text{ \
\ \ \ \ \ \ \ \ \ \ \ \ \ }if\text{ \ }k=m+1,m+2,...\text{ }.%
\end{array}%
\right.  \tag{7.16}
\end{equation}%
$f(z)$ is given by (7.4) gives the sharpness.
\end{theorem}

\begin{proof}
We write 
\begin{equation*}
\frac{1+g(z)}{1-g(z)}=\tfrac{\Phi _{q,m+1}^{\lambda }}{\left( m+1\right)
\left( 1-\alpha \right) }\left[ \frac{f^{^{\prime }}(z)}{f_{m}^{^{\prime
}}(z)}-\left( \tfrac{\Phi _{q,m+1}^{\lambda }-\left( m+1\right) \left(
1-\alpha \right) }{\Phi _{q,m+1}^{\lambda }}\right) \right] ,
\end{equation*}

where 
\begin{equation*}
g(z)=\tfrac{\left( \frac{\Phi _{q,m+1}^{\lambda }}{\left( m+1\right) \left(
1-\alpha \right) }\right) \overset{\infty }{\underset{k=m+1}{\sum }}%
ka_{k}z^{k-1}}{2+\underset{k=2}{\overset{m}{2\sum }}ka_{k}z^{k-1}+\left( 
\frac{\Phi _{q,m+1}^{\lambda }}{\left( m+1\right) \left( 1-\alpha \right) }%
\right) \underset{k=m+1}{\overset{\infty }{\sum }}ka_{k}z^{k-1}}.
\end{equation*}

Now $\left\vert g(z)\right\vert \leq 1$ if and only if 
\begin{equation*}
\underset{k=2}{\overset{m}{\sum }}k\left\vert a_{k}\right\vert +\left( 
\tfrac{\Phi _{q,m+1}^{\lambda }}{\left( m+1\right) \left( 1-\alpha \right) }%
\right) \underset{k=m+1}{\overset{\infty }{\sum }}k\left\vert
a_{k}\right\vert \leq 1.
\end{equation*}%
From (2.1), it is sufficient to show that 
\begin{equation*}
\underset{k=2}{\overset{m}{\sum }}k\left\vert a_{k}\right\vert +\left( 
\tfrac{\Phi _{q,m+1}^{\lambda }}{\left( m+1\right) \left( 1-\alpha \right) }%
\right) \underset{k=m+1}{\overset{\infty }{\sum }}k\left\vert
a_{k}\right\vert \leq \underset{k=2}{\overset{\infty }{\sum }}\frac{\Phi
_{q,k}^{\lambda }}{1-\alpha }\left\vert a_{k}\right\vert ,
\end{equation*}%
which is equivalent to 
\begin{equation*}
\underset{k=2}{\overset{m}{\sum }}\left( \tfrac{\Phi _{q,k}^{\lambda
}-k\left( 1-\alpha \right) }{1-\alpha }\right) \left\vert a_{k}\right\vert +%
\underset{k=m+1}{\overset{\infty }{\sum }}\left( \tfrac{\left( m+1\right)
\Phi _{q,k}^{\lambda }-k\Phi _{q,m+1}^{\lambda }}{\left( m+1\right) \left(
1-\alpha \right) }\right) \left\vert a_{k}\right\vert \geq 0.
\end{equation*}%
To prove the result (7.15), define the function $g(z)$ by

\begin{equation*}
\frac{1+g(z)}{1-g(z)}=\tfrac{\left( m+1\right) \left( 1-\alpha \right) +\Phi
_{q,m+1}^{\lambda }}{\left( m+1\right) \left( 1-\alpha \right) }\left[ \frac{%
f_{m}^{^{\prime }}(z)}{f^{^{\prime }}(z)}-\tfrac{\Phi _{q,m+1}^{\lambda }}{%
\left( m+1\right) \left( 1-\alpha \right) +\Phi _{q,m+1}^{\lambda }}\right] ,
\end{equation*}

and by similar arguments in first part we get desired result.
\end{proof}

\textbf{Remark 4. }$(i)$ \ Putting $\lambda =0$ and $(ii)$ $\lambda =1$ in
Theorem 7.7, we obtain the following sharp results, respectively.

\begin{corollary}
If $f\in \mathcal{S}$, satisfies the condition (2.1) and $\frac{f(z)}{z}\neq
0(0<\left\vert z\right\vert <1),$ then%
\begin{equation}
\func{Re}\left( \frac{f^{^{\prime }}(z)}{f_{m}^{^{\prime }}(z)}\right) \geq
1-\tfrac{\left( m+1\right) \left( 1-\alpha \right) }{\left[ m+1\right]
_{q}(1+\beta )-(\alpha +\beta )},  \tag{7.17}
\end{equation}%
\newline
and 
\begin{equation}
\func{Re}\left( \frac{f_{m}^{^{\prime }}(z)}{f^{^{\prime }}(z)}\right) \geq 
\tfrac{\left[ m+1\right] _{q}(1+\beta )-(\alpha +\beta )}{\left[ m+1\right]
_{q}(1+\beta )-(\alpha +\beta )+\left( m+1\right) \left( 1-\alpha \right) }.
\tag{7.18}
\end{equation}
\end{corollary}

\begin{corollary}
If $f\in \mathcal{S}$, satisfies the condition (2.1) and $\frac{f(z)}{z}\neq
0(0<\left\vert z\right\vert <1),$ then%
\begin{equation}
\func{Re}\left( \frac{f^{^{\prime }}(z)}{f_{m}^{^{\prime }}(z)}\right) \geq
1-\tfrac{\left( m+1\right) \left( 1-\alpha \right) }{\left[ m+1\right] _{q}%
\left[ (m+1)(1+\beta )-(\alpha +\beta )\right] },  \tag{7.19}
\end{equation}%
\newline
and 
\begin{equation}
\func{Re}\left( \frac{f_{m}^{^{\prime }}(z)}{f^{^{\prime }}(z)}\right) \geq 
\tfrac{\left[ m+1\right] _{q}\left[ (m+1)(1+\beta )-(\alpha +\beta )\right] 
}{\left[ m+1\right] _{q}\left[ (m+1)(1+\beta )-(\alpha +\beta )\right]
+\left( m+1\right) \left( 1-\alpha \right) }.  \tag{7.20}
\end{equation}
\end{corollary}

\textbf{Remark 5.} Letting $q\rightarrow 1^{-}$ in Theorems 7.1, 7.4 and
7.7, respectively, we get Theorems 4.1 and 4.2, respectively, for the class $%
\QTR{sl}{S}_{q}^{\lambda }(\alpha ,\beta )$ studied by Rosy et al. [31].

\textbf{Acknowledgements }

The authors express their sincere thanks to the referees for their valuable
comments and suggestions.

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\end{document}