The main aim of this section is to present some classes of multivalent q-starlike functions involving the functions with negative coefficients. A subset of the class \(\mathcal{A} ( \mathfrak{p} )\) which contains all such functions with negative coefficient, that is,
$$ f (z ) =z^{\mathfrak{p}}-\sum_{n=1}^{\infty } \vert a_{n+ \mathfrak{p}} \vert z^{n+\mathfrak{p}} $$
(3.1)
will be denoted here by \(\mathcal{T}\). We also let
$$ \mathcal{TS}_{ (q,j ) }^{\ast } [ \mathfrak{p},{ v},s,X,L ] :=\mathcal{S}_{ ( q,j ) }^{\ast } [ \mathfrak{p},{v},s,X,L ] \cap \mathcal{T} \quad (j=1,2,3). $$
(3.2)
Theorem 3
If \(-1\leqq L< X\leqq 1\), then
$$ \mathcal{TS}_{ ( q,1 ) }^{\ast } [ \mathfrak{p},{ v} ,s,X,L ] \equiv \mathcal{TS}_{ ( q,2 ) }^{\ast } [ \mathfrak{p},{v},s,X,L ] \equiv \mathcal{TS}_{ ( q,3 ) }^{\ast } [ \mathfrak{p},{ v},s,X,L ] . $$
Proof
By the virtue of Theorem 1, it suffices to show that
$$ \mathcal{TS}_{ ( q,1 ) }^{\ast } [ \mathfrak{p},{ v},s,X,L ] \subseteq \mathcal{TS}_{ ( q,3 ) }^{\ast } [ \mathfrak{p},{v},s,X,L ] . $$
Indeed, in the light of Definition 6 for a function \(f\in \mathcal{TS}_{ ( q,1 ) }^{\ast } [ \mathfrak{p},{v},s,X,L ] \), we have
$$ \Re \biggl( \frac{ ( L-1 ) \frac{z^{{v}} ( \mathfrak{D}_{q}^{ ( {v}+s ) }f ) ( z ) }{ ( \mathfrak{D}_{q}^{ ( s ) }f ) ( z ) }- ( X-1 ) }{ ( L+1 ) \frac{z^{{v}} ( \mathfrak{D}_{q}^{ ( {v}+s ) }f ) ( z ) }{ ( \mathfrak{D}_{q}^{ ( s ) }f ) ( z ) }- ( X+1 ) } \biggr) \geqq 0, $$
so that
$$ \Re \biggl( \frac{ ( L-1 ) \frac{z^{{v}} ( \mathfrak{D}_{q}^{ ( {v}+s ) }f ) ( z ) }{ ( \mathfrak{D}_{q}^{ ( s ) }f ) ( z ) }- ( X-1 ) }{ ( L+1 ) \frac{z^{{v}} ( \mathfrak{D}_{q}^{ ( {v}+s ) }f ) ( z ) }{ ( \mathfrak{D}_{q}^{ ( s ) }f ) ( z ) }- ( X+1 ) }-1 \biggr) \geqq -1. $$
After some elementary and simple calculations, we deduce that
$$ \Re \biggl( \frac{2 [ ( \mathfrak{D}_{q}^{ ( s ) }f ) ( z ) -z^{{v}} ( \mathfrak{D}_{q}^{ ( {v}+s ) }f ) ( z ) ] }{ ( L+1 ) z^{{v} } ( \mathfrak{D}_{q}^{ ( {v}+s ) }f ) ( z ) - ( X+1 ) ( \mathfrak{D}_{q}^{ ( s ) }f ) ( z ) } \biggr) \geqq -1, $$
that is,
$$ -2\Re \biggl( \frac{\Upsilon _{1}z^{\mathfrak{p}-s}-\sum_{n=1}^{\infty }\Upsilon _{ ( 2,n ) }a_{n+\mathfrak{p}}z^{n+\mathfrak{p}-s}}{ \Upsilon _{4}z^{\mathfrak{p}-s}-\sum_{n=1}^{\infty }\Upsilon _{ ( 3,n ) }a_{n+\mathfrak{p}}z^{n+\mathfrak{p}-s}} \biggr) \geqq -1. $$
If we now choose that z lies on the real axis, then
$$ \frac{z^{{v}} ( \mathfrak{D}_{q}^{ ( {v}+s ) }f ) ( z ) }{ ( \mathfrak{D}_{q}^{ (s ) }f ) ( z ) } $$
assumes real values. In this case, upon letting \(z\rightarrow 1-\) along the real line, we get
$$ 2\Upsilon _{1}-\sum_{n=1}^{\infty }2 \Upsilon _{ ( 2,n ) } \vert a_{n+\mathfrak{p}} \vert \geqq - \vert \Upsilon _{4} \vert +\sum_{n=1}^{\infty } \vert \Upsilon _{ ( 3,n ) } \vert \vert a_{n+\mathfrak{p}} \vert , $$
(3.3)
where \(\Upsilon _{1}\), \(\Upsilon _{ (2,n ) }\), \(\Upsilon _{ ( 3,n ) }\), and \(\Upsilon _{4}\) are given by (2.4), (2.5), (2.6), and (2.7) respectively. We see that the last expression in (3.3) satisfies the inequality in (2.3). Hence our proof of Theorem 3 is now completed. □
Remark 4
First of all, if we put \({v}=s+1=\mathfrak{p}=1\) in Theorem 3, we get the corresponding result due to Srivastava et al. [33]. Secondly, if we put
$$ X=1-2\alpha \quad ( 0\leqq \alpha < 1 )\quad \text{and}\quad L=-1, $$
Theorem 3 gives the corresponding result that was proved by Khan et al. [8]. Thirdly, if we assign the following values to the parameters in Theorem 3:
$$ X=1-2\alpha\quad ( 0\leqq \alpha < 1 )\quad \text{and}\quad -L= {v}=s+1= \mathfrak{p}=1, $$
we have the following known result.
Corollary 2
(see [35, Theorem 8])
If \(0\leqq \alpha <1\), then
$$ \mathcal{TS}_{(q,1)}^{\ast } ( \alpha ) \equiv \mathcal{TS} _{(q,2)}^{\ast } ( \alpha ) \equiv \mathcal{TS}_{(q,3)}^{\ast } ( \alpha ) . $$
Corollary 3
Let the function f of the form (3.1) be in the class \(\mathcal{TS}_{q}^{\ast } [ j,\mathfrak{p} ,{v},s,X,L ]\) (\(j=1,2,3\)). Then
$$ a_{n+\mathfrak{p}}\leqq \frac{ \vert \Upsilon _{4} \vert -2\Upsilon _{1}}{ ( 2\Upsilon _{ ( 2,n ) }+ \vert \Upsilon _{ ( 3,n ) } \vert ) }. $$
(3.4)
The following function \(f_{t} ( z ) \)
$$ f_{t} (z ) =z^{\mathfrak{p}}- \frac{ \vert \Upsilon _{4} \vert -2\Upsilon _{1}}{ ( 2\Upsilon _{ ( 2,1 ) }+ \vert \Upsilon _{ ( 3,1 ) } \vert ) }z^{ \mathfrak{ p}+1} $$
(3.5)
is best possible, where \(\Upsilon _{1}\), \(\Upsilon _{ ( 2,n ) }\), \(\Upsilon _{ ( 3,n ) }\), and \(\Upsilon _{4}\) are given by (2.4), (2.5), (2.6), and (2.7), respectively.
By means of Theorem 3, it should be understood that Type 1, Type 2, and Type 3 of the multivalent q-starlike functions which involve the Janowski functions are similar. Therefore, for convenience, we state and prove the following distortion theorem by using the notation \(\mathcal{TS}_{q}^{\ast } [j,\mathfrak{p},{v},s,X,L ] \) in which it is tacitly assumed that \(j=1,2,3\).
Theorem 4
If \(f\in \mathcal{TS}_{q}^{\ast } [ j,\mathfrak{p} ,{v},s,X,L ]\) (\(j=1,2,3\)), then
$$ \bigl\vert f ( z ) \bigr\vert \geqq r^{\mathfrak{p}}- \biggl( \frac{ \vert \Upsilon _{4} \vert -2\Upsilon _{1}}{ ( 2\Upsilon _{ ( 2,1 ) }+ \vert \Upsilon _{ ( 3,1 ) } \vert ) } \biggr) r^{\mathfrak{p}+1} \quad (n\in \mathbb{N})\ \bigl\vert z^{ \mathfrak{p}} \bigr\vert =r^{\mathfrak{p}}\ (0< r< 1) $$
(3.6)
and
$$ \bigl\vert f ( z ) \bigr\vert \leqq r^{\mathfrak{p}}+ \biggl( \frac{ \vert \Upsilon _{4} \vert -2\Upsilon _{1}}{ ( 2\Upsilon _{ ( 2,1 ) }+ \vert \Upsilon _{ ( 3,1 ) } \vert ) } \biggr) r^{\mathfrak{p}+1} \quad (n\in \mathbb{N})\ \bigl\vert z^{ \mathfrak{p}} \bigr\vert =r^{\mathfrak{p}}\ (0< r< 1). $$
(3.7)
The equalities in (3.6) and (3.7) are attained for the function \(f (z ) \) given by (3.5) and where \(\Upsilon _{1}\), \(\Upsilon _{ ( 2,n )}\), \(\Upsilon _{ ( 3,n )}\), and \(\Upsilon _{4}\) are given by (2.4), (2.5), (2.6), and (2.7), respectively.
Proof
The following inequality can be easily deduced from Theorem 2:
$$ \bigl( 2\Upsilon _{ ( 2,1 ) }+ \vert \Upsilon _{ ( 3,1 ) } \vert \bigr) \sum_{n=1}^{\infty } \vert a_{n+ \mathfrak{p}} \vert \leqq \sum_{n=1}^{ \infty } \bigl( 2\Upsilon _{ ( 2,n ) }+ \vert \Upsilon _{ ( 3,n ) } \vert \bigr) \vert a_{n+ \mathfrak{p}} \vert < \vert \Upsilon _{4} \vert -2 \Upsilon _{1}, $$
which yields
$$ \bigl\vert f ( z ) \bigr\vert \leqq r^{\mathfrak{p} }+ \sum _{n=1}^{\infty } \vert a_{n+\mathfrak{p}} \vert r^{n+ \mathfrak{p}}\leqq r^{\mathfrak{p}}+r^{\mathfrak{p}+1}\sum _{n=1}^{ \infty } \vert a_{n+\mathfrak{p}} \vert \leqq r^{ \mathfrak{p}}+ \frac{ \vert \Upsilon _{4} \vert -2\Upsilon _{1}}{ ( 2\Upsilon _{ ( 2,1 ) }+ \vert \Upsilon _{ ( 3,1 ) } \vert ) }r^{ \mathfrak{p}+1}. $$
Similarly, we have
$$ \bigl\vert f ( z ) \bigr\vert \geq r^{\mathfrak{p} }- \sum _{n=1}^{\infty } \vert a_{n+\mathfrak{p}} \vert r^{n+ \mathfrak{p}}\geq r^{\mathfrak{p}}-r^{\mathfrak{p}+1}\sum _{n=1}^{ \infty } \vert a_{n+\mathfrak{p}} \vert \geq r^{ \mathfrak{p}}- \frac{ \vert \Upsilon _{4} \vert -2\Upsilon _{1}}{ ( 2\Upsilon _{ ( 2,1 ) }+ \vert \Upsilon _{ ( 3,1 ) } \vert ) }r^{ \mathfrak{p}+1}, $$
where \(\Upsilon _{1}\), \(\Upsilon _{ ( 2,n )}\), \(\Upsilon _{ ( 3,n ) }\), and \(\Upsilon _{4}\) are given by (2.4), (2.5), (2.6), and (2.7), respectively. The proof of Theorem 4 is now completed. □
Remark 5
First of all, if we put \({v}=s+1=\mathfrak{p}=1\) in Theorem 1, we get the corresponding result due to Srivastava et al. [33]. Secondly, if we put
$$ X=1-2\alpha \quad ( 0\leqq \alpha < 1 )\quad \text{and}\quad L=-1, $$
Theorem 4 will give the corresponding result that was proved by Khan et al. [8]. Thirdly, if we put
$$ X=1-2\alpha \quad ( 0\leqq \alpha < 1 ) \quad \text{and}\quad -L= {v}=s+1= \mathfrak{p}=1 $$
in Theorem 4 and let \(q\longrightarrow 1{-}\), we have the following known result.
Corollary 4
(see [23])
If \(f\in \mathcal{TS}^{\ast } ( \alpha ) \), then
$$ r- \biggl( \frac{1-\alpha }{2-\alpha } \biggr) r^{2}\leqq \bigl\vert f ( z ) \bigr\vert \leqq r+ \biggl( \frac{1-\alpha }{2-\alpha } \biggr) r^{2} \quad \bigl( \vert z \vert =r\ (0< r< 1) \bigr). $$
The next theorem (Theorem 5) can be proven similarly as we proved Theorem 4, so we choose to omit the details involved in our proof of Theorem 5.
Theorem 5
If \(f\in \mathcal{TS}_{q}^{\ast } [ j,\mathfrak{p}, {v},s,X,L ]\) (\(j=1,2,3\)), then
$$ \bigl\vert f^{\prime } ( z ) \bigr\vert \geqq \mathfrak{p}r^{ \mathfrak{p}-1}- \biggl( \frac{ ( \mathfrak{p}+1 ) ( \vert \Upsilon _{4} \vert -2\Upsilon _{1} ) }{ ( 2\Upsilon _{ ( 2,1 ) }+ \vert \Upsilon _{ ( 3,1 ) } \vert ) } \biggr) r^{\mathfrak{p}} \quad (n\in \mathbb{N})\ \bigl\vert z^{ \mathfrak{p} } \bigr\vert =r^{\mathfrak{p}}\ (0< r< 1) $$
and
$$ \bigl\vert f^{\prime } ( z ) \bigr\vert \leqq \mathfrak{p}r^{ \mathfrak{p}-1}+ \biggl( \frac{ ( \mathfrak{p}+1 ) ( \vert \Upsilon _{4} \vert -2\Upsilon _{1} ) }{ ( 2\Upsilon _{ ( 2,1 ) }+ \vert \Upsilon _{ ( 3,1 ) } \vert ) } \biggr) r^{\mathfrak{p}} \quad (n\in \mathbb{N})\ \bigl\vert z^{ \mathfrak{ p}} \bigr\vert =r^{\mathfrak{p}}\ (0< r< 1). $$
The function \(f_{t} ( z ) \) given by (3.5) is the extremal function.
If in Theorem 5 we take
$$ X=1-2\alpha\quad ( 0\leqq \alpha < 1 )\quad \text{and} \quad -L= {v}=s+1= \mathfrak{p}=1, $$
and let \(q\longrightarrow 1{-}\), we get the following known result.
Corollary 5
(see [23])
If \(f\in \mathcal{TS}^{\ast } ( \alpha ) \), then
$$ 1- \biggl( \frac{2 ( 1-\alpha ) }{2-\alpha } \biggr) r \leqq \bigl\vert f^{\prime } ( z ) \bigr\vert \leqq 1+ \biggl( \frac{ 2 ( 1-\alpha ) }{2-\alpha } \biggr) r \quad \bigl( \vert z \vert =r\ (0< r< 1) \bigr). $$
Finally, for the class \(\mathcal{TS}_{q}^{\ast } [ j, \mathfrak{p},{v},s,X,L ]\) (\(j=1,2,3\)) of multivalent q-starlike functions with negative coefficients, the results related to the radii of close-to-convexity, starlikeness, and convexity are deduced.
Theorem 6
Let \(f\in \mathcal{TS}_{q}^{\ast } [ j,\mathfrak{p},{v} ,s,X,L ]\) (\(j=1,2,3\)). Then, for \(\vert z \vert \leqq r_{0} ( j, \mathfrak{p},n,X,L, \chi )\), the function f is \(\mathfrak{p}\)-valent close-to-convex of order χ with (\(0\leqq \chi < \mathfrak{p} \)), where
$$ r_{0}=\inf_{n\geq 1} \biggl[ \frac{ ( 2\Upsilon _{ ( 2,n ) }+ \vert \Upsilon _{ ( 3,n ) } \vert ) ( \mathfrak{p}-\chi ) }{ ( \vert \Upsilon _{4} \vert -2\Upsilon _{1} ) ( n+\mathfrak{p} ) } \biggr] ^{\frac{1}{n}}. $$
(3.8)
The function \(f_{t} ( z ) \) given by (3.5) is best possible.
Proof
Using Theorem 2 in conjunction with (3.1), for \(\vert z \vert < r_{0}\), we find that
$$ \biggl\vert \frac{f^{\prime } ( z ) }{z^{\mathfrak{p}-1}}- \mathfrak{p } \biggr\vert < \mathfrak{p}- \chi \quad \bigl( \vert z \vert \leqq r_{0} \bigr) . $$
We have thus completed the proof of Theorem 6. □
Remark 6
If we put
$$ X=1-2\alpha\quad ( 0\leqq \alpha < 1 ) \quad \text{and}\quad L=-1, $$
Theorem 6 gives the corresponding result that was proved by Khan et al. [8].
Theorem 7
Let \(f\in \mathcal{TS}_{q}^{\ast } [ j,\mathfrak{p},{v} ,s,X,L ]\) (\(j=1,2,3\)). Then, for \(\vert z \vert \leqq r_{1} ( j,\mathfrak{p} ,n,X,L, \chi )\), the function f is a \(\mathfrak{p}\)-valent starlike of order χ with (\(0\leqq \chi <\mathfrak{p} \)), where
$$ r_{1}=\inf_{n\geq 1} \biggl[ \frac{ ( 2\Upsilon _{ ( 2,n ) }+ \vert \Upsilon _{ ( 3,n ) } \vert ) ( \mathfrak{p}-\chi ) }{ ( \vert \Upsilon _{4} \vert -2\Upsilon _{1} ) ( n+\mathfrak{p}-\chi ) } \biggr] ^{\frac{1 }{n}}. $$
(3.9)
The result is sharp for the function \(f_{t} ( z ) \) given by (3.5).
Proof
Using arguments similar to those in the proof of Theorem 6, it can be seen that
$$ \biggl\vert \frac{zf^{\prime } ( z ) }{f ( z ) }- \mathfrak{p } \biggr\vert < \mathfrak{p}- \chi\quad \bigl( \vert z \vert \leqq r_{1} \bigr) , $$
which evidently proves Theorem 7. □
Remark 7
If we put
$$ X=1-2\alpha\quad ( 0\leqq \alpha < 1 ) \quad \text{and}\quad L=-1, $$
Theorem 7 gives the corresponding result that was proved by Khan et al. [8].
Corollary 6
Let \(f\in \mathcal{TS}_{q}^{\ast } [ j,\mathfrak{p},{v} ,s,X,L ]\) (\(j=1,2,3\)). Then, for \(\vert z \vert \leqq r_{2} ( j,\mathfrak{p} ,n,X,L, \chi )\), the function f is a \(\mathfrak{p}\)-valent convex of order χ with (\(0\leqq \chi <\mathfrak{p} \)) where
$$ r_{2}=\inf_{n\geq 1} \biggl[ \frac{ ( 2\Upsilon _{ ( 2,n ) }+ \vert \Upsilon _{ ( 3,n ) } \vert ) \mathfrak{p} ( \mathfrak{p}-\chi ) }{ ( \vert \Upsilon _{4} \vert -2\Upsilon _{1} ) ( n+\mathfrak{p} ) (n+ \mathfrak{p}-\chi ) } \biggr] ^{\frac{1}{n}}. $$
(3.10)
The result is sharp for the function \(f_{t} ( z ) \) given by (3.5).