Convolution Properties of Certain Classes of Analytic Functions Defined by Jackson q-Derivative

Abstract

In this paper, the Jackson q-derivative is used to investigate two classes of analytic functions in the open unit disc. The coefficient conditions and inclusion properties of the functions in these classes are established by convolution methods.

1. Introduction

Let $\mathcal{B}$ be the class of functions that are analytic and of the form:

$$f\left(z\right)=z+\sum _{k=2}^{\infty }{a}_k{z}^k,(z\in \mathbb{E}:=\{z\in \mathbb{C}:\left|z\right|<1\}).$$

(1)

The Hadamard product (or convolution) of two functions $f,g\in \mathcal{B}$, denoted by $f\ast g,$ is defined by

$$(f\ast g)\left(z\right)=f\left(z\right)\ast g\left(z\right)=z+\sum _{k=2}^{\infty }{a}_k{b}_k{z}^k,$$

where f is given by (1) and $g\left(z\right)=z+{\displaystyle \sum _{k=2}^{\infty }}{b}_k{z}^k.$ In 1978, Silverman et al. [1] obtained characterizations of convex, starlike and spiral-like functions in terms of convolutions. For each of these classes $\chi$, they determined a function g that depends on $\chi$, such that $\frac{1}{z}(f\ast g)\ne 0$. both fundamental and adequate for f to be in $\chi$. As a result of the work by Silverman et al. [1], many important properties of certain subclasses of analytic functions were studied by various authors (for related works, one may refer to [2,3,4,5,6,7,8]). For q∈$(0,1),$ the Jackson q-derivative of a function f$\in \mathcal{B}$ is given by (see [9,10])

$$D_{q}f\left(z\right)=\left\{\begin{array}{cc}\frac{f\left(qz\right) - f\left(z\right)}{(q-1)z}\hfill & (z\ne 0),\hfill \\ f^{{}^{\prime }}\left(0\right)\hfill & (z=0).\hfill \end{array}\right.$$

(2)

Thus, from (2), we have

$$D_{q}f\left(z\right)=1+\sum _{k=2}^{\infty }{\left[k\right]}_q{a}_k{z}^{k-1},$$

(3)

where

$${\left[n\right]}_q=\frac{1-q^n}{1-q},$$

and, as $q\to 1^{-},$ ${\left[n\right]}_q\to n.$ Here we used

$$D_q\left(z^n\right)=\frac{q^n-1}{q-1}{z}^{n-1}={\left[n\right]}_q{z}^{n-1}(n\in \mathbb{N}:=\{1,2,3,\dots \}),$$

and

$$\underset{q\to 1^{-}}{lim}{D}_q\left(z^n\right)=n{z}^{n-1}=\frac{d}{dz}{z}^n.$$

Moreover, we have the following q-derivative rules

$$D_q\left[f\left(z\right)g\left(z\right)\right]=g\left(z\right)D_{q}f\left(z\right)+f\left(qz\right)D_{q}g\left(z\right)$$

and

$$D_q\frac{f\left(z\right)}{g\left(z\right)}=\frac{g\left(z\right)D_{q}f\left(z\right)-f\left(z\right)D_{q}g\left(z\right)}{g\left(z\right)g\left(qz\right)}.$$

We recall a known differential operator ${\mathsf{\Omega }}_q^n\left(z\right),$ which was introduced by Govindaraj and Sivasubramanian (see [11]), and is also known as the Salagean q-differential operator, defined recursively on $n\in {\mathbb{N}}_0:=\mathbb{N}\cup \left\{0\right\}$ as follows: For $f\in \mathcal{B}$ and q∈$(0,1),$

$${\mathsf{\Omega }}_q^{0}f\left(z\right):=f\left(z\right),{\mathsf{\Omega }}_q^{1}f\left(z\right):=z{D}_{q}f\left(z\right),...,{\mathsf{\Omega }}_q^{n}f\left(z\right):=z{D}_q\left({\mathsf{\Omega }}_q^{n-1}f\left(z\right)\right).$$

(4)

Then, we find that

$${\mathsf{\Omega }}_q^{n}f\left(z\right)=(f\ast G_q^n)\left(z\right)(n\in {\mathbb{N}}_0,f\in \mathcal{B}),$$

(5)

where

$$G_q^n\left(z\right):=z+\sum _{k=2}^{\infty }{\left[k\right]}_q^n{z}^k(n\in {\mathbb{N}}_0,z\in \mathbb{E}).$$

Definition 1.

A function $f\in \mathcal{B}$ is said to be in the class ${\mathcal{S}}_{\lambda ,q}^{*}(A,B)$ if and only if

$$(1-\lambda )\frac{f\left(z\right)}{z}+\lambda D_{q}f\left(z\right)\prec \frac{1+Az}{1+Bz}\left(z\in \mathbb{E}\right),$$

(6)

where q∈$(0,1),0\le \lambda \le 1,-1\le B<A\le 1,D_q$ is the Jackson q-derivative and ≺ denotes the usual subordination (see [12,13,14] ).

Definition 2.

A function $f\in \mathcal{B}$ is said to be in the class ${\mathcal{C}}_{\lambda ,q}(A,B)$ if and only if

$$D_{q}f\left(z\right)+\lambda qz{D}_q\left(D_{q}f\left(z\right)\right)\prec \frac{1+Az}{1+Bz}\left(z\in \mathbb{E}\right),$$

(7)

where $q$∈$(0,1),0\le \lambda \le 1,-1\le B<A\le 1$ and $D_q$ is the Jackson q-derivative.

With the help of the Salagean q-differential operator ${\mathsf{\Omega }}_q^n$ given by (5), we have the following definition.

Definition 3.

A function $f\in \mathcal{B}$ is said to be in the class ${\mathcal{S}}_{\lambda ,q}^{*}(n,A,B)$ if and only if

$$(1-\lambda )\frac{{\mathsf{\Omega }}_q^{n}f\left(z\right)}{z}+\lambda \frac{{\mathsf{\Omega }}_q^{n+1}f\left(z\right)}{z}\prec \frac{1+Az}{1+Bz}\left(z\in \mathbb{E}\right),$$

where q∈$(0,1),$$-1\le B<A\le 1,0\le \lambda \le 1$ and $n\in N_0=\{0,1,2,\dots \}.$

In this paper, we use a technique similar to that given by Silverman et al. [1] in order to obtain certain convolution properties of the two classes ${\mathcal{S}}_{\lambda ,q}^{*}(A,B)$ and ${\mathcal{C}}_{\lambda ,q}(A,B)$. Furthermore, the coefficient conditions and inclusion properties of the functions in these classes are established.

2. Convolution Conditions

Theorem 1.

Let $f\left(z\right)$ be a function of the form (1). Then $f\left(z\right)\in {\mathcal{S}}_{\lambda ,q}^{*}(A,B)$ if and only if

$$\frac{1}{z}\left[f\left(z\right)\ast \left\{\left(1-\lambda \right)\frac{z-C{z}^2}{\left(1-z\right)}+\lambda \frac{z-(qC+(1-qz)C)z^2}{\left(1-z\right)\left(1-qz\right)}\right\}\right]\ne 0\left(z\in \mathbb{E}\right)$$

(8)

where $C=C_{\theta }=\frac{e^{-i\theta }+A}{A-B},$ and $0\le \theta <2\pi .$

Proof. 

Let function f be in the class ${\mathcal{S}}_{\lambda ,q}^{*}(A,B)$ if and only if

$$(1-\lambda )\frac{f\left(z\right)}{z}+\lambda D_{q}f\left(z\right)=\frac{1+A\omega \left(z\right)}{1+B\omega \left(z\right)},$$

which is equivalent to

$$(1-\lambda )\frac{f\left(z\right)}{z}+\lambda D_{q}f\left(z\right)\ne \frac{1+A{e}^{i\theta }}{1+B{e}^{i\theta }},(z\in \mathbb{E},\theta \in \left[0,2\pi \right))$$

which simplifies to

$$\left[(1-\lambda )f\left(z\right)+\lambda z{D}_{q}f\left(z\right)\right]\left(1+B{e}^{i\theta }\right) - z\left(1+A{e}^{i\theta }\right) \ne 0.$$

(9)

Since

$$f\left(z\right)=f\left(z\right)\ast \frac{z}{\left(1-z\right)},$$

and

$$D_{q}f\left(z\right)=\frac{1}{\left(1-z\right)(1-qz)}.$$

(10)

This implies that

$$(1-\lambda )f\left(z\right)+\lambda z{D}_{q}f\left(z\right)=f\left(z\right)\ast \frac{z-\left(1-\lambda \right)q{z}^2}{\left(1-z\right)(1-qz)}.$$

(11)

Using (9) and (11), we get

$$\frac{1}{z}\left[f\left(z\right)\ast \left\{\frac{\left(z-\left(1-\lambda \right)q{z}^2\right)\left(1+B{e}^{i\theta }\right)-z\left(1-z\right)\left(1-qz\right)\left(1+A{e}^{i\theta }\right)}{\left(1-z\right)(1-qz)}\right\}\right]\ne 0$$

(12)

that is, that

$$\frac{1}{z}\left[f\left(z\right)\ast \frac{\begin{array}{c}[-\left(1-\lambda \right)\left(1-qz\right)\left\{\left(A-B\right)z{e}^{i\theta }-z^2\left(1+A{e}^{i\theta }\right)\right\}\\ -\lambda \left\{\left(A-B\right)z{e}^{i\theta }-\left(q\left(1+A{e}^{i\theta }\right) + \left(1-qz\right)\left(1+A{e}^{i\theta }\right)\right)z^2\right\}]\end{array}}{\left(1-z\right)(1-qz)}\right]\ne 0$$

(13)

or, equivalently,

$$\frac{1}{z}\left[f\left(z\right)\ast -\left(1-\lambda \right)\left\{\frac{z-\frac{\left(e^{-i\theta }+A\right)}{\left(A-B\right)}{z}^2}{\left(1-z\right)}\right\}-\lambda \left\{\frac{z-\left(\frac{q\left(e^{-i\theta }+A\right)}{\left(A-B\right)}+\frac{\left(1-qz\right)\left(e^{-i\theta }+A\right)}{\left(A-B\right)}\right)z^2}{\left(1-z\right)(1-qz)}\right\}\right]\ne 0.$$

Then, (12) can be rewritten as the following

$$\frac{1}{z}\left[f\left(z\right)\ast \left\{\left(1-\lambda \right)\frac{z-C{z}^2}{\left(1-z\right)}+\lambda \frac{z-(qC+(1-qz)C)z^2}{\left(1-z\right)\left(1-qz\right)}\right\}\right]\ne 0\left(z\in \mathbb{E}\right).$$

Hence, the first part of Theorem 1 was proven.

Conversely, since assumption (8) holds for $C=0,$ it follows that

$$\left(1-\lambda \right)f\left(z\right)+\lambda z{D}_{q}f\left(z\right)\ne 0\mathrm{for}\mathrm{all}z\in \mathbb{E},$$

hence the function $\phi \left(z\right)=\left(1-\lambda \right)f\left(z\right)+\lambda z{D}_{q}f\left(z\right)$ is analytic in $\mathbb{E}$ (i.e., it is regular in $z_0=0$, with $\phi \left(0\right)=1$).

We obtain from the first part that

$$(1-\lambda )\frac{f\left(z\right)}{z}+\lambda D_{q}f\left(z\right)\ne \frac{1+A{e}^{i\theta }}{1+B{e}^{i\theta }},(z\in \mathbb{E},0\le \theta <2\pi ).$$

(14)

If we denote

$$\mathsf{\Psi }\left(z\right)=\frac{1+Az}{1+Bz},(z\in \mathbb{E}),$$

relation (14) shows that $\phi \left(\mathbb{E}\right)\cap \mathsf{\Psi }(\partial \mathbb{E})=\varphi$. Thus, the simply connected domain $\phi \left(\mathbb{E}\right)$ is included in a connected component of $\mathbb{C}\setminus \mathsf{\Psi }(\partial \mathbb{E}).$ From here, using the fact that $\phi \left(0\right)=\mathsf{\Psi }\left(0\right)$ together with the univalent of the function $\mathsf{\Psi }$, it follows that $\phi \left(z\right)\prec \mathsf{\Psi }\left(z\right)$; that is $f\left(z\right)\in {\mathcal{S}}_{\lambda ,q}^{*}(A,B)$. Thus, the second part of Theorem 1 was proven. □

Taking $\lambda =1$ in Theorem 1, we get the following corollary.

Corollary 1.

Let the function f be of the form (1). Then

$$D_{q}f\left(z\right)\prec \frac{1+Az}{1+Bz}(0<q<1,-1\le B<A\le 1),$$

if and only if

$$\frac{1}{z}\left[f\left(z\right)\ast \frac{z-(qC+(1-qz)C)z^2}{\left(1-z\right)\left(1-qz\right)}\right]\ne 0\left(z\in \mathbb{E}\right)$$

where $C=C_{\theta }=\frac{e^{-i\theta }+A}{A-B},$ and $0\le \theta <2\pi .$

Letting $q\to 1^{-}$ in Theorem 1, we acquire the following corollary.

Corollary 2.

Let the function f be of the form (1). Then,

$$(1-\lambda )\frac{f\left(z\right)}{z}+\lambda f^{{}^{\prime }}\left(z\right)\prec \frac{1+Az}{1+Bz}(-1\le B<A\le 1),$$

if and only if

$$\frac{1}{z}\left[f\left(z\right)\ast \left(\left(1-\lambda \right)\frac{z-C{z}^2}{\left(1-z\right)}+\lambda \frac{z-(C+(1-z)C)z^2)}{{\left(1-z\right)}^2}\right)\right]\ne 0\left(z\in \mathbb{E}\right)$$

where $C=C_{\theta }=\frac{e^{-i\theta }+A}{A-B}$ and $0\le \theta <2\pi .$

Theorem 2.

A function $f\left(z\right)$ of the form (1) is in the class ${\mathcal{S}}_{{}^{\lambda ,q}}^{*}(n,A,B)$ if and only if

$$1-\sum _{k=2}^{\infty }{\left[k\right]}_q^n\left(\frac{e^{-i\theta }+B}{A-B}\right)\left[\left(1-\lambda \right)+\lambda {\left[k\right]}_q\right]a_k{z}^{k-1}\ne 0$$

(15)

for all $0\le \theta <2\pi$ and $z\in \mathbb{E}.$

Proof. 

Note that

$$f\in {\mathcal{S}}_{{}^{\lambda ,q}}^{*}(n,A,B)\iff {\mathsf{\Omega }}_q^{n}f\left(z\right)\in {\mathcal{S}}_{{}^{\lambda ,q}}^{*}(A,B).$$

From Theorem 1, we have $f\in {\mathcal{S}}_{{}^{\lambda ,q}}^{*}(n,A,B)$ if and only if

$$\frac{1}{z}\left[{\mathsf{\Omega }}_q^{n}f\left(z\right)\ast \left\{\left(1-\lambda \right)\frac{z-C{z}^2}{\left(1-z\right)}+\lambda \frac{z-(qC+(1-qz)C)z^2}{\left(1-z\right)\left(1-qz\right)}\right\}\right]\ne 0\left(z\in \mathbb{E}\right)$$

(16)

where $C=C_{\theta }=\frac{e^{-i\theta }+A}{A-B},$ and $0\le \theta <2\pi .$ Now we can easily deduce that

$$\begin{array}{cc}& \frac{1}{z}\left[{\mathsf{\Omega }}_q^{n}f\left(z\right)\ast \left\{\left(1-\lambda \right)\frac{z-C{z}^2}{\left(1-z\right)} + \lambda \frac{z-(qC+(1-qz)C))z^2}{\left(1-z\right)\left(1-qz\right)}\right\}\right]\hfill \\ =& \frac{1}{z}\left[{\mathsf{\Omega }}_q^{n}f\left(z\right)\ast \left(1-\lambda \right)\left\{Cz+\frac{z(1-C)}{\left(1-z\right)}\right\} + \lambda \left\{Cz+\frac{z(1-C)}{\left(1-z\right)\left(1-qz\right)}\right\}\right].\hfill \end{array}$$

(17)

Using

$$\frac{z}{1-z}=z+\sum _{k=2}^{\infty }{z}^k,\frac{z}{\left(1-z\right)\left(1-qz\right)}=z+\sum _{k=2}^{\infty }{\left[k\right]}_q{z}^k,$$

and

$${\mathsf{\Omega }}_q^{n}f\left(z\right)=z+\sum _{k=2}^{\infty }{\left[k\right]}_q^n{a}_k{z}^k,$$

(17) may be written as

$$1+\sum _{k=2}^{\infty }{\left[k\right]}_q^n\left(1-C\right)\left[\left(1-\lambda \right)+\lambda {\left[k\right]}_q\right]a_k{z}^{k-1}$$

(18)

since $C=\frac{e^{-i\theta }+A}{A-B},$ then (18) gives

$$1-\sum _{k=2}^{\infty }{\left[k\right]}_q^n\left(\frac{e^{-i\theta }+B}{A-B}\right)\left[\left(1-\lambda \right)+\lambda {\left[k\right]}_q\right]a_k{z}^{k-1}$$

therefore, (16) becomes

$$1-\sum _{k=2}^{\infty }{\left[k\right]}_q^n\left(\frac{e^{-i\theta }+B}{A-B}\right)\left[\left(1-\lambda \right)+\lambda {\left[k\right]}_q\right]a_k{z}^{k-1}\ne 0.$$

This finishes the proof of Theorem 2. □

Theorem 3.

If $f\left(z\right)\in \mathcal{B}$ satisfies the inequality

$$\sum _{k=2}^{\infty }\left(1+\left|B\right|\right)\left[\left(1-\lambda \right)+\lambda {\left[k\right]}_q\right]{\left[k\right]}_q^n\left|a_k\right|\le \left(A-B\right)$$

(19)

then $f\left(z\right)\in {\mathcal{S}}_{{}^{\lambda ,q}}^{*}(n,A,B).$

Proof. 

Since

$$\left|\frac{e^{-i\theta }+B}{A-B}\right|\le \frac{\left|e^{-i\theta }\right|+\left|B\right|}{\left|A-B\right|}=\frac{1+\left|B\right|}{A-B}$$

then,

$$\begin{array}{ccc}& & \left|1-\sum _{k=2}^{\infty }{\left[k\right]}_q^n\left(\frac{e^{-i\theta }+B}{A-B}\right)\left[\left(1-\lambda \right)+\lambda {\left[k\right]}_q\right]a_k{z}^{k-1}\right|\hfill \\ & \ge & 1-\sum _{k=2}^{\infty }{\left[k\right]}_q^n\left|\frac{e^{-i\theta }+B}{A-B}\right|\left[\left(1-\lambda \right)+\lambda {\left[k\right]}_q\right]\left|a_k\right|{\left|z\right|}^{k-1}\hfill \\ & \ge & 1-\sum _{k=2}^{\infty }{\left[k\right]}_q^n\left(\frac{1+\left|B\right|}{A-B}\right)\left[\left(1-\lambda \right)+\lambda {\left[k\right]}_q\right]\left|a_k\right|>0.\left(z\in \mathbb{E}\right)\hfill \end{array}$$

Thus, from Theorem 2, we have $f\left(z\right)\in {\mathcal{S}}_{{}^{\lambda ,q}}^{*}(n,A,B)$, which ends the proof. □

Theorem 4.

${\mathcal{S}}_{{}^{\lambda ,q}}^{*}$$(n+1,A,B)\subset {\mathcal{S}}_{{}^{\lambda ,q}}^{*}$$(n,A,B).$

Proof. 

Since $f\left(z\right)\in {\mathcal{S}}_{{}^{\lambda ,q}}^{*}(n+1,A,B),$ we see from Theorem 2 that

$$1-\sum _{k=2}^{\infty }{\left[k\right]}_q^{n+1}\left(\frac{e^{-i\theta }+B}{A-B}\right)\left[\left(1-\lambda \right)+\lambda {\left[k\right]}_q\right]a_k{z}^{k-1}\ne 0.$$

(20)

we can write (20) as

$$\left[1+\sum _{k=2}^{\infty }{\left[k\right]}_q{z}^{k-1}\right]\ast \left[1-\sum _{k=2}^{\infty }{\left[k\right]}_q^n\left(\frac{e^{-i\theta }+B}{A-B}\right)\left[\left(1-\lambda \right)+\lambda {\left[k\right]}_q\right]a_k{z}^{k-1}\right]\ne 0.$$

(21)

Since,

$$\begin{array}{cc}& \left[1+\sum _{k=2}^{\infty }{\left[k\right]}_q{z}^{k-1}\right]\ast \left[1+\sum _{k=2}^{\infty }\frac{1}{{\left[k\right]}_q}{z}^{k-1}\right]\hfill \\ =& 1+\sum _{k=2}^{\infty }{z}^{k-1}=\frac{1}{1-z}.\hfill \end{array}$$

(22)

By utilizing the property, if $f\ne 0$ and $g\ast h\ne 0,$ then $f\ast (g\ast h)\ne 0,$ (21) can be written as

$$1-\sum _{k=2}^{\infty }{\left[k\right]}_q^n\left(\frac{e^{-i\theta }+B}{A-B}\right)\left[\left(1-\lambda \right)+\lambda {\left[k\right]}_q\right]a_k{z}^{k-1}\ne 0$$

(23)

which means that $f\left(z\right)\in {\mathcal{S}}_{{}^{\lambda ,q}}^{*}(n,A,B).$ This finishes the proof of Theorem 4. □

Theorem 5.

Let the function f be of the form (1). Then $f\in {\mathcal{C}}_{{}^{\lambda ,q}}(A,B)$ if and only if

$$\frac{1}{z}\left[f\left(z\right)\ast \left\{Cz+\left(1-\lambda \right)\frac{z(1-C)}{\left(1-z\right)(1-qz)}+\lambda \frac{z\left(1-C\right)\left(1+qz\right)}{\left(1-z\right)\left(1-qz\right)\left(1-q^{2}z\right)}\right\}\right]\ne 0$$

(24)

where$z\in \mathbb{E},C=C_{\theta }=\frac{e^{-i\theta }+A}{A-B}$ and $0\le \theta <2\pi .$

Proof. 

Note that

$$f\in {\mathcal{C}}_{{}^{\lambda ,q}}(A,B)\iff z{D}_{q}f\left(z\right)\in {\mathcal{S}}_{{}^{\lambda ,q}}^{*}(A,B).$$

Theorem 1 gives

$$\frac{1}{z}\left[z{D}_{q}f\left(z\right)\ast \left\{\left(1-\lambda \right)\left\{Cz+\frac{z(1-C)}{\left(1-z\right)}\right\}+\lambda \left\{Cz+\frac{z(1-C)}{\left(1-z\right)\left(1-qz\right)}\right\}\right\}\right]\ne 0,$$

(25)

where $z\in \mathbb{E},C=C_{\theta }=\frac{e^{-i\theta }+A}{A-B}$ and $0\le \theta <2\pi .$ Since

$$z{D}_q\left(f\ast g\right)=f\left(z\right)\ast z{D}_{q}g\left(z\right),$$

then, (25) can be written as

$$\frac{1}{z}\left[f\left(z\right)\ast \left\{\left(1-\lambda \right)\left(\frac{z(1-C)}{\left(1-z\right)(1-qz)}+Cz\right) +\lambda \left(\frac{z\left(1-C\right)\left(1+qz\right)}{\left(1-z\right)\left(1-qz\right)\left(1-q^{2}z\right)}+Cz\right)\right\}\right]\ne 0.$$

This finishes the proof of Theorem 5. □

Taking $\lambda =1$ in Theorem 5, we get the following corollary.

Corollary 3.

Let the function $f$ be of the form (1). Then

$$D_{q}f\left(z\right)+qz{D}_q\left(D_{q}f\left(z\right)\right)\prec \frac{1+Az}{1+Bz}(0<q<1,-1\le B<A\le 1),$$

if and only if

$$\frac{1}{z}\left[f\left(z\right)\ast \left(\frac{z\left(1-C\right)\left(1+qz\right)}{\left(1-z\right)\left(1-qz\right)\left(1-q^{2}z\right)}+Cz\right)\right]\ne 0\left(z\in \mathbb{E}\right)$$

where $C=C_{\theta }=\frac{e^{-i\theta }+A}{A-B}$ and $0\le \theta <2\pi .$

Letting $q\to 1^{-}$ in Corollary( 3), we acquire the following corollary.

Corollary 4.

Let the function f be of the form in (1). Then

$$f^{{}^{\prime }}\left(z\right)+z{f}^{{}^{″}}\left(z\right)\prec \frac{1+Az}{1+Bz}(-1\le B<A\le 1),$$

if and only if

$$\frac{1}{z}\left[f\left(z\right)\ast \left(\frac{z\left(1-C\right)\left(1+z\right)}{{\left(1-z\right)}^3}+Cz\right)\right]\ne 0\left(z\in \mathbb{E}\right)$$

where $C=C_{\theta }=\frac{e^{-i\theta }+A}{A-B}$ and $0\le \theta <2\pi .$

3. Conclusions

In this paper, we introduced and studied two new subclasses of analytic functions in the open unit disc using the Jackson q-derivative. A similar technique to that given by Silverman et al. [1] was used to obtain certain convolution properties for these two classes. In addition, the coefficient conditions and inclusion properties of these classes are established. Several special cases have been examined as applications of our main results.