*Background*

Let <sup>A</sup>*p* (*p* ∈ N = {0, 1, 2, . . .}) contain all multivalent functions say *f* that are holomorphic or analytic in a subset D = {*z* : |*z*| < 1} of a complex plane C and having the series form:

$$f(z) = z^p + \sum\_{l=1}^{\infty} a\_{l+p} z^{l+p}, \ (z \in \mathbb{D})\,. \tag{1}$$

For two analytic functions *f* and *g* in D, then *f* is subordinate to *g*, symbolically presented as *f* ≺ *g* or *f* (*z*) ≺ *g* (*z*), if we can find an analytic function *w* with the properties *w* (0) = 0 & |*w* (*z*)| < 1 such that *f*(*z*) = *g*(*w*(*z*)). Also, if *g* is univalent in D, then we have

$$f(z) \preccurlyeq \mathfrak{g}(z) \iff f(0) = \mathfrak{g}(0) \quad \text{and} \quad f(\mathbb{D}) \subset \mathfrak{g}(\mathbb{D}).$$

For given *q* ∈ (0, <sup>1</sup>), the derivative in *q*-analogue of *f* is given by

$$\mathcal{D}\_q f(z) = \frac{f(z) - f\left(qz\right)}{z\left(1 - q\right)}, \ \left(z \neq 0, \ q \neq 1\right). \tag{2}$$

Making (1) and (2), we easily ge<sup>t</sup> that for *n* ∈ N and *z* ∈ D:

$$D\_q \left\{ \sum\_{n=1}^{\infty} a\_{n+p} z^{n+p} \right\} = \sum\_{n=1}^{\infty} \left[ n + p \right]\_q a\_{n+p} z^{n+p-1},\tag{3}$$

where

$$\left[n\right]\_q = \frac{1 - q^n}{1 - q} = 1 + \sum\_{l=1}^{n-1} q^l, \ \left[0, q\right] = 0.$$

For *n* ∈ Z∗ := Z\ {−1, −2, . . .} , the *q*-number shift factorial is given as

$$\begin{aligned} [n]\_q! = \left\{ \begin{array}{ll} 1, \ n = 0, \\ [1]\_q \, [2]\_q \dots \, [n]\_q, \ n \in \mathbb{N}. \end{array} \right. \end{aligned} $$

Also, with *x* > 0, the *q*-analogue of the Pochhammer symbol has the form

$$[\mathbf{x},q]\_{q\mathbf{u}} = \begin{cases} \ 1,\ \mathbf{n}=\mathbf{0},\\ \ [\mathbf{x},q][\mathbf{x}+\mathbf{1},q]\cdot\cdot[\mathbf{x}+\mathbf{n}-\mathbf{1},q],\ \mathbf{n}\in\mathbb{N},\end{cases}$$

and, for *x* > 0, the Gamma function in *q*-analogue is presented as

$$
\Gamma\_q\left(\mathfrak{x} + 1\right) = \left[\mathfrak{x}, q\right] \Gamma\_q\left(t\right) \text{ and } \Gamma\_q\left(1\right) = 1.
$$

We now consider a function

$$\Phi\_p\left(q,\mu+1;z\right) = z^p + \sum\_{n=2}^{\infty} \Lambda\_{n+p} z^{n+p}, \ (\mu > -1, \ z \in \mathbb{D}), \tag{4}$$

with

$$
\Lambda\_{n+p} = \frac{[\mu+1, q]\_{n+p}}{[n+p]\_q!}.\tag{5}
$$

The series defined in (4) converges absolutely in D. Using <sup>Φ</sup>*p* (*q*, *μ*; *z*) with *μ* > −1 and idea of convolution, Arif et al. [20] established a differential operator L*μ*+*p*−<sup>1</sup> *q* : <sup>A</sup>*p* → <sup>A</sup>*p* by

$$\mathcal{L}\_q^{\mu+p-1} f\left(z\right) = \Phi\_\mathbb{P}\left(q\_\prime\mu; z\right) \* f(z) = z^p + \sum\_{n=2}^\infty \Lambda\_{n+p} a\_{n+p} z^{n+p} \,, \left(z \in \mathbb{D}\right) . \tag{6}$$

We also note that

$$\lim\_{q \to 1^{-}} \Phi\_{\mathcal{P}}\left(q, \mu; z\right) = \frac{z^{p}}{\left(1 - z\right)^{\mu + 1}} \text{ and } \lim\_{q \to 1^{-}} \mathcal{L}\_{q}^{\mu + p - 1} f(z) = f(z) \* \frac{z^{p}}{\left(1 - z\right)^{\mu + 1}}.$$

Now, when *q* → 1<sup>−</sup>, the operator defined in (6) becomes the familiar differential operator investigated in [21] and further, setting *p* = 1, we ge<sup>t</sup> the most familiar operator known as Ruscheweyh operator [12] (see also [22,23]). Also, for different types of operators in *q*-analogue, see the works [16,17,19,24–26].

Motivated from the work studied in [3,18,27–29], we establish a family S∗*p* (*q*, *μ*, *A*, *B*) using the operator L*μ*+*p*−<sup>1</sup> *q*as follows:

**Definition 1.** *Suppose that q* ∈ (0, 1) *and* −1 *B* < *A* 1. *Then, f* ∈ <sup>A</sup>*p belongs to the set* S∗*p* (*q*, *μ*, *A*, *<sup>B</sup>*), *if it satisfies*

$$\frac{zD\_q\mathcal{L}\_q^{\mu+p-1}f\left(z\right)}{\left[p\_\prime q\right]\mathcal{L}\_q^{\mu+p-1}f\left(z\right)} \prec \frac{1+Az}{1+Bz},\tag{7}$$

*where the function* 1+*Az* 1+*Bzis known as Janowski function studied in [30].*

Alternatively, 
$$f \in \mathcal{S}\_p^\*\left(q, \mu, A, B\right) \Leftrightarrow \left| \frac{\frac{zD\_q \mathcal{L}\_q^{\mu+p-1} f(z)}{[p, q] \mathcal{L}\_q^{\mu+p-1} f(z)} - 1}{A - B \frac{zD\_q \mathcal{L}\_q^{\mu+p-1} f(z)}{[p, q] \mathcal{L}\_q^{\mu+p-1} f(z)}} \right| < 1. \tag{8}$$

Note: We will assume throughout our discussion, unless otherwise stated,

$$-1 \le B < A \le 1, \ q \in (0, 1), \ p \in \mathbb{N}, \text{ and } \mu > -1.$$

#### **2. A Set of Lemmas**

**Lemma 1.** *[31] Let h* (*z*) = 1 + ∞ ∑ *<sup>n</sup>*=1 *dnzn* ≺ *K* (*z*) = 1 + ∞ ∑ *<sup>n</sup>*=1 *knzn in* D*. If K* (*z*) *is convex univalent in* D, *then,*

$$|d\_n| \le |k\_1| \, \, \, for \, n \ge 1 \,.$$

**Lemma 2.** *Let* W *contain all functions w that are analytic in* D*, which satisfies w* (0) = 0 *&* |*w*(*z*)| < 1 *if the function w* ∈ W*, given by*

$$w(z) = \sum\_{k=1}^{\infty} w\_k z^k \ (z \in \mathbb{D})\ .$$

*Then, for λ* ∈ C, *we have*

$$\left| w\_2 - \lambda w\_1^2 \right| \stackrel{<}{=} \max\left\{ 1; \, |\lambda| \right\},\tag{9}$$

*and*

$$\left| w\_3 + \frac{1}{4} w\_1 w\_2 + \frac{1}{16} w\_1^3 \right| \le 1. \tag{10}$$

*These results are the best possible.*

> For the first and second part, see references [32,33], respectively.

#### **3. Main Results and Their Consequences**

**Theorem 1.** *Let f* ∈ <sup>A</sup>*p have the series form* (1) *and satisfy the inequality given by*

$$\sum\_{n=1}^{\infty} \wedge\_{n+p} \left( \left[ n+p, q \right] \left( 1-B \right) - \left[ p, q \right] \left( 1-A \right) \right) \left| a\_{n+p} \right| \le \left[ p, q \right] \left( A-B \right). \tag{11}$$

*Then, f* ∈ S∗*p* (*q*, *μ*, *A*, *<sup>B</sup>*).

**Proof.** To show *f* ∈ S∗*p* (*q*, *μ*, *A*, *<sup>B</sup>*), we just need to show the relation (8). For this, we consider

$$\left| \frac{\frac{zD\_{q}\mathcal{L}\_{q}^{\mu+p-1}f(z)}{[p,q]\mathcal{L}\_{q}^{\mu+p-1}f(z)} - 1}{A - B\frac{zD\_{q}\mathcal{L}\_{q}^{\mu+p-1}f(z)}{[p,q]\mathcal{L}\_{q}^{\mu+p-1}f(z)}} \right| = \left. \left| \frac{zD\_{q}\mathcal{L}\_{q}^{\mu+p-1}f\left(z\right) - [p,q]\mathcal{L}\_{q}^{\mu+p-1}f\left(z\right)}{A\left[p,q\right]\mathcal{L}\_{q}^{\mu+p-1}f\left(z\right) - BzD\_{q}\mathcal{L}\_{q}^{\mu+p-1}f\left(z\right)} \right| \right|.$$

Using (6), and with the help of (11) and (3), we have

$$\begin{array}{c} \leq & \left| \frac{[p,q]z^{p}+\sum\_{n=1}^{\infty}\wedge\_{n+p}a\_{n+p}[n+p,q]z^{n+p}-[p,q]\left(z^{p}+\sum\_{n=1}^{\infty}\wedge\_{n+p}a\_{n+p}z^{n+p}\right)}{A[p,q]\left(z^{p}+\sum\_{n=1}^{\infty}\wedge\_{n+p}a\_{n+p}z^{n+p}\right)-B\left([p,q]z^{p}+\sum\_{n=1}^{\infty}\wedge\_{n+p}a\_{n+p}[n+p,q]z^{n+p}\right)} \right| \\ \leq & \left| \frac{\sum\_{n=1}^{\infty}\wedge\_{n+p}a\_{n+p}[(n+p,q]-[p,q])z^{n+p}}{(A-B)[p,q]z^{p}+\sum\_{n=1}^{\infty}\wedge\_{n+p}a\_{n+p}(A[p,q]-B[n+p,q])z^{n+p}} \right| \\ \leq & \frac{\sum\_{n=1}^{\infty}\wedge\_{n+p}|a\_{n+p}|\left([n+p,q]-[p,q]\right)|z|^{n+p}}{(A-B)[p,q]|z|^{p}-\sum\_{n=1}^{\infty}\wedge\_{n+p}|a\_{n+p}|\left(A[p,q]-B[n+p,q]\right)|z|^{n+p}} \\ \leq & \frac{\sum\_{n=1}^{\infty}\wedge\_{n+p}|a\_{n+p}|\left([n+p,q]-[p,q]\right)}{(A-B)[p,q]-\sum\_{n=1}^{\infty}\wedge\_{n+p}|a\_{n+p}|\left([A[p,q]-B[n+p,q]\right)\right)} < 1\_{\epsilon} \end{array}$$

where we have used the inequality (11) and this completes the proof.

Varying the parameters *μ*, *b*, *A*, and *B* in the last Theorem, we ge<sup>t</sup> the following known results discussed earlier in [34].

**Corollary 1.** *Let f* ∈ A *be given by* (1) *and satisfy the inequality*

$$\sum\_{n=2}^{\infty} \left( \left[ n, q \right] \left( 1 - B \right) - 1 + A \right) \left| a\_n \right| \stackrel{<}{=} A - B.$$

*Then, the function f* ∈ S∗*q* [*<sup>A</sup>*, *<sup>B</sup>*].

By choosing *q* → 1− in the last corollary, we ge<sup>t</sup> the known result proved by Ahuja [22] and, furthermore, for *A* = 1 − *α* and *B* = −1, we obtain the result for the family S∗ (*ξ*) which was proved by Silverman [35].

**Theorem 2.** *Let f* ∈ S∗*p* (*q*, *μ*, *A*, *B*) *be of the form* (1). *Then,*

$$\left| a\_{p+1} \right| \le \frac{\psi\_1 \left( A - B \right)}{\wedge\_{1+p}},\tag{12}$$

*and for n* - 2,

$$\left| a\_{n+p} \right| \le \frac{\left( A - B \right) \psi\_n}{\wedge\_{n+p}} \prod\_{t=1}^{n-1} \left( 1 + \frac{\left[ p, q \right] \left( A - B \right)}{\left( \left[ p + t, q \right] - \left[ p, q \right] \right)} \right), \tag{13}$$

*where*

$$\psi\_n := \psi\_n(p, q) = \frac{[p, q]}{([n + p, q] - [p, q])}.\tag{14}$$

**Proof.** If *f* ∈ S∗*p* (*q*, *μ*, *A*, *<sup>B</sup>*), then by definition we have

$$\frac{z D\_q \mathcal{L}\_q^{\mu+p-1} f\left(z\right)}{\left[p\_\prime q\right] \mathcal{L}\_q^{\mu+p-1} f\left(z\right)} = \frac{1 + Aw(z)}{1 + Bw(z)}.\tag{15}$$

Let us put

Then, by Lemma 1, we ge<sup>t</sup>

$$p(z) = 1 + \sum\_{n=1}^{\infty} d\_n z^n = \frac{1 + Aw(z)}{1 + Bw(z)}.$$
 
$$|d\_n| \le A - B. \tag{16}$$

Now, from (15) and (6), we can write

$$z^p + \sum\_{n=1}^{\infty} \frac{[n+p,p]}{[p,q]} \Lambda\_{n+p} \, a\_{n+p} z^{n+p} = \left(1 + \sum\_{n=1}^{\infty} d\_n z^n\right) \left(z^p + \sum\_{n=1}^{\infty} \Lambda\_{n+p} \, a\_{n+p} z^{n+p}\right). \tag{17}$$

Equating coefficients of *zn*+*<sup>p</sup>* on both sides,

$$a \wedge\_{n+p} \left( [n+p,q] - [p,q] \right) a\_{n+p} = \begin{array} [p,q] \wedge\_{n+p-1} a\_{n+p-1} d\_1 + \cdots + [p,q] \wedge\_{1+p} a\_{1+p} d\_{n-1} \end{array}$$

 −

Taking absolute on both sides and then using (16), we have

$$\left| \bigwedge\_{n+p} \left( [n+p,q] - [p,q] \right) \left| a\_{n+p} \right| \right| \le \left[ p,q \right] \left( A - B \right) \left( 1 + \sum\_{k=1}^{n-1} \wedge\_{k+p} \left| a\_{k+p} \right| \right),$$

and this further implies

$$\left| \left| a\_{\rm u+p} \right| \right| \le \frac{\left( A - B \right) \psi\_n}{\wedge\_{n+p}} \left( 1 + \sum\_{k=1}^{n-1} \wedge\_{k+p} \left| a\_{k+p} \right| \right), \tag{18}$$

,

where *ψn* is given by (14). So, for *n* = 1, we have from (18)

$$\left| a\_{p+1} \right| \le \frac{\left( A - B \right) \psi\_1}{\wedge\_{1+p}}.$$

and this shows that (12) holds for *n* = 1. To prove (13), we apply mathematical induction. Therefore, for *n* = 2, we have from (12):

$$\left|a\_{p+2}\right| \stackrel{<}{=} \frac{(A-B)\,\psi\_2}{\wedge\_{2+p}} \left(1+\wedge\_{1+p} \left|a\_{1+p}\right|\right)\,\iota$$

using (12), we have

$$\left| a\_{p+2} \right| \stackrel{<}{=} \frac{\left( A - B \right) \psi\_2}{\wedge\_{2+p}} \left( 1 + \left( A - B \right) \psi\_1 \right) \wedge$$

which clearly shows that (13) holds for *n* = 2. Let us assume that (13) is true for *n m* − 1, that is,

$$\left| \left| a\_{m-1+p} \right| \right| \stackrel{<}{=} \frac{\left( A - B \right) \psi\_{m-1}}{\wedge\_{m+p-1}} \prod\_{t=1}^{m-2} \left( 1 + \left( A - B \right) \psi\_t \right) \dots$$

Consider

$$\begin{split} \left| \left| a\_{m+p} \right| \right| &\leq \frac{(A-B)\,\psi\_{m}}{\wedge\_{m+p}} \left( 1 + \sum\_{k=1}^{m-1} \wedge\_{k+p} \left| a\_{k+p} \right| \right) \\ &= \frac{(A-B)\,\psi\_{m}}{\wedge\_{m+p}} \left\{ 1 + (A-B)\,\psi\_{1} + \dots + (A-B)\,\psi\_{m-1} \prod\_{t=1}^{m-2} \left( 1 + (A-B)\,\psi\_{t} \right) \right\} \\ &= \frac{(A-B)\,\psi\_{m}}{\wedge\_{m+p}} \prod\_{t=1}^{m-1} \left( 1 + \frac{[p,q] \left( A-B\right)}{\left( [p+t,q] - [p,q] \right)} \right) , \end{split}$$

this implies that the given result is true for *n* = *m*. Hence, using mathematical induction, we achieve the inequality (13).

**Theorem 3.** *Let f* ∈ S∗*p* (*q*, *μ*, *A*, *<sup>B</sup>*)*, and be given by* (1). *Then, for λ* ∈ C

$$\left| a\_{p+2} - \lambda a\_{p+1}^2 \right| \le \frac{(A-B)\psi\_2}{\Lambda\_{p+2}} \left\{ 1; \ |v| \right\} \,\,\, \epsilon$$

*where υ is given by*

$$\upsilon = \left( B - (A - B)\psi\_1 \right) + \frac{\Lambda\_{p+2}\psi\_1^2}{\Lambda\_{p+1}^2\psi\_2}(A - B)\lambda. \tag{19}$$

**Proof.** Let *f* ∈ S∗*p* (*q*, *μ*, *A*, *<sup>B</sup>*), and consider the right-hand side of (15), we have

$$\frac{1+Aw(z)}{1+Bw(z)} = \left(1+A\sum\_{k=1}^{\infty}w\_k z^k\right)\left(1+B\sum\_{k=1}^{\infty}w\_k z^k\right)^{-1}$$

where

$$w(z) = \sum\_{k=1}^{\infty} w\_k z^k \rho$$

and after simple computations, we can rewrite

$$\frac{1+Aw(z)}{1+Bw(z)} = 1 + (A-B)w\_1 z + (A-B) \left\{ w\_2 - Bw\_1^2 \right\} z^2 + \dots \tag{20}$$

Now, for the left hand side of (15), we have

$$\frac{zD\_q \mathcal{L}\_q^{\mu+p-1} f(z)}{[p,q]\_\* \mathcal{L}\_q^{\mu+p-1} f(z)} = \left( 1 + \sum\_{n=1}^{\infty} \frac{[n+p,q]}{[p,q]} \Lambda\_{n+p} a\_{n+p} z^n \right) \left( 1 + \sum\_{n=1}^{\infty} \Lambda\_{n+p} a\_{n+p} z^n \right)^{-1}$$

$$= 1 + \frac{\Lambda\_{1+p}}{\psi\_1} a\_{1+p} z + \left( \frac{\Lambda\_{2+p} a\_{2+p}}{\psi\_2} - \frac{\Lambda\_{1+p}^2 a\_{1+p}^2}{\psi\_1} \right) z^2 + \dots \tag{21}$$

From (20) and (21), we have

$$\begin{array}{rcl} \Lambda\_{p+1} & = & \frac{\psi\_1}{\Lambda\_{p+1}} \left( A - B \right) w\_{1\prime} \end{array} \tag{22}$$

$$a\_{p+2} = \begin{array}{c} \left(A - B\right)\psi\_2\\ \hline \Lambda\_{p+2} \end{array} \left\{ w\_2 + \left(\left(A - B\right)\psi\_1 - B\right)w\_1^2 \right\}.\tag{23}$$

Now, consider

$$\begin{split} \left| a\_{p+2} - \lambda a\_{p+1}^2 \right| &= \left| \frac{(A-B)\mathfrak{\boldsymbol{\varmu}}\_2}{\Lambda\_{p+2}} \left\{ w\_2 + ((A-B)\mathfrak{\boldsymbol{\upmu}}\_1 - B)w\_1^2 \right\} - \lambda \frac{\mathfrak{\boldsymbol{\varmu}}\_1^2}{\Lambda\_{p+1}^2} (A-B)^2 w\_1^2 \right| \\ &= \left| \frac{(A-B)\mathfrak{\boldsymbol{\varmu}}\_2}{\Lambda\_{p+2}} \right| w\_2 - \left\{ (B-(A-B)\mathfrak{\boldsymbol{\upmu}}\_1) + \frac{\Lambda\_{p+2} \mathfrak{\boldsymbol{\upmu}}\_1^2}{\Lambda\_{p+1}^2 \mathfrak{\boldsymbol{\upmu}}\_2} (A-B) \lambda \right\} w\_1^2 \end{split}$$

using Lemma 2, we have

$$\left| a\_{p+2} - \lambda a\_{p+1}^2 \right| \le \frac{(A-B)\psi\_2}{\Lambda\_{p+2}} \left\{ 1; \ |v| \right\}\_{\Lambda}$$

where *υ* is given by

$$\upsilon = \left( B - (A - B)\psi\_1 \right) + \frac{\Lambda\_{p+2}\psi\_1^2}{\Lambda\_{p+1}^2\psi\_2}(A - B)\lambda.$$

This completes the proof.

**Theorem 4.** *Let f* ∈ S∗*p* (*q*, *μ*, *A*, *B*) *and be given by* (1). *Then,*

$$\left| a\_{p+3} - \frac{q+2}{q^2+q+1} \frac{\Lambda\_{1+p} \Lambda\_{2+p}}{\Lambda\_{3+p}} a\_{p+2} a\_{p+1} + \frac{1}{[3,q]} \frac{\Lambda\_{1+p}^3}{\Lambda\_{3+p}} a\_{p+1}^3 \right| \leq (A-B) \left\{ \frac{4\left(2B-1\right)^2 + 1}{8\Lambda\_{3+p}} \right\} \psi\_{3+p}$$

*where ψn and* <sup>∧</sup>*n*+*p are defined by* (14) *and* (5)*, respectively.*

**Proof.** From the relations (20) and (21), we have

$$\left(a\_{p+3} - \frac{q+2}{q^2+q+1} \frac{\Lambda\_{1+p}\Lambda\_{2+p}}{\Lambda\_{3+p}} a\_{p+2}a\_{p+1} + \frac{1}{[3,q]} \frac{\Lambda\_{1+p}^3}{\Lambda\_{3+p}} a\_{p+1}^3\right) = \frac{(A-B)\psi\_3}{\Lambda\_{3+p}} \left\{ w\_3 - 2Bw\_1w\_2 + B^2w\_1^3 \right\},$$

equivalently, we have

$$\begin{split} & \left| \left( a\_{p+3} - \frac{q+2}{q^2+q+1} \frac{\Lambda\_{1+p}\Lambda\_{2+p}}{\Lambda\_{3+p}} a\_{p+2} a\_{p+1} + \frac{1}{[3,q]} \frac{\Lambda\_{1+p}^3}{\Lambda\_{3+p}} a\_{p+1}^3 \right) \right| \\ & \qquad = \frac{(A-B)\psi\_3}{\Lambda\_{3+p}} \left| \left( w\_3 + \frac{1}{4} w\_1 w\_2 + \frac{1}{16} w\_1^3 \right) - \frac{16B^2 - 1}{16} \left( w\_2 - w\_1^2 \right) + \frac{16B^2 - 32B - 5}{16} w\_2 \right| \\ & \qquad \le \frac{(A-B)\psi\_3}{\Lambda\_{3+p}} \left\{ 1 + \frac{16B^2 - 1}{16} + \frac{16B^2 - 32B - 5}{16} \right\} \\ & \qquad \le \frac{(A-B)\psi\_3}{\Lambda\_{3+p}} \left\{ \frac{16B^2 - 16B + 5}{8} \right\}, \end{split}$$

where we have used (9) and (10). This completes the proof.

**Theorem 5.** *Let f* ∈ <sup>A</sup>*p be given by* (1). *Then, the function f is in the class* S∗*p* (*q*, *μ*, *A*, *<sup>B</sup>*), *if and only if*

$$\frac{\epsilon^{j\theta} \left(B - \left[p\_{\prime}q\right]A\right)}{z} \left[\mathcal{L}\_{q}^{\mu+p-1} f\left(z\right) \* \left(\frac{\left(N+1\right)z^{p} - qLz^{p+1}}{\left(1-z\right)\left(1-qz\right)}\right)\right] \neq 0,\tag{24}$$

*for all*

$$\begin{array}{rcl} N &=& N\_{\theta} = \frac{\left( \left[ p, q \right] - 1 \right) e^{-i\theta}}{\left( \left[ p, q \right] A - B \right)}, \\\\ L &=& L\_{\theta} = \frac{\left( e^{-i\theta} + \left[ p, q \right] A \right)}{\left( \left[ p, q \right] A - B \right)}, \end{array} \tag{25}$$

*and also for N* = 0, *L* = 1.

**Proof.** Since the function *f* ∈ S∗*p* (*q*, *μ*, *A*, *B*) is analytic in D, it implies that L*μ*+*p*−<sup>1</sup> *q f* (*z*) = 0 for all *z* ∈ D∗ = D\{0}—that is

$$\frac{e^{i\theta}\left(B - \left[p\_{\prime}q\right]A\right)}{z} \mathcal{L}\_q^{\mu+p-1} f\left(z\right) \neq 0 \quad (z \in \mathbb{D})\ \_\mu$$

and this is equivalent to (24) for *N* = 0 and *L* = 1. From (7), according to the definition of the subordination, there exists an analytic function *w* with the property that *w* (0) = 0 and |*w*(*z*)| < 1 such that

$$\frac{z D\_q \mathcal{L}\_q^{\mu+p-1} f\left(z\right)}{\left[p\_\prime q\right] \mathcal{L}\_q^{\mu+p-1} f\left(z\right)} = \frac{1 + A\omega\left(z\right)}{1 + B\omega\left(z\right)} \left(z \in \mathbb{D}\right).$$

which is equivalent for *z* ∈ D, 0 *θ* < 2*π*

$$\frac{zD\_q \mathcal{L}\_q^{\mu+p-1} f\left(z\right)}{\left[p\_\prime q\right] \mathcal{L}\_q^{\mu+p-1} f\left(z\right)} \neq \frac{1 + Ae^{i\theta}}{1 + Be^{i\theta}}\tag{26}$$

and further written in a more simplified form

$$\left(1+B\epsilon^{i\theta}\right)zD\_q\mathcal{L}\_q^{\mu+p-1}f\left(z\right)-\left[p,q\right]\left(1+A\epsilon^{i\theta}\right)\mathcal{L}\_q^{\mu+p-1}f\left(z\right)\neq 0.\tag{27}$$

Now, using the following convolution properties in (27)

$$\mathcal{L}\_q^{\mu+p-1} f \begin{pmatrix} z \end{pmatrix} \* \frac{z^p}{(1-z)} = \mathcal{L}\_q^{\mu+p-1} f \begin{pmatrix} z \end{pmatrix} \quad \text{and} \quad \mathcal{L}\_q^{\mu+p-1} f \begin{pmatrix} z \end{pmatrix} \* \frac{z^p}{(1-z)(1-qz)} = z D\_q \mathcal{L}\_q^{\mu+p-1} f \begin{pmatrix} z \end{pmatrix}.$$

then, simple computation gives

$$\frac{1}{z} \left[ \mathcal{L}\_q^{\mu+p-1} f \left( z \right) \* \left( \frac{\left( 1 + Bc^{i\theta} \right) z^p}{\left( 1 - z \right) \left( 1 - qz \right)} - \frac{\left[ p \, \slash \right. \left( 1 + Ac^{i\theta} \right) z^p}{\left( 1 - z \right)} \right) \right] \neq 0 \,,$$

or equivalently

$$\frac{\left(B - \left[p,q\right]A\right)\mathfrak{e}^{i\theta}}{z} \left[\mathcal{L}\_q^{\mu+p-1}f\left(z\right)\*\left(\frac{\left(N+1\right)z^p - Lqz^{p+1}}{\left(1-z\right)\left(1-qz\right)}\right)\right] \neq 0,$$

which is the required direct part.

Assume that (11) holds true for *Lθ* − 1 = *Nθ* = 0, it follows that

$$\frac{c^{i\theta}\left(B - \left[p\_\prime q\right]A\right)}{z} \mathcal{L}\_q^{\mu+p-1} f\left(z\right) \neq 0,\text{ for all } z \in \mathbb{D}.$$

Thus, the function *h* (*z*) = *zDq*L*μ*+*p*−<sup>1</sup> *q f*(*z*) [*p*,*q*]L*μ*+*p*−<sup>1</sup> *q f*(*z*) is analytic in D and *h* (0) = 1. Since we have shown that (27) and (11) are equivalent, therefore we have

$$\frac{zD\_{\mathfrak{q}}\mathcal{L}\_{q}^{\mu+p-1}f\left(z\right)}{\left[p,q\right]\mathcal{L}\_{q}^{\mu+p-1}f\left(z\right)} \neq \frac{1+A\epsilon^{i\theta}}{1+B\epsilon^{i\theta}} \quad (z \in \mathbb{D})\,. \tag{28}$$

Suppose that

$$H(z) = \frac{1 + Az}{1 + Bz}, \ z \in \mathbb{D}.$$

Now, from relation (28) it is clear that *H* (*∂*D) ∩ *h* (D) = *φ*. Therefore, the simply connected domain *h* (D) is contained in a connected component of C\*H* (*∂*D). The univalence of the function *h*, together with the fact that *H* (0) = *h* (0) = 1, shows that *h* ≺ *H*, which shows that *f* ∈ S∗*p* (*q*, *μ*, *A*, *<sup>B</sup>*).

We now define an integral operator for the function *f* ∈ <sup>A</sup>*p* as follows:

**Definition 2.** *Let f* ∈ <sup>A</sup>*p*. *Then,* L : <sup>A</sup>*p* → <sup>A</sup>*p is called the q-analogue of Benardi integral operator for multivalent functions defined by* L (*f*) = *<sup>F</sup>η*,*<sup>p</sup> with η* > <sup>−</sup>*p*, *where <sup>F</sup>η*,*<sup>p</sup> is given by*

$$F\_{\eta, p}(z) \quad = \quad \frac{[\eta + p\_\prime q]}{z^{\eta}} \int\_0^z t^{\eta - 1} f(t) d\_q t\_\prime \tag{29}$$

$$=\left.z^{p}+\sum\_{n=1}^{\infty}\frac{\left[\eta+p,q\right]}{\left[\eta+p+n,q\right]}a\_{n+p}z^{n+p},\ \left(z\in\mathbb{D}\right).\tag{30}$$

We easily obtain that the series defined in (30) converges absolutely in D. Now, if *q* → 1, then the operator *<sup>F</sup>η*,*<sup>p</sup>* reduces to the integral operator studied in [29] and further by taking *p* = 1, we obtain the *q*-Bernardi integral operator introduced in [36]. If *q* → 1 and *p* = 1, we obtain the familiar Bernardi integral operator [37].

**Theorem 6.** *If f is of the form* (1)*, it belongs to the family* S∗*p* (*q*, *μ*, *A*, *B*) *and*

$$F\_{\eta,p}\left(z\right) = z^p + \sum\_{n=1}^{\infty} b\_{n+p} z^{n+p} \,\_{\prime} \tag{31}$$

*where <sup>F</sup>η*,*<sup>p</sup> is the integral operator given by* (29), *then*

$$|b\_{p+1}| \stackrel{<}{=} \frac{[\eta+p,q]}{[\eta+p+1,q]} \frac{\psi\_1 \left(A-B\right)}{\wedge\_{1+p}}.$$

*and for n* - 2

$$\left| \left| b\_{p+n} \right| \right| \le \frac{\left[ \eta + p, q \right]}{\left[ \eta + p + n, q \right]} \frac{\left( A - B \right) \psi\_n}{\wedge\_{n+p}} \prod\_{t=1}^{n-1} \left( 1 + \frac{\left[ p, q \right] \left( A - B \right)}{\left( \left[ p + t, q \right] - \left[ p, q \right] \right)} \right) \zeta$$

*where ψn and* <sup>∧</sup>*n*+*p are defined by* (14) *and* (5)*, respectively.*

**Proof.** The proof follows easily by using (30) and Theorem 2.

**Theorem 7.** *Let f* ∈ S∗*p* (*q*, *μ*, *A*, *B*) *and be given by* (1). *In addition, if <sup>F</sup>η*,*<sup>p</sup> is the integral operator is defined by* (29) *and is of the form* (31), *then for σ* ∈ C

$$\left| b\_{p+2} - \sigma b\_{p+1}^2 \right| \stackrel{<}{=} \frac{[\eta + p, \eta]}{[\eta + p + 2, \eta]} \frac{(A - B)\psi\_2}{\Lambda\_{p+2}} \left\{ 1; \ |v| \right\} \,,$$

*where*

$$\upsilon = \left( B - (A - B)\psi\_1 \right) + \frac{\Lambda\_{p+2}\psi\_1^2}{\Lambda\_{p+1}^2\psi\_2}(A - B)\frac{[\eta + p, q][\eta + p + 2, q]}{[\eta + p + 1, q]^2}\sigma. \tag{32}$$

**Proof.** From (30) and (31), we easily have

$$\begin{array}{rcl} b\_{p+1} &=& \frac{[\eta+p,q]}{[\eta+p+1,q]} a\_{p+1},\\ b\_{p+2} &=& \frac{[\eta+p,q]}{[\eta+p+2,q]} a\_{p+2}.\end{array}$$

Now,

$$\left| b\_{p+2} - \sigma b\_{p+1}^2 \right| = \frac{\left[ \eta + p, q \right]}{\left[ \eta + p + 2, q \right]} \left| a\_{p+2} - \sigma \frac{\left[ \eta + p, q \right] \left[ \eta + p + 2, q \right]}{\left( \left[ \eta + p + 1, q \right] \right)^2} a\_{p+1}^2 \right| \dots$$

By using (22) and (23), we have

$$\left| b\_{p+2} - \sigma b\_{p+1}^2 \right| = \frac{[\eta + p, \eta]}{[\eta + p + 2, \eta]} \frac{(A - B)}{\Lambda\_{p+2}} \left| w\_2 - \nu w\_1^2 \right| \,, $$

where *υ* is given by (32). Applying (9), we ge<sup>t</sup>

$$\left| b\_{p+2} - \sigma b\_{p+1}^2 \right| \le \frac{[\eta + p, q]}{[\eta + p + 2, q]} \frac{(A - B)}{\Lambda\_{p+2}} \left\{ 1, |v| \right\} \dots$$

Hence, we have the required result.

#### **4. Future Work**

The idea presented in this paper can easily be implemented to define some more subfamilies of analytic and univalent functions connected with different image domains [38–40].
