1. Introduction

Let $\mathcal{A}$ denotes the class of all function $f(z)$ which are analytic in the open unit disk $E=\{z\in \mathbb{C} :\left \vert z\right \vert < 1\}$ and normalized by $f(0)=0$ and $f^{^{\prime }}(0)=1,$ so each $f\in A$ has the Maclaurin's series expansion of the form:

$f(z)=z+\sum\limits_{n=2}^{\infty }a_{n}z^{n}.$

(1.1)

A function $f:E\rightarrow \mathbb{C}$ is called univalent on $E$ if $f(z_{1})=f(z_{2})$ for all $z_{1}=z_{2},$ $% z_{1}, z_{2}\in E.$ Let $\mathcal{S}\subset A$ be the class of all functions which are univalent in $E$ (see ^[3]). Recall $D\subset \mathbb{C}$ is said to be a starlike with respect to the point $d_{0}\in D$ if and only if the line segment joining $d_{0}$ to every other point $d\in D$ lies entirely in $D$, while the set $D$ is said to be convex if and only if it is starlike with respect to each of its points. By $S^{\ast }$ and $K$ we means the subclasses of $S$ composed of starlike and convex functions. A function $f\in A$ is said to be starlike of order $\alpha,$ $0\leq \alpha < 1$, if

$\Re\left( \frac{zf^{^{\prime }}(z)}{f(z)}\right) >\alpha, z\in E.$

A function $f\in A$ is said to be convex of order $\alpha,$ $0\leq \alpha < 1$, if

$\Re\left( \frac{\left( zf^{^{\prime }}(z)\right) ^{^{\prime }}}{% f^{^{\prime }}(z)}\right) >\alpha, z\in E.$

In 1991, Goodman ^[4] introduced the class $\mathcal{UCV}$ of uniformly convex functions which was extensively studied by Ronning and independently by Ma and Minda ^[1,2]. A more convenient characterization of class $\mathcal{UCV}$ was given by Ma and Minda as:

$f(z)\in \mathcal{UCV}\Longleftrightarrow f(z)\in \mathcal{A}\text{ and }% \Re\left \{ 1+\frac{zf^{^{\prime \prime }}(z)}{f^{^{\prime }}(z)}% \right \} >\left \vert \frac{zf^{^{\prime \prime }}(z)}{f^{^{\prime }}(z)}% \right \vert, \ \ \ \ z\in E.$

In 1999, Kanas and Wisniowska ^[5,6] introduced the class $k-$uniformly convex functions, $k\geq 0,$ denoted by $k-\mathcal{UCV}$ and a related class $k-\mathcal{ST}$ as:

$f\in k-\mathcal{UCV}\Longleftrightarrow zf^{^{\prime }}\in k-\mathcal{ST}% \Longleftrightarrow f\in A\text{ and }\Re\left \{ \frac{\left( zf^{^{\prime }}(z)\right) ^{^{\prime }}}{f^{^{\prime }}(z)}\right \} >\left \vert \frac{zf^{^{\prime \prime }}(z)}{f^{^{\prime }}(z)}\right \vert, % \ z\in E.$

The class $k-\mathcal{UCV}$ was discussed earlier in ^[7], see also ^[8] with same extra restriction and without geometrical interpretation by Bharati et.al ^[8]. In 1985, Nasr et al., studied a natural extension of classical starlikness in order terminology. We say that a function $f(z)\in \mathcal{A}$ is in the class $\mathcal{S}_{k, \gamma }^{\ast },$ $k\geq 0,$ $\gamma \in \mathbb{C} \backslash \{0\},$ if and only if

$\Re\left \{ \frac{1}{\gamma }\left( \frac{zf^{^{\prime }}(z)}{f(z)}% -1\right) \right \} >k\left \vert \frac{1}{\gamma }\left( \frac{zf^{^{\prime }}(z)}{f(z)}-1\right) \right \vert, \ \ \ \ z\in E.$

Several author investigated the properties of the class, $\mathcal{S}% _{k, \gamma }^{\ast }$ and their generalizations in several directions for detail study see ^{[4,6,9,10,11,12,13]}. The convolution or Hadamard product of two function $f$ and $g$ is denoted by $f\ast g$ is defined as

$(f\ast g)z)=\sum\limits_{n=0}^{\infty }a_{n}b_{n}z^{n},$

where $f(z)$ is given by (1.1) and $g(z)=\sum_{n=2}^{\infty }b_{n}z^{n}, \ \ \ \ \ (z\in E).$

If $f(z)$ and $g(z)$ are analytic in $E$, we say that $f(z)$ is subordinate to $g(z)$, written as $f(z)\prec g(z)$, if there exists a Schwarz function $% w(z)$, which is analytic in $E$ with $w(0)=0$ and $|w(z)| < 1$ such that $% f(z)=g(w(z))$. Furthermore, if the function $g(z)$ is univalent in $E$, then we have the following equivalence, see ^[3,14].

$f(z)\prec g(z)\Longleftrightarrow f(0)=g(0)\text{ and }f(E)\subset g(E).% \ \ \ \ z\in E.$

Note that the $q$-difference operator plays an important role in the theory of hypergeometric series and quantum theory, number theory, statistical mechanics, etc. At the beginning of the last century studies on $q$-difference equations appeared in intensive works especially by Jackson ^[33], Carmichael ^[32], Mason ^[34], Adams ^[31] and Trjitzinsky ^[35]. Research work in connection with function theory and $q$-theory together was first introduced by Ismail et al. ^[36]. Till now only non-significant interest in this area was shown although it deserves more attention.

Many differential and integral operators can be written in term of convolution, for details we refer ^[21]. It is worth mentioning that the technique of convolution helps researchers in further investigation of geometric properties of analytic functions.

For any non-negative integer $n$, the $q$-integer number $n$ denoted by $[n]_{q}$, is defined by

$\lbrack n]_{q}=\frac{1-q^{n}}{1-q}, [0]_{q}=0.$

For non-negative integer $n$ the $q$-number shift factorial is defined by

$\lbrack n]_{q}!=[1]_{q}[2]_{q}[3]_{q}...[n]_{q}, \ \ \ \left( [0]_{q}!=1\right) .$

We note that when $q\rightarrow 1,$ $[n]!$ reduces to classical definition of factorial. In general, for a non-integer number $t$, $[t]_{q}$ is defined by $[t]_{q}=\frac{1-q^{t}}{1-q},$ $[0]_{q}=0.$ Throughout in this paper, we will assume $q$ to be a fixed number between 0 and 1

The $q$-difference operator related to the $q$-calculus was introduced by Andrews et al. (see in ^[30] CH 10). For $f\in A,$ the $q$-derivative operator or $q$-difference operator is defined as.

$\partial _{q}f(z)=\frac{f(qz)-f(z)}{z(q-1)}, \ \ z\in E, z\neq, q\neq 1.$

It can easily be seen that for $n\in N=\{1, 2, 3, ...\}$ and $z\in E.$

$\partial _{q}z^{n}=[n]_{q}z^{n-1}, \ \ \partial _{q}\left \{ \sum\limits_{n=1}^{\infty }a_{n}z^{n}\right \} =\sum\limits_{n=1}^{\infty }[n]_{q}a_{n}z^{n-1}.$

Recently, Govindaraj and Sivasubramanian defined Salagean q-differential operator ^[28] as:

Let $f\in A$, let Salagean $q$-differential operator

$S_{q}^{0}f(z)=f(z), \ S_{q}^{1}f(z)=z\partial _{q}f(z), \ \ \ \ \ \ \ \, S_{q}^{m}f(z)=z\partial _{q}\left( S_{q}^{m-1}f(z)\right) .$

A simple calculation implies

$S_{q}^{m}f(z)=f(z)\ast G_{q, m}(z)$

(1.2)

$G_{q, m}(z)=z+\sum\limits_{n=2}^{\infty }\left[n\right] _{q}^{m}z^{n},$

(1.3)

Making use of (1.2) and (1.3) the power series of $S_{q}^{m}f(z)$ for $f$ of the form (1.1) is given by

$S_{q}^{m}f(z)=z+\sum\limits_{n=2}^{\infty }\left[n\right] _{q}^{m}a_{n}z^{n}$

(1.4)

Note that

$Lim_{q\rightarrow 1}G_{q, m}(z)=z+\sum\limits_{n=2}^{\infty }n^{m}z^{n}$

$Lim_{q\rightarrow 1}S_{q}^{m}f(z)=z+\sum\limits_{n=2}^{\infty }n^{m}a_{n}z^{n}$

which is the familiar Salagean derivative ^[29].

Taking motivation from the work shahid et.al ^[23], we introduce new subclass $k-\mathcal{US(}q, \gamma, m\mathcal{)},$ of analytic functions with the theory of $q$-calculus by using Salagean $q$-differential operator.

Definition 1.1. Let $f(z)\in \mathcal{A}$. Then $f(z)$ is in the class $k-\mathcal{ US}(q, \gamma, m),$ $\gamma \in C\backslash \{0\}$, if it satisfies the condition

$\Re\left \{ 1+\frac{1}{\gamma }\left( \frac{z\partial _{q}S_{q}^{m}f(z)% }{S_{q}^{m}f(z)}-1\right) \right \} >k\left \vert \frac{1}{\gamma }\left( \frac{z\partial _{q}S_{q}^{m}f(z)}{S_{q}^{m}f(z)}-1\right) \right \vert, % \ \ \ z\in E.$

By taking specific values of parameters, we obtain many important subclasses studied by various authors in earlier papers. Here we inlist some of them.

(1) For $m=0,$ $q\rightarrow 1,$ and $\gamma =\frac{1}{1-\beta },$ $\beta \in C\backslash \{1\},$ the class $k-\mathcal{US(}q, \gamma, m\mathcal{)}$ reduce into the class $\mathcal{SD}(k, \beta)$ studied by Shams et.al ^[24].

(2) For $m=0,$ $q\rightarrow 1,$ and $\gamma =\frac{2}{1-\beta },$ $\beta \in C\backslash \{1\},$ the class $k-\mathcal{US(}q, \gamma, m\mathcal{)}$ reduces into the class $\mathcal{KD}(k, \beta),$ studied by Owa et.al ^[26].

(3) For $k=1,$ $m=0,$ $q\rightarrow 1,$ and $\gamma =\frac{1}{1-\beta },$ $% \beta \in C\backslash \{1\},$ the class $k-\mathcal{US(}q, \gamma, m\mathcal{)% }$ reduce into the class $\mathcal{S}_{p}(\beta)$ studied by Ali et.al ^[27].

(4) For $k=1,$ $m=0,$ $q\rightarrow 1,$ and $\gamma =\frac{2}{1-\beta },$ $% \beta \in C\backslash \{1\},$ the class $k-\mathcal{US(}q, \gamma, m\mathcal{)% }$ reduces into the class $\mathcal{K}_{p}(\beta),$ studied by Ali et.al ^[27].

(5) For $m=0,$ $q\rightarrow 1,$ the class $k-\mathcal{US(}q, \gamma, m% \mathcal{)}$ reduce into the class $\mathcal{K-ST}$, introduced by Kanas and Wisniowska ^[5].

(6) For $k=0,$ $m=0,$ $q\rightarrow 1,$ and $\gamma =\frac{1}{1-\beta },$ $% \beta \in C\backslash \{1\},$ the class $k-\mathcal{US(}q, \gamma, m\mathcal{)% }$ reduce into the class $\mathcal{S}^{\ast }(\beta)$ $,$ well-known class of starlike of order respectively.

Geometric Interpretation

A function $f(z)\in \mathcal{A}$ is in the class $k-\mathcal{US(}q, \gamma, m% \mathcal{)}$ if and only if $\frac{z\partial _{q}S_{q}^{m}f(z)}{S_{q}^{m}f(z)% }$ takes all the values in the conic domain $\Omega _{k, \gamma }=p_{k, \gamma }(E)$, such that

$\Omega _{k, \gamma }=\gamma \Omega _{k}+(1-\alpha ),$

where

$\Omega _{k}=\left \{ u+iv:u>k\sqrt{\left( u-1\right) ^{2}+v^{2}}\right \} .$

Since $p_{k, \gamma }(z)$ is convex univalent, so above definition can be written as

$\frac{z\partial _{q}S_{q}^{m}f(z)}{S_{q}^{m}f(z)}\prec p_{k, \gamma }(z),$

(1.5)

where

$p_{k, \gamma }(z)=\left \{ \begin{array}{c} \frac{1+z}{1-z}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for }k=0, \\ 1+\frac{2\gamma }{\pi ^{2}}\left( \log \frac{1+\sqrt{z}}{1-\sqrt{z}}\right) ^{2}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for }k=1, \\ 1+\frac{2\gamma }{1-k^{2}}\sinh ^{2}\left \{ \left( \frac{2}{\pi }\arccos k\right) \arctan h\sqrt{z}\right \}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for}\ 0<k<1, \\ 1+\frac{\gamma }{k^{2}-1}\sin \left( \frac{\pi }{2R(t)}\int_{0}^{\frac{u(z)}{% \sqrt{t}}}\frac{1}{\sqrt{1-x^{2}}\sqrt{1-(tx)^{2}}}dx\right) +\frac{\gamma }{% 1-k^{2}}, \ \ \ \ \ \ \ \text{}\ \ \text{for }k>1.% \end{array}% \right.$

(1.6)

The boundary $\partial \Omega _{k, \gamma }$ of the above set becomes the imaginary axis when $k=0,$ while a hyperbola when $0 < k < 1.$ For $k=1$ the boundary $\partial \Omega _{k, \gamma }$ becomes a parabola and it is an ellipse when $k>1$ and in this case where

$u(z)=\frac{z-\sqrt{t}}{1-\sqrt{t}z}, \ \ z\in E,$

and $t\in (0, 1)$ is chosen such that $k=\cosh \, (\pi K^{\prime }(t)/(4K(t)))$. Here $K(t)$ is Legender's complete elliptic integral of first kind and $K^{\prime }(t)=K(\sqrt{1-t^{2}}$) and $K^{\prime }\left(t\right)$ is the complementary integral of $K\left(t\right)$ for details see ^[5,6,14,17]. Moreover, $p_{k, \gamma }(E)$ is convex univalent in $E$, see ^[5,6]. All of these curves have the vertex at the point $\frac{k+\gamma }{k+1}.$

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2. Set of Lemmas

Each of the following lemmas will be needed in our present investigation.

Lemma 2.1. [18]. Let $p(z)=\sum_{n=1}^{\infty }p_{n}z^{n}\prec F(z)=\sum_{n=1}^{\infty }d_{n}z^{n}$ in $E.$ If $F(z)$ is convex univalent in $E$ then

$\left \vert p_{n}\right \vert \leq \left \vert d_{1}\right \vert , \ \ n\geq 1.$

(2.1)

Lemma 2.2. [19]. Let $k\in \lbrack 0, \infty)$ be fixed and let $p_{k, \gamma }$ be defined (1.6). If

$p_{k, \gamma }(z)=1+Q_{1}z+Q_{2}z^{2}+...$

(2.2)

$Q_{1}=\left \{ \begin{array}{ll} \frac{2\gamma A^{2}}{1-k^{2}}, & 0\leq k<1 \\ \frac{8\gamma }{\pi ^{2}}, & k=1, \\ \frac{\pi ^{2}\gamma }{4\left( 1+t\right) \sqrt{t}K^{2}(t)\left( k^{2}-1\right) }, & k>1, % \end{array}% \right.$

(2.3)

$Q_{2}=\left \{ \begin{array}{ll} \frac{A^{2}+2}{3}Q_{1}, & 0\leq k<1 \\ \frac{2}{3}Q_{1}, & k=1, \\ \frac{4K^{2}(t)\left( t^{2}+6t+1\right) -\pi ^{2}}{24K^{2}\left( t\right) \left( 1+t\right) \sqrt{t}}Q_{1}, & k>1, % \end{array}% \right.$

(2.4)

where $A=\frac{2\cos ^{-1}k}{\pi },$ and $t\in (0, 1)$ is chosen such that $\ k=\cosh \left(\frac{\pi K^{^{\prime }}(t)}{K(t)}\right)$, $K(t)$ is the Legendre's complete elliptic integral of the first kind.

Lemma 2.3. [20]. Let $p(z)=1+\sum_{n=1}^{\infty }c_{n}z^{n}\in P,$ let $% p(z)$ be analytic in $E$ and satisfy $Re\{p(z)\}>0$ for $z$ in $E$, then the following sharp estimate holds

$\left \vert c_{2}-\mu c_{1}^{2}\right \vert \leq 2\max \left \{ 1, \left \vert 2\mu -1\right \vert \right \}, \ \ \forall \mu \in \mathbb{C} .$

(2.5)

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3. Main Results

In this section, we will prove our main results.

Theorem 3.1. Let $f(z)\in k-\mathcal{US(}q, \gamma, m\mathcal{)}.$ Then

$S_{q}^{m}f(z)\prec z\exp \int_{0}^{z}\frac{p_{k, \gamma }(w(\xi ))-1}{\zeta }% d\xi,$

(3.1)

where $w(z)$ is analytic in $E$ with $w(0)=0$ and $\left \vert w(z)\right \vert < 1.$ Moreover, for $\left \vert z\right \vert =\rho,$ we have

$\exp \left( \int_{0}^{1}\frac{p_{k, \gamma }(-\rho )-1}{\rho }d\rho \right) \leq \left \vert \frac{S_{q}^{m}f(z)}{z}\right \vert \leq \exp \left( \int_{0}^{1}\frac{p_{k, \gamma }(\rho )-1}{\rho }d\rho \right),$

(3.2)

where $p_{k, \gamma }(z)$ is defined by (1.6).

Proof. If $f(z)\in k-\mathcal{US(}q, \gamma, m\mathcal{)}$ then using the identity (1.5), we obtain

$\frac{z\partial _{q}S_{q}^{m}f(z)}{S_{q}^{m}f(z)}-\frac{1}{z}=\frac{% p_{k, \gamma }(w(z))-1}{z}.$

(3.3)

For some function $w(z)$ is analytic in $E$ with $w(0)=0$ and $\left \vert w(z)\right \vert < 1.$ Integrating (3.3) and after some simplification we have

$S_{q}^{m}f(z)\prec z\exp \int_{0}^{z}\frac{p_{k, \gamma }(w(\xi ))-1}{\zeta }% d\xi .$

(3.4)

This proves (3.1). Noting that the univalent function $p_{k, \gamma }(z)$ maps the disk $|z| < \rho$ $(0 < \rho \leq 1)$ onto a region which is convex and symmetric with respect to the real axis, we see

$p_{k, \gamma }(-\rho \left \vert z\right \vert )\leq \Re\left \{ p_{k, \gamma }(w(\rho z)\right \} \leq p_{k, \gamma }(\rho \left \vert z\right \vert ) \ \ \ (0<\rho \leq 1, \ \ z\in E).$

(3.5)

Using (3.4) and (3.5) gives

$\int_{0}^{1}\frac{p_{k, \gamma }(-\rho \left \vert z\right \vert )-1}{\rho }% d\rho \leq \Re\int_{0}^{1}\frac{p_{k, \gamma }(w\left( \rho \left( z\right) \right) -1}{\rho }d\rho \leq \int_{0}^{1}\frac{p_{k, \gamma }(\rho \left \vert z\right \vert )-1}{\rho }d\rho,$

for $z\in E$. Consequently, subordination (3.4) leads us to

$\int_{0}^{1}\frac{p_{k, \gamma }(-\rho \left \vert z\right \vert )-1}{\rho }% d\rho \leq \log \left \vert \frac{S_{q}^{m}f(z)}{z}\right \vert \leq \int_{0}^{1}\frac{p_{k, \gamma }(\rho \left \vert z\right \vert )-1}{\rho }% d\rho$

$p_{k, \gamma }(-\rho )\leq p_{k, \gamma }(-\rho \left \vert z\right \vert ), % \text{ }p_{k, \gamma }(\rho \left \vert z\right \vert )\leq p_{k, \gamma }(\rho )$

implies that

$\exp \int_{0}^{1}\frac{p_{k, \gamma }(-\rho )-1}{\rho }d\rho \leq \left \vert \frac{S_{q}^{m}f(z)}{z}\right \vert \leq \exp \int_{0}^{1}\frac{p_{k, \gamma }(\rho )-1}{\rho }d\rho .$

this completes the proof.

Theorem 3.2. If $f(z)\in k-\mathcal{US(}q, \gamma, m\mathcal{)}.$ Then

$\left \vert a_{2}\right \vert \leq \frac{\delta }{[2]_{q}^{m}\left \{ [2]_{q}-1\right \} },$

(3.6)

and

$\left \vert a_{n}\right \vert \leq \frac{\delta }{[n]_{q}^{m}\left \{ [n]_{q}-1\right \} }\prod \limits_{j=1}^{n-2}\left( 1+\frac{\delta }{% [j+1]_{q}-1}\right), \ \ \ \ \ \ for \;n=3, 4, ....$

(3.7)

where $\delta =\left \vert Q_{1}\right \vert$ with $Q_{1}$ is given by (2.3).

Proof. Let

$\frac{z\partial _{q}S_{q}^{m}f(z)}{S_{q}^{m}f(z)}=p(z).$

(3.8)

where $p(z)$ is analytic in $E$ and $p(0)=1.$Let $p(z)=1+\sum_{n=1}^{\infty }c_{n}z^{n}$ and $S_{q}^{m}f(z)$ is given by (1.4). Then (3.8) becomes

$z+\sum\limits_{n=2}^{\infty }[n]_{q}^{m+1}a_{n}z^{n}=\left( \sum\limits_{n=0}^{\infty }c_{n}z^{n}\right) \left( z+\sum\limits_{n=2}^{\infty }[n]_{q}^{m}a_{n}z^{n}\right) .$

Now comparing the coefficients of $z^{n}$, we obtain

$n]_{q}^{m+1}a_{n}=[n]_{q}^{m}a_{n}+\sum\limits_{j=1}^{n-1}[j]_{q}^{m}a_{j}c_{n-j}.$

which implies

$a_{n}=\frac{1}{[n]_{q}^{m}\left \{ [n]_{q}-1\right \} }\sum\limits_{j=1}^{n-1}[j]_{q}^{m}a_{j}h_{n-j}.$

Using the results that $\left \vert c_{n}\right \vert \leq \left \vert Q_{1}\right \vert$ given in (^[17]), we have

$\left \vert a_{n}\right \vert \leq \frac{Q_{1}}{[n]_{q}^{m}\left \{ [n]_{q}-1\right \} }\sum\limits_{j=1}^{n-1}[j]_{q}^{m}\left \vert a_{j}\right \vert .$

Let us take $\delta =\left \vert Q_{1}\right \vert.$ Then we have

$\left \vert a_{n}\right \vert \leq \frac{\delta }{[n]_{q}^{m}\left \{ [n]_{q}-1\right \} }\sum\limits_{j=1}^{n-1}[j]_{q}^{m}\left \vert a_{j}\right \vert .$

(3.9)

For $n=2$ in (3.9), we have

$\left \vert a_{2}\right \vert \leq \frac{\delta }{[2]_{q}^{m}\left \{ [2]_{q}-1\right \} },$

(3.10)

which shows that (3.7) holds for $n=2$. To prove (3.7) we use principle of mathematical induction, for this, consider the case $n=3$

$\left \vert a_{3}\right \vert \leq \frac{\delta }{[3]_{q}^{m}\left \{ [3]_{q}-1\right \} }\{1+[2]_{q}^{m}\left \vert a_{2}\right \vert \}.$

Using (3.10), we have

$\left \vert a_{3}\right \vert \leq \frac{\delta }{[3]_{q}^{m}\left \{ [3]_{q}-1\right \} }\{1+\frac{\delta }{[2]_{q}-1}\}.$

which shows that (3.7) holds for $n=3$. Let us assume that (3.7) is true for $n\leq t$, that is,

$\left \vert a_{t}\right \vert \leq \frac{\delta }{[t]_{q}^{m}\left \{ [t]_{q}-1\right \} }\prod \limits_{j=1}^{t-2}\left( 1+\frac{\delta }{% [j+1]_{q}-1}\right), \ \ \ \ \ \ \text{for }n=3, 4, ....$

consider

$\begin{eqnarray*} \left \vert a_{t+1}\right \vert &\leq &\frac{\delta }{[t+1]_{q}^{m}\left \{ [t+1]_{q}-1\right \} }\left \{ 1+[2]_{q}^{m}\left \vert a_{2}\right \vert +[3]_{q}^{m}\left \vert a_{3}\right \vert +[4]_{q}^{m}\left \vert a_{4}\right \vert +...[t]_{q}^{m}\left \vert a_{t}\right \vert \right \} \\ && \\ &\leq &\frac{\delta }{[t+1]_{q}^{m}\left \{ [t+1]_{q}-1\right \} }\left \{ \begin{array}{c} 1+\frac{\delta }{[2]_{q}-1}+\frac{\delta }{[3]_{q}-1}\left( 1+\frac{\delta }{% [2]_{q}-1}\right) +... \\ \\ +\frac{\delta }{[t]_{q}-1}\prod\limits_{j=1}^{t-2}\left( 1+\frac{\delta }{[j+1]_{q}-1}\right)% \end{array}% \right \} \\ && \\ &=&\frac{\delta }{[t+1]_{q}^{m}\left \{ [t+1]_{q}-1\right \} }\prod\limits_{j=1}^{t-1}\left( 1+\frac{\delta }{[j+1]_{q}-1}\right) . \end{eqnarray*}$

which proves the assertion of theorem $n=t+1.$ Hence (3.7) holds for all $n,$ $n\geq 3.$

This completes the proof.

Theorem 3.3. Let $0\leq k < \infty$ be fixed and let $f(z)\in k-\mathcal{US(}% q, \gamma, m\mathcal{)}$ with the form (1.1) then for a complex number $\mu$

$\left \vert a_{3}-\mu a_{2}^{2}\right \vert \leq \frac{d_{1}}{2\left[3% \right] _{q}^{m}\left \{ [3]_{q}-1\right \} }\max \left[1, \left \vert 2v-1\right \vert \right],$

(3.11)

where

$v=\frac{1}{2}\left \{ 1-\frac{d_{2}}{d_{1}}-d_{1}\left( \frac{1}{\left \{ [2]_{q}-1\right \} }-\mu \frac{\left[3\right] _{q}^{m}\left \{ [3]_{q}-1\right \} }{2\left[2\right] _{q}^{m}\left \{ [2]_{q}-1\right \} }% \right) \right \} .$

(3.12)

$Q_{1}$ and $Q_{2}$ are given by (2.3) and (2.4).

Proof. Let $f(z)\in k-\mathcal{US(}q, \gamma, m\mathcal{)},$ then there exists Schwarz function $w(z)$, with $w(0)=0$ and $|w(z)| < 1$ such that

$\frac{z\partial _{q}S_{q}^{m}f(z)}{S_{q}^{m}f(z)}=p_{k, \gamma }(w(z)) \ \ z\in E.$

(3.13)

Let $p(z)\in \mathcal{P}$ be a function defined as

$p(z)=\frac{1+w(z)}{1-w(z)}=1+c_{1}z+c_{2}z^{2}+...$

This gives

$w(z)=\frac{c_{1}}{2}z+\frac{1}{2}(c_{2}-\frac{c_{1}^{2}}{2})z^{2}+...$

and

$p_{k, \gamma }(w(z))=1+\frac{Q_{1}c_{1}}{2}z+\left \{ \frac{Q_{2}c_{1}^{2}}{4}% +\frac{1}{2}(c_{2}-\frac{c_{1}^{2}}{2})Q_{1}\right \} z^{2}+...$

(3.14)

$\frac{z\partial _{q}S_{q}^{m}f(z)}{S_{q}^{m}f(z)}=1+\left[2\right] _{q}^{m}\left \{ [2]_{q}-1\right \} a_{2}z+\left \{ \left[3\right] _{q}^{m}\left \{ [3]_{q}-1\right \} a_{3}-\left( \left[2\right] _{q}^{m}\right) ^{2}\left \{ [2]_{q}-1\right \} a_{2}^{2}\right \} z^{2}$

(3.15)

Using (3.14) in (3.13) and coparing with (3.15), we obtain

$a_{2}=\frac{Q_{1}c_{1}}{2\left[2\right] _{q}^{m}\left \{ [2]_{q}-1\right \} }.$

and

$a_{3}=\frac{1}{\left[3\right] _{q}^{m}\left \{ [3]_{q}-1\right \} }\left \{ \frac{Q_{1}c_{2}}{2}+\frac{c_{1}^{2}}{4}\left( Q_{2}-Q_{1}+\frac{Q_{1}^{2}}{% \left \{ [2]_{q}-1\right \} }\right) \right \} .$

For any complex number $\mu$ and after some calculation we have

$a_{3}-\mu a_{2}^{2}=\frac{Q_{1}}{2\left[3\right] _{q}^{m}\left \{ [3]_{q}-1\right \} }\left \{ c_{2}-vc_{1}^{2}\right \},$

(3.16)

where

$v=\frac{1}{2}\left \{ 1-\frac{Q_{2}}{Q_{1}}-Q_{1}\left( \frac{1}{\left \{ [2]_{q}-1\right \} }-\mu \frac{\left[3\right] _{q}^{m}\left \{ [3]_{q}-1\right \} }{2\left[2\right] _{q}^{m}\left \{ [2]_{q}-1\right \} }% \right) \right \} .$

Using a lemm (2.5) on (3.16) we have the required results.

Theorem 3.4. If a function $f(z)\in \mathcal{A}$ has the form (1.1) satisfies the condition

$\sum\limits_{n=2}^{\infty }\left \{ \left \{ [n]_{q}-1\right \} (k+1)+\left \vert \gamma \right \vert \right \} \left \vert \left[n\right] _{q}^{m}\right \vert \left \vert a_{n}\right \vert \leq \left \vert \gamma \right \vert$

(3.17)

then $f(z)\in k-\mathcal{US(}q, \gamma, m\mathcal{)}.$

Proof. Let we note that

$\begin{eqnarray} \left \vert \frac{z\partial _{q}S_{q}^{m}f(z)}{S_{q}^{m}f(z)}-1\right \vert &=&\left \vert \frac{z\partial _{q}S_{q}^{m}f(z)-S_{q}^{m}f(z)}{S_{q}^{m}f(z)% }\right \vert =\left \vert \frac{\sum\limits_{n=2}^{\infty }[n]_{q}^{m}\left \{ [n]_{q}-1\right \} a_{n}z^{n}}{z+\sum\limits_{n=2}^{\infty }[n]_{q}^{m}a_{n}z^{n}}% \right \vert \notag \\ &\leq &\frac{\sum\limits_{n=2}^{\infty }\left \vert [n]_{q}^{m}\left \{ [n]_{q}-1\right \} \right \vert \left \vert a_{n}\right \vert }{% 1-\sum\limits_{n=2}^{\infty }\left \vert [n]_{q}^{m}\right \vert \left \vert a_{n}\right \vert }. \end{eqnarray}$

(3.18)

From (3.17) it follows that

$1-\sum\limits_{n=2}^{\infty }\left \vert [n]_{q}^{m}\right \vert \left \vert a_{n}\right \vert >0.$

To show that $f(z)\in k-\mathcal{US(}q, \gamma, m\mathcal{)}$ it is suffies that

$\left \vert \frac{k}{\gamma }\left( \frac{z\partial _{q}S_{q}^{m}f(z)}{% S_{q}^{m}f(z)}-1\right) \right \vert -\Re\left \{ \frac{1}{\gamma }% \left( \frac{z\partial _{q}S_{q}^{m}f(z)}{S_{q}^{m}f(z)}-1\right) \right \} \leq 1.$

From (3.18), we have

$\begin{eqnarray*} &&\left \vert \frac{k}{\gamma }\left( \frac{z\partial _{q}S_{q}^{m}f(z)}{% S_{q}^{m}f(z)}-1\right) \right \vert -\Re\left \{ \frac{1}{\gamma }% \left( \frac{z\partial _{q}S_{q}^{m}f(z)}{S_{q}^{m}f(z)}-1\right) \right \} \\ &\leq &\frac{k}{\left \vert \gamma \right \vert }\left \vert \frac{z\partial _{q}S_{q}^{m}f(z)}{S_{q}^{m}f(z)}-1\right \vert +\frac{1}{\left \vert \gamma \right \vert }\left \vert \frac{z\partial _{q}S_{q}^{m}f(z)}{S_{q}^{m}f(z)}% -1\right \vert \\ &\leq &\frac{(k+1)}{\left \vert \gamma \right \vert }\left \vert \frac{% z\partial _{q}S_{q}^{m}f(z)}{S_{q}^{m}f(z)}-1\right \vert =\left \vert \frac{% z\partial _{q}S_{q}^{m}f(z)-S_{q}^{m}f(z)}{S_{q}^{m}f(z)}\right \vert \\ &\leq &\frac{(k+1)}{\left \vert \gamma \right \vert }\frac{% \sum\limits_{n=2}^{\infty }\left \vert [n]_{q}^{m}\left \{ [n]_{q}-1\right \} \right \vert \left \vert a_{n}\right \vert }{1-\sum\limits_{n=2}^{\infty }\left \vert [n]_{q}^{m}\right \vert \left \vert a_{n}\right \vert } \\ &\leq &1. \end{eqnarray*}$

Because from (3.8).

When $q\rightarrow 1,$ $m=0,$ $\gamma =1-\alpha,$ with $0\leq \alpha < 1$, then we have the following known result, proved by Shams et-al. in ^[24].

Corollary 3.1. A function $f\in A$ and of the form (1.1) is in the class $k-\mathcal{US}% \left(1-2\alpha \right)$, if it satisfies the condition

$\sum\limits_{n=2}^{\infty }\left \{ n(k+1)-(k+\alpha )\right \} \left \vert a_{n}\right \vert \leq 1-\alpha$

where $0\leq \alpha < 1$ and $k\geq 0$.

When $q\rightarrow 1,$ $m=0,$ $\gamma =1-\alpha,$ with $0\leq \alpha < 1$ and $k=0$, then we have the following known result, proved by Selverman in ^[25]

Corollary 3.2. A function $f\in A$ and of the form (1.1) is in the class $0-\mathcal{US}% \left(1-\alpha, \right)$, if it satisfies the condition

$\sum\limits_{n=2}^{\infty }\left \{ n-\alpha \right \} \left \vert a_{n}\right \vert \leq 1-\alpha, \ \ \ \ \ \ 0\leq \alpha <1.$

Theorem 3.5. Let $f(z)\in k-\mathcal{US}(q, \gamma, m).$ Then $f(E)$ contains an open disk of radius

$\frac{\lbrack 2]^{m}\left \{ [2]_{q}-1\right \} }{2[2]_{q}^{m}\left \{ [2]_{q}-1\right \} +\delta }.$

where $Q_{1}$ is given by (2.3)

Proof. Let $w_{0}\neq 0$ be a complex number such that $f(z)\neq w_{0}$ for $z\in E.$ Then

$f_{1}(z)=\frac{w_{0}f(z)}{w_{0}-f(z)}=z+\left( a_{2}+\frac{1}{w_{0}}\right) z^{2}+...$

since $f_{1}(z)$ is univalent, so

$\left \vert a_{2}+\frac{1}{w_{0}}\right \vert \leq 2.$

know using (3.6), we have

$\left \vert \frac{1}{w_{0}}\right \vert \leq \frac{2[2]_{q}^{m}\left \{ [2]_{q}-1\right \} +\delta }{[2]_{q}^{m}\left \{ [2]_{q}-1\right \} },$

hence we have.

$\left \vert w_{0}\right \vert \geq \frac{\lbrack 2]_{q}^{m}\left \{ [2]_{q}-1\right \} }{2[2]_{q}^{m}\left \{ [2]_{q}-1\right \} +\delta }.$

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Acknowledgments

The authors wish to thank the referee for the helpful suggestions and comments.

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Conflict of Interest

No potential conflict of interest was reported by the authors.

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