# Coefficient inequalities for certain subclasses of multivalent functions associated with conic domain

## Abstract

A number of families of q-extensions of analytic functions in the open unit disk $$\mathbb{U}$$ have been defined by means of basic (or q-)calculus and considered from many distinctive prospectives and viewpoints. In this paper, we generalize and study certain subclasses of analytic functions involving higher-order q-derivative operators. We settle characteristic equations for these presumably new classes and also study numerous coefficient inequalities. For the results obtained in this presentation, we also carry out appropriate connections with those in multiple other concerning works on this subject.

## Introduction and definitions

By $$\mathcal{A}(p)$$ we denote the class of functions with series representation

$$f(z)=z^{p}+\sum_{n=1}^{\infty }a_{n+p}z^{n+p}\quad \bigl(p\in \mathbb{N}=\{1,2,\ldots\}\bigr),$$
(1.1)

which is analytic and p-valent in the open unit disk

$$\mathbb{U}= \bigl\{ z:z\in {\mathbb{C} } \text{ and } \vert z \vert < 1 \bigr\} .$$

In particular, we denote

$$\mathcal{A}:=\mathcal{A}(1).$$

Moreover, we denote by $$\mathcal{S}\subset \mathcal{A}$$ the class of all univalent functions in the unit disk $$\mathbb{U}$$.

For two analytic functions $$f_{j}$$, $$j=1,2$$, in $$\mathbb{U}$$, a function $$f_{1}$$ is said to be subordinate to the function $$f_{2}$$ and write as

$$f_{1}\prec f_{2}\quad \text{or}\quad f_{1} ( z ) \prec f_{2} ( z )$$

if in $$\mathbb{U}$$, we can find an analytic Schwarz function w with

$$w ( 0 ) =0\quad \text{and}\quad \bigl\vert w ( z ) \bigr\vert < 1$$

such that

$$f_{1} ( z ) =f_{2} \bigl( w ( z ) \bigr) .$$

Further, if the function $$f_{2}$$ is univalent in $$\mathbb{U}$$ then following equivalence relation holds true

$$f_{1}(z)\prec f_{2}(z)\quad (z\in \mathbb{U})\quad \Rightarrow\quad f_{1}(0)=f_{2}(0) \quad \text{and}\quad f_{1}( \mathbb{U})\subset f_{2}(\mathbb{U}).$$

The noteworthy class of Carathéodory functions $$\mathcal{P}$$ consists of all analytic functions ψ in $$\mathbb{U}$$ normalized by

$$\psi ( z ) =1+\sum_{n=1}^{\infty }\psi _{n}z^{n}$$
(1.2)

and satisfying

$$\Re \bigl( \psi ( z ) \bigr) >0 \quad ( \forall z\in \mathbb{U} ) .$$

For a convex function f, it is always true that the image of f under $$\mathbb{U}$$ and all circles lying within $$\mathbb{U}$$ centered at the origin are convex arcs. But justification is required whether the characteristic still holds for circles with center at any other point, say ξ. Goodman [4, 5] answered negatively and defined uniformly convex and starlike functions that have this nice characteristic. Analytically, he defined uniformly convex and starlike functions, respectively, as

$$\Re \biggl\{ 1+\frac{(z-\xi )f^{\prime \prime }(z)}{f^{\prime }(z)} \biggr\} >0 \quad (z\in \mathbb{U})$$

and

$$\Re \biggl\{ 1+\frac{(z-\xi )f^{\prime }(z)}{f^{(}z)-f^{(}\xi )} \biggr\} >0 \quad (z\in \mathbb{U}).$$

We denote the former by $$\mathcal{U}\mathcal{CV}$$ and the later by $$\mathcal{U}\mathcal{S}^{*}\mathcal{T}$$. It is natural to ask whether classical Alexander’s result holds for these two classes, but there are counterexamples , which show that the relation is not true for these classes. Rønning , using $$\mathcal{U}\mathcal{CV}$$, introduced the class

$$\mathcal{S}^{*}\mathcal{T}= \bigl\{ f\in \mathcal{A}:f(z)=zg^{\prime }(z), g(z)\in \mathcal{U}\mathcal{CV} \bigr\}$$

and succeeded in proving that neither $$\mathcal{S}^{*}\mathcal{T}\not \subset \mathcal{U}\mathcal{S}^{*}\mathcal{T}$$ nor $$\mathcal{U}\mathcal{S}^{*}\mathcal{T}\not \subset \mathcal{S}^{*}\mathcal{T}$$. Ultimately, Rønning  and Ma and Minda  introduced the following one-variable characterization of these classes.

### Definition 1

Let $$f\in \mathcal{A}$$. Then $$f\in \mathcal{U}\mathcal{CV}$$ if

$$\Re \biggl\{ \frac{ (zf^{\prime }(z) )^{\prime }}{f^{\prime }(z)} \biggr\} > \biggl\vert \frac{zf^{\prime \prime }(z)}{f^{\prime }(z)} \biggr\vert \quad (z\in \mathbb{U}).$$

### Definition 2

Let $$f\in \mathcal{A}$$. Then $$f\in \mathcal{S}^{*}\mathcal{T}$$ if

$$\Re \biggl\{ \frac{zf^{\prime }(z)}{f(z)} \biggr\} > \biggl\vert \frac{zf^{\prime }(z)}{f(z)}-1 \biggr\vert \quad (z\in \mathbb{U}).$$

This led to the basis for conic domains introduced by Kanas and Wiśniowska [10, 11] as

$$\varOmega _{k}= \bigl\{ u+iv:u>k\sqrt{ ( u-1 ) ^{2}+v^{2}} \bigr\} ,\quad k\geqq 0.$$
(1.3)

These domains represent the right half-plane, a parabola, a hyperbola, and an ellipse for $$k = 0$$, $$k = 1$$, $$0 < k < 1$$, and $$k > 1$$, respectively.

The role of an extremal function is done by the function $$p_{k}(z)$$ given by

$$p_{k}(z)= \textstyle\begin{cases} \frac{1+z}{1-z} & (k=0), \\ 1+\frac{2}{\pi ^{2}} ( \log \frac{1+\sqrt{z}}{1-\sqrt{z}} ) ^{2} & (k=1), \\ 1+\frac{2}{1-k^{2}} \sinh ^{2} \{ ( \frac{2}{\pi } \arccos k ) \arctan h\sqrt{z} \} & (0\leq k< 1), \\ 1+\frac{1}{k^{2}-1} [1+\sin ( \frac{\pi }{2K(\kappa )} \int _{0}^{\frac{u(z)}{\sqrt{\kappa }}} \frac{dt}{\sqrt{1-t^{2}}\sqrt{1-\kappa ^{2}t^{2}}} ) ] & (k>1) ,\end{cases}$$
(1.4)

where

$$u(z)=\frac{\sqrt{\kappa }-z}{\sqrt{\kappa }z-1}\quad ( \forall z\in \mathbb{U} ),$$

and $$\kappa \in (0,1)$$ is chosen so that

$$k=\cosh \bigl(\pi K^{\prime }(\kappa )/\bigl(4K(\kappa )\bigr)\bigr),$$

where $$K(\kappa )$$ denotes the first-kind Legendre complete elliptic integral, and its derivative is given by

$$K^{\prime }(\kappa )=K\bigl(\sqrt{1-\kappa ^{2}}\bigr)$$

and called the complementary integral of $$K ( t )$$. If we set the function

$$p_{k} ( z ) =1+\delta _{k}z+\cdots ,$$

subsequently, by  it is obvious that from (1.4) we have

$$\delta _{k}= \textstyle\begin{cases} \frac{8 ( \arccos k ) ^{2}}{\pi ( 1-k^{2} ) } & ( 0\leq k< 1 ), \\ \frac{8}{\pi ^{2}} & ( k=1 ), \\ \frac{\pi ^{2}}{4k^{2} ( 1+t ) \sqrt{t}R^{2} ( t ) } & ( k>1 ) . \end{cases}$$
(1.5)

Subjected to the above-mentioned conic domain, we define some elementary classes.

### Definition 3

Let f be a function from the functional class $$\mathcal{A}$$. Then $$f\in k$$-$$\mathcal{U} \mathcal{CV}$$ if

$$\Re \biggl\{ \frac{ (zf^{\prime }(z) )^{\prime }}{f^{\prime }(z)} \biggr\} >k \biggl\vert \frac{zf^{\prime \prime }(z)}{f^{\prime }(z)} \biggr\vert \quad ( \forall z\in \mathbb{U} \text{ and }k\geq 0 ) .$$

### Definition 4

A normalized analytic function f belongs to the class k-$$\mathcal{S}^{*}\mathcal{T}$$ if

$$\Re \biggl\{ \frac{zf^{\prime }(z)}{f(z)} \biggr\} >k \biggl\vert \frac{zf^{\prime }(z)}{f(z)}-1 \biggr\vert \quad ( \forall z\in \mathbb{U}\text{ and }k \geq 0 ) .$$

### Definition 5

Let f be a function from the functional class $$\mathcal{A}$$. Then $$f\in k$$-$$\mathcal{UQ}$$ if there exists a function $$g\in k$$-$$\mathcal{U} \mathcal{CV}$$ such that

$$\Re \biggl\{ \frac{ (zf^{\prime }(z) )^{\prime }}{g^{\prime }(z)} \biggr\} >k \biggl\vert \frac{zf^{\prime \prime }(z)}{g^{\prime }(z)} \biggr\vert \quad ( \forall z\in \mathbb{U} \text{ and }k\geq 0 ) .$$

### Definition 6

A normalize analytic function f belongs to the class k-$$\mathcal{UK}$$ if there exists a function $$g\in k$$-$$\mathcal{S}^{*}\mathcal{T}$$ such that

$$\Re \biggl\{ \frac{zf^{\prime }(z)}{g(z)} \biggr\} >k \biggl\vert \frac{zf^{\prime }(z)}{g(z)}-1 \biggr\vert \quad ( \forall z\in \mathbb{U}\text{ and }k \geq 0 ) .$$

Now we recall some firm footing concept details and definitions of the q-difference calculus, which play a vital role in our presentation. Unless otherwise notified, we presume that $$0< q<1$$ and $$p\in \mathbb{N}= \{ 1,2,3,\ldots \}$$. For a nonnegative number λ, the q-number $$[ \lambda ] _{q}$$ is defined by

$$[ \lambda ] _{q}=\sum_{j=0}^{j-1}q^{j}=1+q+q^{2}+ \cdots +q^{j-1}, \qquad [ 0 ] _{q}=0.$$

In general, for $$\lambda \in \mathbb{C}$$, we have $$[ \lambda ] _{q}=\frac{1-q^{\lambda }}{1-q}$$. The q-factorials $$[ j ] _{q}!$$ are defined by $$ _{q}!=0$$ and $$[j ] _{q}!=\prod_{k=1}^{j} [ k ] _{q}$$. It is straightforward to observe that $$\lim_{q\rightarrow 1-} [\lambda ]_{q}=\lambda$$ and $$\lim_{q\rightarrow 1-} [j]_{q}!=j!$$.

### Definition 7

(See  and )

For a function f from class $$\mathcal{A}$$, the q-derivative (or q-difference) operator $$D_{q}$$ in a subset of complex numbers $$\mathbb{C}$$ is defined by

$$( D_{q}f ) ( z ) = \textstyle\begin{cases} \frac{f ( z ) -f ( qz ) }{ ( 1-q ) z} & ( z\neq0 ), \\ f^{\prime } ( 0 ) & ( z=0 ) ,\end{cases}$$
(1.6)

provided that $$f^{\prime } ( 0 )$$ exists.

We observe from Definition 7 that

$$\lim_{q\rightarrow 1-} ( D_{q}f ) ( z ) = \lim _{q\rightarrow 1-} \frac{f ( z ) -f ( qz ) }{ ( 1-q ) z}=f^{\prime } ( z )$$

for a differentiable function f in a subset of $$\mathbb{C}$$. Further, by (1.1) and (1.6) we obtain

\begin{aligned}& \bigl( D_{q}^{ ( 1 ) }f \bigr) ( z ) = [ p ] _{q}z^{p-1}+\sum_{n=1}^{\infty } [ n+p ] _{q}a_{n+p}z^{n+p-1}, \end{aligned}
(1.7)
\begin{aligned}& \bigl( D_{q}^{ ( 2 ) }f \bigr) ( z ) = [ p ] _{q} [ p-1 ] _{q}z^{p-2}+\sum _{n=1}^{ \infty } [ n+p ] _{q} [ n+p-1 ] _{q}a_{n+p}z^{n+p-2}, \end{aligned}
(1.8)
\begin{aligned}& \textstyle\begin{array}{@{}ccc@{}} \quad \qquad \quad \ \ \cdot & \qquad \qquad \cdot & \qquad \qquad \cdot \\ \quad \qquad \quad \ \ \cdot &\qquad \qquad \cdot & \qquad \qquad \cdot \\ \quad \qquad \quad \ \ \cdot & \qquad \qquad \cdot & \qquad \qquad \cdot \end{array}\displaystyle \\& \bigl( D_{q}^{ ( p ) }f \bigr) ( z ) = [ p ] _{q}!+\sum_{n=1}^{\infty } \frac{ [ n+p ] _{q}!}{ [ n ] _{q}!}a_{n+p}z^{n}, \end{aligned}
(1.9)

where $$( D_{q}^{ ( p ) }f ) ( z )$$ is the pth q-derivative of $$f ( z )$$.

Recently, the studies of q-calculus have inspired an intense interest of researchers because of its advantages in many areas of mathematics and physics. The significance of the operator $$D_{q}$$ is quite obvious by its applications in the study of several subclasses of analytic functions. Initially, in 1990, Ismail et al.  gave the idea of q-extension of the class of starlike functions; nevertheless, a foothold usage of the q-calculus in the context of geometric function theory was effectively invoked by Srivastava . After that, wonderful studies have been done by numerous mathematicians offering a momentous part in the advancement of geometric function theory. In particular, the study the q-Mittag-Leffler functions for close-to-convex functions was done by Srivastava and Bansal  (see also ). In , they also considered the functional class of q-starlike functions related to conic region $$\sigma _{k}$$, where the estimate of the third Hankel determinant has been settled in  (see also ). Recently, Srivastava et al. (see, e.g., [15, 31, 34, 35] published a set of papers, in which they concentrated on the class of q-starlike functions related to the Janowski functions from different aspects. For some more recent investigations on q-calculus, we refer to [13, 16, 27, 29, 33]. In this paper, we mainly generalize the work presented in Srivastava et al. .

### Definition 8

(See )

A function $$f\in \mathcal{A}$$ belongs to the functional class $$\mathcal{S}_{q}^{\ast }$$ if

$$f ( 0 ) =f^{\prime } ( 0 ) -1=0$$
(1.10)

and

$$\biggl\vert \frac{z}{f ( z ) } ( D_{q}f ) z- \frac{1}{1-q} \biggr\vert \leq \frac{1}{1-q}.$$
(1.11)

In view of the last inequality, it is obvious that, in the limiting case $$q\rightarrow 1-$$,

$$\biggl\vert w-\frac{1}{1-q} \biggr\vert \leq \frac{1}{1-q},$$

the above closed disk is merely the right-half plane, and the class $$\mathcal{S}_{q}^{\ast }$$ of q-starlike functions turns into the prominent class $$\mathcal{S}^{\ast }$$. Analogously, by the principle of subordination we may express relations (1.10) and (1.11) as follows (see ):

$$\frac{z}{f ( z ) } ( D_{q}f ) ( z ) \prec \widehat{p} ( z )\quad \biggl( \widehat{p} ( z ) =\frac{1+z}{1-qz} \biggr) .$$

The notation $$\mathcal{S}_{q}^{\ast }$$was first used by Sahoo et al. .

### Remark 1

In defining the class $${\mathcal{C}}_{q}$$ of q-convex functions, the most important Alexander theorem  for functions $$f\in {\mathcal{A}}$$ was used by Baricz and Swaminathan  as

$$f ( z ) \in {\mathcal{C}}_{q}\quad \iff\quad z ( D_{q}f ) ( z ) \in {\mathcal{S}}_{q}^{\ast }.$$

The generalization of $$\mathcal{P} (p_{k} )$$ to k-$$\mathcal{P}_{q}$$ presented in Definition 9 is due to Srivastava et al. . He used the conic domain and the earlier discussed q-calculus as follows.

### Definition 9

A function $$\psi \in \mathcal{P}$$ belongs to the class k-$$\mathcal{P }_{q}$$ if the following relation holds:

$$\psi ( z ) \prec \frac{2p_{k} ( z ) }{ ( 1+q ) + ( 1-q ) p_{k} ( z ) }$$

with the function $$p_{k} ( z )$$ given in equation (1.4).

It is interesting that in geometric characteristics the function $$\psi \in k$$-$$\mathcal{P}_{q}$$ takes all values from the domain $$\varOmega _{k,q}$$, $$k\geq 0$$, analytically given by

$$\varOmega _{k,q}= \biggl\{ w:\Re \biggl( \frac{ ( 1+q ) w}{ ( q-1 ) w+2} \biggr) >k \biggl\vert \frac{ ( 1+q ) w}{ ( q-1 ) w+2}-1 \biggr\vert \biggr\} ,$$

which represent q-analogues of generalized conic regions.

Note the following easily observable facts about the class k-$$\mathcal{P}_{q}$$.

### Remark 2

Firstly, we see that

$$k\text{-}\mathcal{P}_{q}\subseteq \mathcal{P} \biggl[ \frac{2k}{2k+1+q} \biggr] ,$$

where $$\mathcal{P} [ \frac{2k}{2k+1+q} ]$$ is the famous class of functions with real parts greater than $$\frac{2k}{2k+1+q}$$. Secondly, we have

$$\lim_{q\longrightarrow 1-}k\text{-}\mathcal{P}_{q}= \mathcal{P} ( p_{k} ) ,$$

where $$\mathcal{P} ( p_{k} )$$ is the familiar class given by Wisniowska and Kanas . Thirdly, by taking the limit we obtain

$$\lim_{q\longrightarrow 1-}0\text{-}\mathcal{P}_{q}= \mathcal{P} ,$$

where $$\mathcal{P}$$ is the class of functions given by (1.2).

Using the q-differential operator, various new classes have been defined. Hence it is natural to give the following definition.

### Definition 10

A function $$f\in \mathcal{A}$$ is said to belong to the class k-$$\mathcal{S}_{q}^{\ast } ( p )$$ if

$$\Re \biggl( \frac{ ( 1+q ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}f ) ( z ) }}{ ( q-1 ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}f ) ( z ) }+2} \biggr) >k \biggl\vert \frac{ ( 1+q ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}f ) ( z ) }}{ ( q-1 ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}f ) ( z ) }+2}-1 \biggr\vert$$

or, equivalently,

$$\frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}f ) ( z ) }\in k\text{-}\mathcal{P}_{q}.$$

### Remark 3

First of all, it is straightforward to see that

$$0\text{-}\mathcal{S}_{q}^{\ast } ( 1 ) = \mathcal{S}_{q}^{ \ast },$$

where the functional class $$\mathcal{S}_{q}^{\ast }$$ was considered and analyzed by Ismail et al. . Secondly, we easily observe that

$$\lim_{q\longrightarrow 1-}k\text{-}\mathcal{S}_{q}^{\ast } ( 1 ) =k\text{-}\mathcal{S}^{\ast }=k\text{-} \mathcal{S}^{*}\mathcal{T},$$

where the class k-$$\mathcal{S}^{*}\mathcal{T}$$ was presented and studied by Kanas and Wiśniowska . Thirdly,

$$k\text{-}\mathcal{S}_{q}^{\ast } ( 1 ) =k\text{-} \mathcal{S}_{q}^{\ast },$$

where the function class k-$$\mathcal{S}_{q}^{\ast }$$ was initially considered and studied by Srivastava et al. . Finally,

$$\lim_{q\longrightarrow 1-}0\text{-}\mathcal{S}_{q}^{\ast } ( 1 ) =\mathcal{S}^{\ast },$$

where $$\mathcal{S}^{\ast }$$ is the essential class of starlike functions.

### Definition 11

Just as in Remark 1, by the Alexander theorem  the class k-$$\mathcal{C}_{q}$$ can be defined by the following relation:

$$f ( z ) \in k\text{-}\mathcal{C}_{q} ( p ) \quad \iff\quad \frac{z}{p!} \bigl( D_{q}^{ ( p ) }f \bigr) ( z ) \in k \text{-}\mathcal{S}_{q}^{\ast } ( p ) .$$

### Definition 12

Any function $$f\in \mathcal{A} ( p )$$ is said to belong to the class k-$$\mathcal{K}_{q} ( p )$$ if

$$\Re \biggl( \frac{ ( 1+q ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}g ) ( z ) }}{ ( q-1 ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}g ) ( z ) }+2} \biggr) >k \biggl\vert \frac{ ( 1+q ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}g ) ( z ) }}{ ( q-1 ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}g ) ( z ) }+2}-1 \biggr\vert$$

or, equivalently,

$$\frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}g ) ( z ) }\in k\text{-}\mathcal{P}_{q}$$

for some $$g\in k$$-$$\mathcal{S}_{q}^{\ast } ( p )$$.

### Definition 13

In similar manner as in Remark 1, by using the idea of Alexander’s relation  we define the class k-$$\mathcal{C}_{q}^{\ast } ( p )$$ by the following relation:

$$f ( z ) \in k\text{-}\mathcal{C}_{q}^{\ast } ( p ) \quad \iff\quad \frac{z}{p!} \bigl( D_{q}^{ ( p ) }f \bigr) ( z ) \in k\text{-}\mathcal{K}_{q} ( p ) .$$

### Remark 4

First of all, we can see that

$$0\text{-}\mathcal{K}_{q} ( 1 ) =\mathcal{K}_{q},$$

where $$\mathcal{K}_{q} ( p )$$ is the function class defined and examined by Raghavendar et al. . Secondly, in the limit case, we have

$$\lim_{q\longrightarrow 1-}k\text{-}\mathcal{K}_{q} ( 1 ) =k \text{-}\mathcal{K}=k\text{-}\mathcal{UK}\quad \text{and}\quad \lim_{q\longrightarrow 1-}k\text{-} \mathcal{C}_{q}^{\ast } ( 1 ) =k\text{-} \mathcal{C}^{\ast }=k\text{-}\mathcal{UQ},$$

where k-$$\mathcal{UK} ( p )$$ and k-$$\mathcal{UQ} ( p )$$ are the function classes introduced and studied by Acu . Thirdly, we have

$$\lim_{q\longrightarrow 1-}0\text{-}\mathcal{K}_{q} ( 1 ) = \mathcal{K}\quad \text{and}\quad \lim_{q\longrightarrow 1-}0 \text{-}\mathcal{C}_{q}^{\ast } ( 1 ) = \mathcal{C}^{\ast },$$

where $$\mathcal{C}^{\ast }$$ and $$\mathcal{K}$$ are the function classes of quasi-convex and close-to-convex functions; for details, see [12, 19].

## A set of lemmas

Each of the lemmas given further will be helpful in demonstrating our main results.

### Lemma 1

()

Letψbe a function of the form

$$\psi ( z ) =1+\sum_{n=1}^{\infty }\psi _{n}z^{n}$$

subordinate to a function$$\mathfrak{H}$$of the form

$$\mathfrak{H} ( z ) =1+\sum_{n=1}^{\infty }C_{n}z^{n}.$$

In particular, when$$\mathfrak{H}$$is univalent in the unit disk$$\mathbb{U}$$and$$\mathcal{H}(\mathbb{U})$$is convex, then

$$\vert \psi _{n} \vert \leq \vert C_{1} \vert \quad ( n\in \mathbb{N} ) .$$

### Lemma 2

Suppose that the sequence $$\{ a_{k} \} _{k=0}^{\infty }$$ is defined by

$$a_{p}=1$$

and

$$a_{n+p}= \frac{\delta _{k} ( q+1 ) [n+1]_{q}!}{2[n+p]_{q}! ( [ n+1 ] _{q}-1 ) } \sum_{l=1}^{n-1} \frac{[p+l]_{q}!}{[l+1]_{q}!}a_{p+l}.$$
(2.1)

Then

$$a_{n+p-1}=\prod_{j=2}^{n} \frac{[j]_{q}\{2([j]_{q}-1)+\delta _{k}(q+1)\}}{2\{[j]_{q}-1\}[j+p-1]_{q}}.$$

### Proof

By (2.1) we easily get

$$\frac{{}[ n+p]_{q}!}{[n+1]_{q}!} \bigl( [ n+1 ] _{q}-1 \bigr) a_{n+p}=\sum_{l=1}^{n} \frac{[n+p-l]_{q}!}{[n+1-l]_{q}!}a_{n+p-l}c_{l}$$
(2.2)

and

$$\frac{[ n+p-1]_{q}!}{[n+1]_{q}!} \bigl( [ n+1 ] _{q}-1 \bigr) a_{n+p-1}=\sum_{l=1}^{n} \frac{[n+p-1-l]_{q}!}{[n+1-l]_{q}!}a_{n+p-1-l}c_{l}.$$
(2.3)

Combining (2.2) and (2.3), we obtain

$$\frac{a_{n+p}}{a_{n+p-1}}= \frac{[n+1]_{q}\{2([n]_{q}-1)+\delta _{k}(1+q)\}}{2\{ \{[n+1]_{q}-1\}[n+p]_{q}}.$$

Similarly, we deduce the following result:

$$a_{n+p-1}=\frac{a_{n+p-1}}{a_{n+p-2}}\cdot \frac{a_{n+p-2}}{a_{n+p-3}}\cdot\ldots\cdot \frac{a_{p+2}}{a_{p+1}}\cdot \frac{a_{p+1}}{a_{p}}\cdot a_{p}.$$

This completes the proof of Lemma 2. □

## Main results

In this section, we prove our main results. We assume that

$$k\geq 0\quad \text{and}\quad q\in ( 0,1 ) .$$

### Theorem 1

Letfbe ap-valently analytic function of the form (1.1). Thenfbelongs to the classk-$$\mathcal{S}_{q}^{\ast }$$if it satisfies the condition

$$\sum_{n=1}^{\infty }\frac{[n+p]_{q}!}{[n+1]_{q}!} \varLambda _{3} \vert a_{n+p} \vert < ( 1+q ),$$
(3.1)

where

$$\varLambda _{3}=2 ( k+1 ) q\bigl( [ n+1 ] _{q}-1\bigr)+ \bigl\vert ( q-1 ) [ n+1 ] _{q}+2 \bigr\vert .$$
(3.2)

### Proof

If (3.1) holds, then it suffices to establish the inequality

$$k \biggl\vert \frac{ ( 1+q ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{D_{q}^{(p-1)}f ( z ) }}{ ( q-1 ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{D_{q}^{(p-1)}f ( z ) }+2}-1 \biggr\vert -\Re \biggl[ \frac{ ( 1+q ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{D_{q}^{(p-1)}f ( z ) }}{ ( q-1 ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{D_{q}^{(p-1)}f ( z ) }+2}-1 \biggr] < 1.$$

Since

\begin{aligned} & k \biggl\vert \frac{ ( 1+q ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}f ) ( z ) }}{ ( q-1 ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}f ) ( z ) }+2}-1 \biggr\vert -\Re \biggl[ \frac{ ( 1+q ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}f ) ( z ) }}{ ( q-1 ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}f ) ( z ) }+2}-1 \biggr] \\ &\quad \leq ( k+1 ) \biggl\vert \frac{ ( 1+q ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}f ) ( z ) }}{ ( q-1 ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}f ) ( z ) }+2}-1 \biggr\vert \\ &\quad =2 ( k+1 ) \biggl\vert \frac{z ( D_{q}^{(p)}f ) ( z ) - ( D_{q}^{(p-1)}f ) ( z ) }{ ( q-1 ) z ( D_{q}^{(p)}f ) ( z ) +2 ( D_{q}^{(p-1)}f ) ( z ) } \biggr\vert \\ &\quad =2 ( k+1 ) \biggl\vert \frac{\sum_{n=1}^{\infty }\frac{[n+p]_{q}!}{[n+1]_{q}!} ( [n+1]_{q}-1 ) a_{n+p}z^{n+1}}{ ( q+1 ) [p]_{q}!+\sum_{n=1}^{\infty }\frac{[n+p]_{q}!}{[n+1]_{q}!} \{ ( q-1 ) [ n+1 ] _{q}+2 \} a_{n+p}z^{n+1}} \biggr\vert \\ &\quad \leq 2 \frac{\sum_{n=1}^{\infty }\frac{[n+p]_{q}!}{[n+1]_{q}!}([n+1]_{q}-1) \vert a_{n+p} \vert }{ ( q+1 ) +\sum_{n=2}^{\infty }\frac{[n+p]_{q}!}{[n+1]_{q}!} \vert ( q-1 ) [ n+1 ] _{q}+2 \vert \vert a_{n+p} \vert }, \end{aligned}
(3.3)

the upper bound of the relation given by (3.3) is unity if

$$\sum_{n=1}^{\infty }\frac{[n+p]_{q}!}{[n+1]_{q}!} \varLambda _{3}\cdot \vert a_{n+p} \vert < ( 1+q ) ,$$

where $$\varLambda _{3}$$ is given by (3.2). Consequently, proof is completed. □

If in Theorem 3, we put $$p=1$$ and $$q\longrightarrow 1-$$, then we get the following result.

### Corollary 1

(See )

Any function$$f\in \mathcal{A}$$of the form (1.1) belongs to the classk-$$\mathcal{S}^{*}\mathcal{T}$$if it satisfies the inequality

$$\sum_{n=1}^{\infty } \bigl\{ (k+1)n+1 \bigr\} \vert a_{n} \vert < 1.$$

### Theorem 2

A function$$f\in \mathcal{A} ( p )$$of the form (1.1) belongs to the function classk-$$\mathcal{C}_{q} ( p )$$if

$$\sum_{n=2}^{\infty }\frac{[n+p]_{q}!}{[n]_{q}!} \varLambda _{3} \vert a_{n+p} \vert < ( 1+q ) ,$$

where$$\varLambda _{3}$$is defined in (3.2).

### Proof

We omit the details of the proof, since it easily follows by applying Theorem 1 in conjunction with Definition 11. □

### Theorem 3

A function$$f\in \mathcal{A} (p )$$having series expansion (1.1) belongs to the classk-$$\mathcal{K}_{q}^{\ast } (p )$$if

$$\sum_{n=2}^{\infty } \bigl\{ 2 ( k+1 ) \varLambda _{1}+ \varLambda _{2} \bigr\} < ( 1+q ) ,$$
(3.4)

where

$$\varLambda _{1}= \bigl\vert b_{n+p}- [ n+1 ] _{q}a_{n+p} \bigr\vert$$
(3.5)

and

$$\varLambda _{2}= \bigl\vert ( 1-q ) [ n+1 ] _{q}a_{n+p}-2b_{n+p} \bigr\vert .$$
(3.6)

### Proof

Assuming that (3.4) holds, it suffices to check that

$$k \biggl\vert \frac{ ( 1+q ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}g ) ( z ) }}{ ( q-1 ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}g ) ( z ) }+2}-1 \biggr\vert -\Re \biggl[ \frac{ ( 1+q ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}g ) ( z ) }}{ ( q-1 ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}g ) ( z ) }+2}-1 \biggr] < 1.$$

We have

\begin{aligned} & k \biggl\vert \frac{ ( 1+q ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}g ) ( z ) }}{ ( q-1 ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}g ) ( z ) }+2}-1 \biggr\vert -\Re \biggl[ \frac{ ( 1+q ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}g ) ( z ) }}{ ( q-1 ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}g ) ( z ) }+2}-1 \biggr] \\ &\quad \leq ( k+1 ) \biggl\vert \frac{ ( 1+q ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}g ) ( z ) }}{ ( q-1 ) \frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}g ) ( z ) }+2}-1 \biggr\vert \\ &\quad =2 ( k+1 ) \biggl\vert \frac{z ( D_{q}^{(p)}f ) ( z ) -2 ( D_{q}^{(p-1)}g ) ( z ) }{ ( q-1 ) z ( D_{q}^{(p)}f ) ( z ) +2 ( D_{q}^{(p-1)}g ) ( z ) } \biggr\vert \\ &\quad =2 ( k+1 ) \biggl\vert \frac{\sum_{n=1}^{\infty }\frac{[n+p]_{q}!}{[n+1]_{q}!} \{ [ n+1 ] _{q}a_{n+p}-b_{n+p} \} z^{n+1}}{ ( q+1 ) [p]_{q}!z+\sum_{n=1}^{\infty }\frac{[n+p]_{q}!}{[n+1]_{q}!} \{ ( q-1 ) [ n+1 ] _{q}a_{n+p}+2b_{n+p} \} z^{n+1}} \biggr\vert \\ &\quad \leq \frac{2 ( k+1 ) \sum_{n=1}^{\infty }\frac{[n+p]_{q}!}{[n+1]_{q}!} \vert [ n+1 ] _{q}a_{n+p}-b_{n+p} \vert }{ ( q+1 ) [p]_{q}!-\sum_{n=2}^{\infty }\frac{[n+p]_{q}!}{[n+1]_{q}!} \vert ( 1-q ) [ n+1 ] _{q}a_{n+p}-2b_{n+p} \vert }. \end{aligned}
(3.7)

The last expression in (3.7) is bounded above by 1 if

$$\sum_{n=1}^{\infty } \bigl\{ 2 ( k+1 ) \varLambda _{1}+ \varLambda _{2} \bigr\} < ( 1+q ) ,$$

where $$\varLambda _{1}$$ and $$\varLambda _{2}$$ are given by (3.5) and (3.6), respectively, which completes the proof. □

### Theorem 4

A functionffrom the class$$\mathcal{A} (p )$$having series expansion (1.1) belongs to the classk-$$\mathcal{C}_{q}^{\ast } (p )$$if

$$\sum_{n=1}^{\infty } [ n+1 ] _{q} \bigl\{ 2 ( k+1 ) \varLambda _{1}+\varLambda _{2} \bigr\} < ( 1+q ),$$

where$$\varLambda _{1}$$and$$\varLambda _{2}$$are respectively presented in (3.5) and (3.6).

### Proof

The proof of Theorem 4 follows easily by using Theorem 3 and Definition 13. □

### Theorem 5

Let$$f\in k$$-$$\mathcal{S}_{q}^{\ast } (p )$$be of the form (1.1). Then

$$\vert a_{n+p-1} \vert \leq \prod_{j=2}^{n} \frac{[j]_{q}\{2([j-1]_{q}-1)+(q+1)\delta _{k}\}}{2 \{ [j]_{q}-1 \} [j+p-1]_{q}}\quad \bigl( n\in \mathbb{N}\setminus \{ 1 \} \bigr) .$$
(3.8)

### Proof

For $$f\in k$$-$$\mathcal{S}_{q}^{\ast } ( p )$$, by definition we obtain

$$\frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}f ) ( z ) }=\psi ( z ) ,$$
(3.9)

where

$$\psi ( z ) \prec \frac{2p_{k} ( z ) }{ ( 1+q ) + ( 1-q ) p_{k} ( z ) }.$$
(3.10)

If

$$p_{k} ( z ) =1+\delta _{k}z+\cdots ,$$

then after some appropriate computations, condition (3.10) can be written as

$$\psi ( z ) \prec 1+\frac{ ( 1+q ) }{2}\delta _{k}z+ \cdots ,$$
(3.11)

and if

$$\psi ( z ) =1+\sum_{n=1}^{\infty }\psi _{n}z^{n},$$
(3.12)

then by application of Lemma 1 in conjunction with (3.11) and (3.12) we have

$$\vert \psi _{n} \vert \leq \frac{ ( 1+q ) }{2} \vert \delta _{k} \vert \quad ( n\in \mathbb{N} ) .$$
(3.13)

Now from (3.9) we set

$$z \bigl( D_{q}^{(p)}f \bigr) ( z ) = \bigl( D_{q}^{(p-1)}f \bigr) ( z ) \psi ( z ) ,$$

which implies that

\begin{aligned}{} [ p]_{q}!z+\sum_{n=1}^{\infty } \frac{ [ n+p ] _{q}!}{[n+1]_{q}!}a_{n+p}z^{n+1} =& \Biggl( [p]_{q}!z+\sum_{n=2}^{\infty } \frac{ [ n+p ] _{q}!}{[n+1]_{q}!}a_{n+p}z^{n+1} \Biggr) \\ &{}\cdot \Biggl( 1+\sum_{n=1}^{\infty }\psi _{n}z^{n} \Biggr) . \end{aligned}

Now the comparison of the corresponding coefficients of $$z^{n}$$ gives

$$\frac{{}[ n+p]_{q}!}{[n+1]_{q}!} \bigl( [ n+1 ] _{q}-1 \bigr) a_{n+p}=\sum_{l=1}^{n} \frac{[n+p-l]_{q}!}{[n+1-l]_{q}!}a_{n+p-l}c_{l}, \quad a_{p}=1.$$

Equivalently,

$$\vert a_{n+p} \vert \leq \frac{{}[ n+1]_{q}!}{[n+p]_{q}! ( [ n+1 ] _{q}-1 ) }\sum _{l=1}^{n} \frac{[n+p-l]_{q}!}{[n+1-l]_{q}!} \vert a_{n+p-l} \vert \vert c_{l} \vert , \quad a_{p}=1.$$

Moreover, by (3.13) we have

$$\vert a_{n+p} \vert \leq \frac{\delta _{k} ( q+1 ) [n+1]_{q}!}{2[n+p]_{q}! ( [ n+1 ] _{q}-1 ) }\sum _{l=1}^{n-1}\frac{[p+l]_{q}!}{[l+1]_{q}!} \vert a_{p+l} \vert ,\quad a_{p}=1.$$
(3.14)

Now, using Lemma 2, we have

$$\vert a_{n+p-1} \vert \leq \prod_{j=2}^{n} \frac{[j]_{q}\{2([j]_{q}-1)+\delta _{k}(q+1)\}}{2\{[j]_{q}-1\}[j+p-1]_{q}}.$$
(3.15)

□

Specifically, for instance, setting $$p=1$$ and letting $$q\longrightarrow 1-$$, we obtain the estimate on the nth coefficient of the class k-$$\mathcal{S}^{*}\mathcal{T}$$, settled by Wisniowska and Kanas as follows.

### Corollary 2

(See )

For an analytic function$$f\in k$$-$$\mathcal{S}^{*}\mathcal{T}$$, we have

$$\vert a_{n} \vert \leq \prod_{j=0}^{n-2} \frac{j(j-2)+\delta _{k}}{(j-1)(j)}\quad \bigl( n\in \mathbb{N}\setminus \{ 1 \} \bigr).$$

### Theorem 6

Let$$f\in k$$-$$\mathcal{C}_{q} ( p )$$be of the form (1.1). Then

$$\vert a_{n+p-1} \vert \leq \frac{1}{ [ n+p ] _{q}}\prod _{j=2}^{n} \frac{[j]_{q}\{2([j-1]_{q}-1)+(q+1)\delta _{k}\}}{2 \{ [j]_{q}-1 \} [j+p-1]_{q}} \quad \bigl( n \in \mathbb{N}\setminus \{ 1 \} \bigr) .$$

### Proof

The proof of Theorem 6 follows easily by using Definition 11 and Theorem 5. □

### Theorem 7

Let$$f\in k$$-$$\mathcal{K}_{q}^{\ast }$$be of the form (1.1). Then

\begin{aligned} \vert a_{n+p-1} \vert & \leq \frac{ [ n ] _{q}!}{[n+p]_{q}!} \Biggl( \prod _{j=0}^{n-2} \frac{[j]_{q}\{2([j-1]_{q}-1)+\delta _{k}(q+1)\}}{2\{[j]_{q}-1\}[j+p-1]_{q}} \Biggr) \\ &\quad {} + \frac{ ( q+1 ) \vert \delta _{k} \vert }{2 [ n ] _{q}}\sum_{j=1}^{n-1} \prod_{j=0}^{n-2} \frac{ \vert \delta _{k} ( q+1 ) +2q [ j ] _{q} \vert }{2q [ j+1 ] _{q}}\quad \bigl( n\in \mathbb{N} \setminus \{ 1 \} \bigr) . \end{aligned}
(3.16)

### Proof

By definition, for a function f belonging to k-$$\mathcal{K}_{q} ( p )$$, we have that

$$\frac{z ( D_{q}^{(p)}f ) ( z ) }{ ( D_{q}^{(p-1)}g ) ( z ) }=\psi ( z ) ,$$
(3.17)

where

$$g ( z ) =z^{p}+\sum_{n=1}^{\infty }b_{n+p}z^{n+p},$$

and that

$$\psi ( z ) \prec \frac{2p_{k} ( z ) }{ ( 1+q ) + ( 1-q ) p_{k} ( z ) }.$$
(3.18)

If

$$p_{k} ( z ) =1+\delta _{k}z+\cdots ,$$

then after some convenient computations, condition (3.18) can be written as

$$\psi ( z ) \prec 1+\frac{ ( 1+q ) }{2}\delta _{k}z+ \cdots ,$$
(3.19)

and if

$$\psi ( z ) =1+\sum_{n=1}^{\infty }\psi _{n}z^{n},$$
(3.20)

then applying Lemma 1 in conjunction with (3.19) and (3.20), we get

$$\vert \psi _{n} \vert \leq \frac{ ( 1+q ) }{2} \vert \delta _{k} \vert \quad ( n\in \mathbb{N} ) .$$
(3.21)

Next, equation (3.17) may be written as

$$z \bigl( D_{q}^{(p)}f \bigr) ( z ) = \bigl( D_{q}^{(p-1)}g \bigr) ( z ) \psi ( z ) ,$$

and using the series form, we get

\begin{aligned} {}[ p]_{q}!z+\sum_{n=1}^{\infty } \frac{ [ n+p ] _{q}!}{[n+1]_{q}!}a_{n+p}z^{n+1} =& \Biggl( [p]_{q}!z+\sum_{n=2}^{\infty } \frac{ [ n+p ] _{q}!}{[n+1]_{q}!}b_{n+p}z^{n+1} \Biggr) \\ &{}\cdot \Biggl( 1+\sum_{n=1}^{\infty }\psi _{n}z^{n} \Biggr) . \end{aligned}

Next, the comparison of the corresponding coefficients of $$z^{n}$$ yields

$$\frac{{}[ n+p]_{q}!}{[n+1]_{q}!} \bigl( [ n+1 ] _{q}a_{n+p}-b_{n+p} \bigr) =\sum_{l=1}^{n} \frac{[n+p-l]_{q}!}{[n+1-l]_{q}!}a_{n+p-l}c_{l}\quad ( a_{p}=1 ) .$$

This implies that

$$\frac{{}[ n+p]_{q}!}{[n+1]_{q}!} [ n+1 ] _{q} \vert a_{n+p} \vert \leq \vert b_{n+p} \vert +\sum_{j=1}^{n-1} \vert a_{n+p-l} \vert \vert c_{l} \vert \quad ( a_{1}=1 ) .$$

Moreover, using Theorem 5 and (3.21), we get

\begin{aligned} \vert a_{n+p} \vert & \leq \frac{ [ n ] _{q}!}{[n+p]_{q}!} \Biggl( \prod_{j=0}^{n-2} \frac{[j]_{q}\{2([j-1]_{q}-1)+\delta _{k}(q+1)\}}{2\{[j]_{q}-1\}[j+p-1]_{q}} \Biggr) \\ & \quad {}+ \frac{[n+1]_{q}! ( q+1 ) \vert \delta _{k} \vert }{2 [ n+1 ] _{q}[n+p]_{q}!}\sum_{j=1}^{n-1} \prod_{j=0}^{n-2}\frac{[j]_{q}\{2([j-1]_{q}-1)+\delta _{k}(q+1)\}}{2\{[j]_{q}-1\}[j+p-1]_{q}}. \end{aligned}

Thus we have proved the statement of Theorem 7. □

Putting $$p=1$$ and letting $$q\longrightarrow 1-$$ in Theorem 7, we obtain the estimates of the nth coefficients of the functions from the class k-$$\mathcal{UK}$$, given by Noor et al.

### Corollary 3

(See )

Let$$f\in k$$-$$\mathcal{UK}$$be of the form (1.1). Then

$$\vert a_{n} \vert \leq \frac{ ( \vert \delta _{k} \vert ) _{n-1}}{n!}+ \frac{ \vert \delta _{k} \vert }{n}\sum_{j=1}^{n-1} \frac{ ( \vert \delta _{k} \vert ) _{j-1}}{ ( j-1 ) !}\quad \bigl( n\in \mathbb{N}\setminus \{ 1 \} \bigr) .$$

Further, setting

$$k=0=p-1$$

in Theorem 7, then $$\delta _{k}=2$$, and letting $$q\longrightarrow 1-$$, we get the known result by Kaplan et al.

### Corollary 4

()

Let$$f\in \mathcal{K}$$be an analytic function. Then

$$\vert a_{n} \vert \leq n \quad \bigl( n\in \mathbb{N}\setminus \{ 1 \} \bigr) .$$

### Theorem 8

Let$$f\in k$$-$$\mathcal{C}_{q}^{\ast } ( p )$$with series expansion (1.1). Then

\begin{aligned} \vert a_{n} \vert & \leq \frac{ [ n ] _{q}^{2}!}{[n+p]_{q}^{2}!} \Biggl( \prod _{j=0}^{n-2} \frac{[j]_{q}\{2([j-1]_{q}-1)+\delta _{k}(q+1)\}}{2\{[j]_{q}-1\}[j+p-1]_{q}} \Biggr) +\frac{ ( q+1 ) \vert \delta _{k} \vert }{2[n+p]_{q}!} \\ &\quad {} \cdot \sum_{j=1}^{n-1}\prod _{j=0}^{n-2} \frac{[j]_{q}\{2([j-1]_{q}-1)+\delta _{k}(q+1)\}}{2\{[j]_{q}-1\}[j+p-1]_{q}}\quad \bigl( n\in \mathbb{N}\setminus \{ 1 \} \bigr) . \end{aligned}
(3.22)

### Proof

Using Theorem 7 and Definition 13 immediately yields the proof. □

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### Acknowledgements

The work here is supported by UKM Grant: FRGS/1/2019/STG06/UKM/01/1.

Not applicable.

## Funding

UKM Grant: FRGS/1/2019/STG06/UKM/01/1.

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