# Properties of neighborhood for certain classes associated with complex order and m-q-p-valent functions with higher order

## Abstract

In this paper, by using q-calculus (Jackson’s q-derivative) $$D_{q,p}$$ we defined new operator $$D_{\lambda ,q,p}^{n}f^{(m)}(z)$$. After that, we used this operator to introduce two new subclasses of multivalent analytic functions with complex order. Also, we obtained coefficients estimates and consequent inclusion relationships involving the $$N_{j,\delta ,m}^{p,q}(f)$$-neighborhood of these classes

## Introduction

Let $${\mathcal {A}}_{j}(p)$$ denote the class of functions in the form:

\begin{aligned} f(z)=z^{p}+\sum _{k=j+p}^{\infty }a_{k}z^{k}\; \;(j,p\in {{\mathbb {N}}} =\{1,2,\ldots \}), \end{aligned}
(1.1)

which are analytic and p-valent in the open is open unit disk$${\mathbb {U}}= \{z:\left| z\right| <1\}.$$We note that $${\mathcal {A}}_{1}(p)= {\mathcal {A}}(p)$$(see [13, 30]) and $${\mathcal {A}}_{1}(1)={\mathcal {A}}.$$ Also let T(pj) denote the subclass of $${\mathcal {A}}_{j}(p)$$which can express in the form:

\begin{aligned} f(z)=z^{p}-\sum _{k=j+p}^{\infty }a_{k}z^{k}\;\;(a_{k}\ge 0,\ j,p\in {\mathbb {N}} ). \end{aligned}
(1.2)

In recent years, the topic of q-calculus had attracted the attention of several researchers (see, for example, [2, 15, 16, 23, 34, 43,44,45]). Quantum calculus is the modern name for the investigation of calculus without limits. The quantum calculus or q-calculus began with Jackson in the early twentieth century, but this kind of calculus had already been worked out by Euler and Jacobi. In the general run, the q-calculus is used in various fields of Mathematics and Physics. Also, q-calculus appeared the connection between Mathematics and Physics. It had a lot of applications in different mathematical areas such as number theory, combinatorics, orthogonal polynomials, basic hyper-geometric functions and other sciences quantum theory, mechanics and the theory of relativity. Several convolutional and fractional calculus q-operators were defined by many researchers. The generalization of derivative and integral in q-calculus is known as q-analogue derivative and q-analogue integral. Recently, many authors used the q- analogue derivative and q-analogue integral to generalize many classes and many operators in Geometric Function Theory (see, for example, [14, 33, 40, 42]).

For a function $$f(z)\in A(p)$$ given by (1.1) (with $$j=1)$$ Jackson’s q-derivative (or q-difference) $$D_{q,p}$$ $$(0<q<1)$$is defined as follows:

\begin{aligned} D_{q,p}f(z)=\left\{ \begin{array}{ccc} \frac{f(z)-f(qz)}{(1-q)z} &{} &{} for\ z\ne 0, \\ &{} &{} \\ f^{\prime }(0)\ \ &{} &{} for\ z=0, \end{array} \underset{\ \ \ \ \ }{\overset{\ \ \ \ }{\underset{}{(f\in \mathcal {A(}p))}}} \right. \end{aligned}
(1.3)

provided that $$f^{\prime }(0)$$exists. From (1.1) (with $$j=1$$) and (1.3), we deduce that

\begin{aligned} D_{q,p}f(z)=\left[ p\right] _{q}z^{p-1}+\sum _{k=p+1}^{\infty }\left[ k\right] _{q}a_{k}z^{k-1}, \end{aligned}
(1.4)

such that q-integer number k $$\left[ k\right] _{q}$$ is defined by

\begin{aligned} \left[ k\right] _{q}=\frac{1-q^{k}}{1-q}=1+\sum \limits _{k=1}^{n-1}q^{k},\ 0<q<1,\ \left[ 0\right] _{q}=0. \end{aligned}
(1.5)

We observe that

\begin{aligned} \lim _{q\rightarrow 1^{-}}D_{q,p}f(z)=\lim _{q\rightarrow 1^{-}}\frac{ f(z)-f(qz)}{(1-q)z}=f^{\prime }(z), \end{aligned}

for a function fwhich is differentiable in a given subset of $${\mathbb {C}}$$. For all $$f(z)\in T(p,j),$$we find ( see  )

\begin{aligned} f^{(m)}(z)&=\theta (p,m)z^{p-m}\nonumber \\&\quad -\sum \limits _{k=j+p}^{\infty }\theta (k,m)a_{k}z^{k-m}\ (p,j\in {\mathbb {N}},\ m\in {\mathbb {N}}_{0}={\mathbb {N}} \cup \{0\},\ p>m), \end{aligned}
(1.6)

where $$\theta (p,m)$$is defined by

\begin{aligned} \theta (p,m)=\frac{p!}{(p-m)!}=\left\{ \begin{array}{ll} 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ , &{} m=0, \\ p(p-1)\ldots (p-m+1), &{} m\ne 0. \end{array} \right. \ \end{aligned}
(1.7)

For $$f\in T(p,j),$$ we introduce the operator $$D_{\lambda ,q,p}^{n}f^{(m)}:T(p,j)\rightarrow T(p,j)\ (\lambda \ge 0,\ n,m\in {\mathbb {N}} _{0},0<q<1,\ j,p\in {\mathbb {N}} ,\ p>m)$$as follows:

\begin{aligned} D_{\lambda ,q,p}^{n}f^{(m)}(z)= & {} f^{(m)}(z),\\ D_{\lambda ,q,p}^{1}f^{(m)}(z)= & {} D_{\lambda ,q,p}\left( D_{\lambda ,q,p}^{0}f^{(m)}(z)\right) =(1-\lambda )f^{(m)}(z)+\frac{\lambda }{\left[ p-m \right] _{q}}zD_{q,p}(f^{(m)}(z)) \\= & {} \theta (p,m)z^{p\mathbf {-}m}-\sum \limits _{k=j+p}^{\infty }\theta (k,m) \left[ \tfrac{\left[ p-m\right] _{q}+\lambda \left( \left[ k-m\right] _{q}- \left[ p-m\right] _{q}\right) }{\left[ p-m\right] _{q}}\right] a_{k}z^{k-m},\\&D_{\lambda ,q,p}^{2}f^{(m)}(z) =D_{\lambda ,q,p}\left( D_{\lambda ,q,p}^{1}f^{(m)}(z)\right) =(1-\lambda )D_{\lambda ,q,p}^{1}f^{(m)}(z)+\frac{ \lambda }{\left[ p-m\right] _{q}}zD_{q,p}(D_{\lambda ,q,p}^{1}f^{(m)}(z)) \\= & {} \theta (p,m)z^{p\mathbf {-}m}-\sum \limits _{k=j+p}^{\infty }\theta (k,m) \left[ \tfrac{\left[ p-m\right] _{q}+\lambda \left( \left[ k-m\right] _{q}- \left[ p-m\right] _{q}\right) }{\left[ p-m\right] _{q}}\right] ^{2}a_{k}z^{k-m} \vdots \end{aligned}
\begin{aligned} D_{\lambda ,q,p}^{n}f^{(m)}(z)&=D_{\lambda ,q,p}\left( D_{\lambda ,q,p}^{n-1}f^{(m)}(z)\right) \nonumber \\&=(1-\lambda )D_{\lambda ,q,p}^{n-1}f^{(m)}(z)+ \frac{\lambda }{\left[ p-m\right] _{q}}zD_{q,p}(D_{\lambda ,q,p}^{n-1}f^{(m)}(z))(n\in {\mathbb {N}} ) \end{aligned}
(1.8)

From (1.2) and (1.8), we can obtain

\begin{aligned} =\theta (p,m)z^{p\mathbf {-}m}-\sum \limits _{k=j+p}^{\infty }\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k}z^{k-m}, \end{aligned}

where

\begin{aligned} \Psi _{q,p}^{n,m}(k,\lambda )&=\left[ \frac{\left[ p-m\right] _{q}+\lambda \left( \left[ k-m\right] _{q}-\left[ p-m\right] _{q}\right) }{\left[ p-m \right] _{q}}\right] ^{n} \nonumber \\&\quad (\lambda \ge 0,\ m,n\in {\mathbb {N}} _{0},0<q<1,\ j,p\in {\mathbb {N}} ,p>m). \end{aligned}
(1.9)

We note that

1. (1)

$$D_{\lambda ,q,p}^{n}f^{(0)}(z)=I_{q,p}^{n}(\lambda )f(z),$$(Aouf and Madian , with $$\varrho =0$$]);

2. (2)

$$\lim _{q\rightarrow 1^{-}}D_{1,q,p}^{n}f^{(m)}(z)=D_{p}^{n}f^{(m)}(z),$$(Aouf [8, 9]);

3. (3)

$$\lim _{q\rightarrow 1^{-}}D_{1,q,p}^{n}f^{(0)}(z)=D_{p}^{n}f(z)$$(see [11, 18], Cătaş , with $$l=0$$] and );

4. (4)

$$\lim _{q\rightarrow 1^{-}}D_{1,q,1}^{n}f^{(0)}(z)=D^{n}f(z)$$(see ([26, 27], with $$l=0$$));

5. (5)

$$\lim _{q\rightarrow 1^{-}}D_{\lambda ,q,1}^{n}f^{(0)}(z)=D_{\lambda }^{n}f(z)$$(see [1, 17, 21]);

6. (6)

$$D_{1,q,1}^{n}f^{(0)}(z)=D_{q}^{n}f(z)$$(see  ), $$\lim _{q\rightarrow 1^{-}}D_{q}^{n}f(z)=D^{n}f(z)$$(see Sălă gean  see also [10, 12]);

7. (7)

$$\lim _{q\rightarrow 1^{-}}D_{1,q,p}^{n}f^{(m)}(z)=D_{p}^{n}f^{(m)}(z)$$(see Aouf et al. );

8. (8)

$$\lim _{q\rightarrow 1^{-}}D_{\lambda ,q,p}^{n}f^{(m)}(z)=I_{\lambda ,p}^{n}f^{(m)}(z)$$

\begin{aligned} =\left\{ \begin{array}{c} f\in T(p,j):I_{\lambda ,p}^{n}f^{(m)}(z)=\theta (p,m)z^{p-m}-\sum \limits _{k=j+p}^{\infty }\left( \left[ \frac{p-m+\lambda \left( k-p\right) }{ p-m}\right] \right) ^{n}\theta (k,m)a_{k}z^{k-m}, \\ n,m\in {\mathbb {N}} _{0},\ j,p\in {\mathbb {N}} ,\lambda \ge 0,\ p>m \end{array} \right\} ; \end{aligned}
9. (9)

$$D_{1,q,p}^{n}f^{(m)}(z)=I_{q,p}^{n}f^{(m)}(z)$$

\begin{aligned} =\left\{ \begin{array}{c} f\in T(p,j):I_{q,p}^{n}f^{(m)}(z)=\theta (p,m)z^{p-m}-\sum \limits _{k=j+p}^{\infty }\left( \frac{\left[ k-m\right] _{q}}{\left[ p-m \right] _{q}}\right) ^{n}\theta (k,m)a_{k}z^{k-m}, \\ n,m\in {\mathbb {N}} _{0},0<q<1,\ j,p\in {\mathbb {N}} ,p>m \end{array} \right\} ; \end{aligned}
10. (10)

$$D_{1,q,p}^{n}f^{(0)}(z)=D_{q,p}^{n}f(z)$$

\begin{aligned} =\left\{ \begin{array}{c} f\in T(p,j):D_{q,p}^{n}f(z)=z^{p}-\sum \limits _{k=j+p}^{\infty }\left( \frac{ \left[ k\right] _{q}}{\left[ p\right] _{q}}\right) ^{n}a_{k}z^{k}, \\ n\in {\mathbb {N}} _{0},\ j,p\in {\mathbb {N}} ,\ 0<q<1 \end{array} \right\} . \end{aligned}

Now by using $$D_{\lambda ,q,p}^{n}f^{(m)}(z),$$we defined the classes $$F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )$$and $$G_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )$$in below definitions:

### Definition 1

Assume$$f(z)\in T(p,j),$$then $$f(z)\in F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )$$if it satisfies the following inequality:

\begin{aligned} \left| \frac{1}{b}\left[ \dfrac{(1-\sigma )zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))+\sigma zD_{q,p}(zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z)))}{(1-\sigma )D_{\lambda ,q,p}^{n}f^{(m)}(z)+\sigma zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))}-\left[ p-m\right] _{q}\right] \right| <\beta \end{aligned}
\begin{aligned}&(b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\lambda \ge 0,\ 0\le \sigma \le 1,\ 0<\beta \le 1,p>m). \end{aligned}
(1.10)

We observe that:

1. (1)

$$\lim _{q\rightarrow 1^{-}}F_{q,p}^{n,m}(j,1,\sigma ,b,\beta )=S_{j}(n,p,m,\sigma ,b,\beta )$$ see Aouf et al. ;

2. (2)

$$F_{q,p}^{n,0}(j,\lambda ,\sigma ,b,\beta )=S_{q}^{n}(j,\lambda ,p,\sigma ,b,\beta )$$see Aouf and madian , with $$\varrho =0$$];

3. (3)

$$\lim _{q\rightarrow 1^{-}}F_{q,p}^{0,0}(j,\lambda ,\sigma ,b,\beta )=S_{j}(p,\sigma ,b,\beta )$$ see Aouf and Mostafa , with $$b_{k}=1$$;

4. (4)

$$\lim _{q\rightarrow 1^{-}}F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )=F_{p}^{n,m}(j,\lambda ,\sigma ,b,\beta )\;$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \tfrac{z(I_{\lambda ,p}^{n}f^{(m)}(z))^{\prime }+\sigma z^{2}(I_{\lambda ,p}^{n}f^{(m)}(z))^{\prime \prime }}{(1-\sigma )I_{\lambda ,p}^{n}f^{(m)}(z)+\sigma z(I_{\lambda ,p}^{n}f^{(m)}(z))^{\prime }}-\left( p-m\right) \right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\lambda \ge 0,\ 0\le \sigma \le 1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}
5. (5)

$$F_{q,p}^{n,m}(j,1,\sigma ,b,\beta )=F_{q,p}^{n,m}(j,\sigma ,b,\beta )$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \tfrac{(1-\sigma )zD_{q,p}(I_{q,p}^{n}f^{(m)}(z))+\sigma zD_{q,p}(zD_{q,p}(I_{q,p}^{n}f^{(m)}(z)))}{(1-\sigma )I_{q,p}^{n}f^{(m)}(z)+\sigma zD_{q,p}(I_{q,p}^{n}f^{(m)}(z))}-\left[ p-m \right] _{q}\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\ 0\le \sigma \le 1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}
6. (6)

$$\ F_{q,p}^{n,0}(j,1,\sigma ,b,\beta )=F_{q,p}^{n}(j,\sigma ,b,\beta )$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \tfrac{(1-\sigma )zD_{q,p}(D_{q,p}^{n}f(z))+\sigma zD_{q,p}(zD_{q,p}(D_{q,p}^{n}f(z)))}{ (1-\sigma )D_{q,p}^{n}f(z)+\sigma zD_{q,p}(D_{q,p}^{n}f(z))}-\left[ p\right] _{q}\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\ 0\le \sigma \le 1,\ 0<\beta \le 1\overset{}{}\right\} ; \end{aligned}
7. (7)

$$\lim _{q\rightarrow 1^{-}}F_{q,p}^{n,m}(j,\lambda ,0,b,\beta )=SF_{p}^{n,m}(j,\lambda ,\sigma ,b,\beta )$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \tfrac{z(I_{\lambda ,p}^{n}f^{(m)}(z))^{\prime }}{I_{\lambda ,p}^{n}f^{(m)}(z)}-\left( p-m\right) \right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\lambda \ge 0,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}
8. (8)

$$F_{q,p}^{n,m}(j,1,0,b,\beta )=SF_{q,p}^{n,m}(j,b,\beta )\;$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \tfrac{ zD_{q,p}(I_{q,p}^{n}f^{(m)}(z))}{I_{q,p}^{n}f^{(m)}(z)}-\left[ p-m\right] _{q}\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}
9. (9)

$$\ F_{q,p}^{n,m}(j,\lambda ,0,b,\beta )=S_{q,p}^{n,m}(j,\lambda ,b,\beta )$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \dfrac{ zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))}{D_{\lambda ,q,p}^{n}f^{(m)}(z)}- \left[ p-m\right] _{q}\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}}^{*},m,n\in {\mathbb {N}}_{0},p,j\in {\mathbb {N}},\ 0<q<1,\lambda \ge 0,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}
10. (10)

$$\lim _{q\rightarrow 1^{-}}F_{q,p}^{n,m}(j,\lambda ,1,b,\beta )=KF_{p}^{n,m}(j,\lambda ,b,\beta )\;$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ 1+\tfrac{z(I_{\lambda ,p}^{n}f^{(m)}(z))^{\prime \prime }}{(I_{\lambda ,p}^{n}f^{(m)}(z))^{\prime } }-\left( p-m\right) \right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\lambda \ge 0,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}
11. (11)

$$F_{q,p}^{n,m}(j,1,1,b,\beta )=KF_{q,p}^{n,m}(j,b,\beta )\;$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \tfrac{ D_{q,p}(zD_{q,p}(I_{q,p}^{n}f^{(m)}(z)))}{D_{q,p}(I_{q,p}^{n}f^{(m)}(z))}- \left[ p-m\right] _{q}\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}
12. (12)

$$\ F_{q,p}^{n,m}(j,\lambda ,1,b,\beta )=K_{q,p}^{n,m}(j,\lambda ,b,\beta )$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \dfrac{ D_{q,p}(zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z)))}{D_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))}-\left[ p-m\right] _{q}\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\lambda \ge 0,\ 0<\beta \le 1,p>m\overset{}{}\right\} . \end{aligned}

### Definition 2

Assume $$f(z)\in T(p,j),\;$$if it satisfies (1.11), then$$f(z)\in G_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )\;$$

\begin{aligned} \left| \frac{1}{b}\left\{ (1-\sigma )\frac{D_{\lambda ,q,p}^{n}f^{(m)}(z)}{z^{p-m}}+\sigma \frac{D_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))}{[p-m]_{q}z^{p-m-1}}-\theta (p,m)\right\} \right| <\beta \end{aligned}
\begin{aligned} (b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\lambda \ge 0,\ 0\le \sigma \le 1,\ 0<\beta \le 1,p>m). \end{aligned}
(1.11)

We note that:

1. (1)

$$\lim _{q\rightarrow 1^{-}}G_{q,p}^{n,m}(j,1,\sigma ,b,\beta )=K_{j}(n,p,m,\sigma ,b,\beta )\;$$ see Aouf et al. ;

2. (2)

$$G_{q,p}^{n,0}(j,\lambda ,\sigma ,b,\beta )=K_{q}^{n}(j,\lambda ,p,\sigma ,b,\beta )\;$$see Aouf and Madian , with $$\varrho =0$$];

3. (3)

$$\lim _{q\rightarrow 1^{-}}G_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )=G_{p}^{n,m}(j,\lambda ,\sigma ,b,\beta )\;$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ (1-\sigma )\frac{ I_{\lambda ,p}^{n}f^{(m)}(z)}{z^{p-m}}+\sigma \frac{(I_{\lambda ,p}^{n}f^{(m)}(z))^{\prime }}{(p-m)z^{p-m-1}}-\theta (p,m)\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}}^{*},m,n\in {\mathbb {N}}_{0},p,j\in {\mathbb {N}} ,\lambda \ge 0,\ 0\le \sigma \le 1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}
4. (4)

$$\ G_{q,p}^{n,m}(j,1,\sigma ,b,\beta )=G_{q,p}^{n,m}(j,\sigma ,b,\beta )$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left\{ (1-\sigma )\frac{ I_{q,p}^{n}f^{(m)}(z)}{z^{p-m}}+\sigma \frac{D_{q,p}(I_{q,p}^{n}f^{(m)}(z))}{ [p-m]_{q}z^{p-m-1}}-\theta (p,m)\right\} \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}}^{*},m,n\in {\mathbb {N}}_{0},p,j\in {\mathbb {N}},\ 0<q<1,0\le \sigma \le 1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}
5. (5)

$$\ G_{q,p}^{n,0}(j,1,\sigma ,b,\beta )=G_{q,p}^{n}(j,\sigma ,b,\beta )$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left\{ (1-\sigma )\frac{ D_{q,p}^{n}f(z)}{z^{p}}+\sigma \frac{D_{q,p}(D_{q,p}^{n}f(z))}{[p]_{q}z^{p-1} }-1\right\} \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}}^{*},n\in {\mathbb {N}}_{0},p,j\in {\mathbb {N}},\ 0<q<1,\ 0\le \sigma \le 1,\ 0<\beta \le 1\overset{}{}\right\} ; \end{aligned}
6. (6)

$$\lim _{q\rightarrow 1^{-}}G_{q,p}^{n,m}(j,\lambda ,1,b,\beta )=L_{p}^{n,m}(j,\lambda ,b,\beta )\;$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \frac{(I_{\lambda ,p}^{n}f^{(m)}(z))^{\prime }}{(p-m)z^{p-m-1}}-\theta (p,m)\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\lambda \ge 0,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}
7. (7)

$$\ G_{q,p}^{n,m}(j,1,1,b,\beta )=M_{q,p}^{n,m}(j,b,\beta )$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \frac{ D_{q,p}(I_{q,p}^{n}f^{(m)}(z))}{[p-m]_{q}z^{p-m-1}}-\theta (p,m)\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}
8. (8)

$$\ G_{q,p}^{n,0}(j,1,1,b,\beta )=G_{q,p}^{n}(j,b,\beta )$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \frac{ D_{q,p}(D_{q,p}^{n}f(z))}{[p]_{q}z^{p-1}}-1\right] \right|<\beta \right. \\&\left. b\in {\mathbb {C}} ^{*},n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\ 0\le \sigma \le 1,\ 0<\beta \le 1\overset{}{}\right\} ; \end{aligned}
9. (9)

$$\lim _{q\rightarrow 1^{-}}G_{q,p}^{n,m}(j,\lambda ,0,b,\beta )=O_{p}^{n,m}(j,\lambda ,b,\beta )\;$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \frac{I_{\lambda ,p}^{n}f^{(m)}(z)}{z^{p-m}}-\theta (p,m)\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\lambda \ge 0,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}
10. (10)

$$\ G_{q,p}^{n,m}(j,1,0,b,\beta )=R_{q,p}^{n,m}(j,b,\beta )$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \frac{ I_{q,p}^{n}f^{(m)}(z)}{z^{p-m}}-\theta (p,m)\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}
11. (11)

$$\ G_{q,p}^{n,0}(j,1,0,b,\beta )=P_{q,p}^{n}(j,b,\beta )$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \frac{D_{q,p}^{n}f(z)}{ z^{p}}-1\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\ 0<\beta \le 1\overset{}{}\right\} ; \end{aligned}
12. (12)

$$G_{q,p}^{n,m}(j,\lambda ,1,b,\beta )=G_{q,p}^{n,m}(j,\lambda ,b,\beta )\;$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \frac{D_{q,p}(D_{,\lambda ,q,p}^{n}f^{(m)}(z))}{[p-m]_{q}z^{p-m-1}}-\theta (p,m)\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\lambda \ge 0,\ 0<q<1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}
13. (13)

$$G_{q,p}^{n,m}(j,\lambda ,0,b,\beta )=GL_{p}^{n,m}(j,\lambda ,b,\beta )$$

\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \frac{I_{\lambda ,p}^{n}f^{(m)}(z)}{z^{p-m}}-\theta (p,m)\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\lambda \ge 0,\ 0<\beta \le 1,p>m\overset{}{}\right\} . \end{aligned}

Now, as a results of Authors articles see ([3,4,5,6,7] [29, 31, 34, 35, 38]), we define the neighborhood $$(j,\delta )$$for $$f\in T(p,j)$$by

\begin{aligned} N_{j,\delta }^{p}(f)= & {} \left\{ g:g\in T(p,j),\ g(z)\right. \nonumber \\&\quad \left. =z^{p}-\sum \limits _{k=j+p}^{\infty }b_{k\ }z^{k}\ \text {and }\sum \limits _{k=j+p}^{\infty }k\vert a_{k}-b_{k\ }\vert \le \delta \right\} . \end{aligned}
(1.12)

In specially, if

\begin{aligned} h(z)=z^{p}\ (p\in {\mathbb {N}} ), \end{aligned}
(1.13)

we obtain

\begin{aligned} N_{j,\delta }^{p}(h)=\left\{ g:g\in T(p,j),\ g(z)=z^{p}-\sum \limits _{k=j+p}^{\infty }b_{k\ }z^{k}\ \text {and }\sum \limits _{k=j+p}^{\infty }k\left| b_{k\ }\right| \le \delta \right\} . \end{aligned}
(1.14)

Now, we define the $$\left( q,j,\delta ,m\right) -$$neighborhood for $$f\in T(p,j)$$ by

\begin{aligned} N_{j,\delta ,m}^{p,q}(f)= & {} \left\{ g:g\in T(p,j),\ g(z)=z^{p}\right. \nonumber \\&\quad \left. -\sum \limits _{k=j+p}^{\infty }b_{k\ }z^{k}\ \text {and }\sum \limits _{k=j+p}^{\infty }\left[ k-m\right] _{q}\left| a_{k}-b_{k\ }\right| \le \delta \right\} . \end{aligned}
(1.15)

In particular, if $$h(z)\;$$given by (1.12), we immediately have

\begin{aligned} N_{j,\delta ,m}^{p,q}(h)= & {} \left\{ g:g\in T(p,j),\ g(z)=z^{p}\right. \nonumber \\&\quad \left. -\sum \limits _{k=j+p}^{\infty }b_{k\ }z^{k}\ \text {and }\sum \limits _{k=j+p}^{\infty }\left[ k-m\right] _{q}\left| b_{k\ }\right| \le \delta \right\} . \end{aligned}
(1.16)

We note that

1. (i)

$$\ N_{j,\delta ,0}^{p,q}(f)=N_{j,\delta }^{p,q}(f)\;$$and$$\ N_{j,\delta ,0}^{p,q}(h)=N_{j,\delta }^{p,q}(h)\;$$(see [14, 20]);

2. (ii)

$$\ \lim _{q\rightarrow 1^{-}}N_{j,\delta ,0}^{p,q}(f)=N_{j,\delta }^{p}(f)\;$$and$$\ \lim _{q\rightarrow 1^{-}}N_{j,\delta ,0}^{p,q}(h)=N_{j,\delta }^{p}(h)\;$$(see  and Aouf et al.  ).

## Preliminaries

On the other hand, we assume through the article that, $$b\in {\mathbb {C}} ^{*},\ n,m\in {\mathbb {N}} _{0},\ p,j\in {\mathbb {N}} ,\ \lambda \ge 0,\ 0<q<1,\ 0\le \sigma \le 1$$ ,$$\ 0<\beta \le 1,p>m$$ and$$\ \Psi _{q,p}^{n,m}(k,\lambda )\;$$is given by (1.9). To prove the main outcomes in the article we need Lemmas 1 and 2 below.

### Lemma 1

Let$$f\in T(p,j)$$ is given by (1.2), then$$f\in F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta ),$$

if and only if

\begin{aligned} \sum \limits _{k=j+p}^{\infty }{(}\left[ k-m\right] _{q}{+\beta }\left| b\right| {-}\left[ p-m\right] _{q}{ )[1+\sigma (}\left[ k-m\right] _{q}{-1)]\theta (k,m)\Psi }_{q,p}^{n,m} {(k,\lambda )a}_{k}\nonumber \\ {\le }{\beta }\left| b\right| \theta (p,m){ [1+\sigma (}\left[ p-m\right] _{q}{-1)].} \end{aligned}
(2.1)

### Proof

If $$f\in F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta ).$$ Then we have

\begin{aligned} \mathrm{Re}\left\{ \tfrac{(1-\sigma )zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))+\sigma zD_{q,p}(zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z)))}{(1-\sigma )D_{\lambda ,q,p}^{n}f^{(m)}(z)+\sigma zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))}-\left[ p-m\right] _{q}\right\} >-\beta \left| b\right| \ (z\in {\mathbb {U}} ), \end{aligned}
(2.2)

or, equivalently,

\begin{aligned} \mathrm{Re}\left\{ \tfrac{-\sum \limits _{k=j+p}^{\infty }(\left[ k-m\right] _{q}- \left[ p-m\right] _{q})[1+\sigma (\left[ k-m\right] _{q}-1)]\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k\ }z^{k-p}}{[1+\sigma (\left[ p-m\right] _{q}-1)]\theta (p,m)-\sum \limits _{k=j+p}^{\infty }[1+\sigma (\left[ k-m \right] _{q}-1)]\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k}\ z^{k-p}} \right\} >-\beta \left| b\right| . \end{aligned}
(2.3)

By setting $$\left| {\small z}\right|$$ $$=r (0\le r<1)$$in (2.3), the term in the denominator of the left hand side of (2.3) is positive for $$0\le r<1$$. Therefore, by Putting $$r\longrightarrow 1$$ through real values, (2.3) helps us to the desired assertion of Lemma 1.

Conversely, assume $$\left| z\right| =1$$and apply the hypothesis (2.1), from (2.3) we have

\begin{aligned}&\left| \dfrac{(1-\sigma )zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))+\sigma zD_{q,p}(zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z)))}{(1-\sigma )D_{\lambda ,q,p}^{n}f^{(m)}(z)+\sigma zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))}-\left[ p-m\right] _{q}\right| \\&=\left| \frac{\sum \limits _{k=j+p}^{\infty }(\left[ k-m\right] _{q}- \left[ p-m\right] _{q})[1+\sigma (\left[ k-m\right] _{q}-1)]\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k\ }z^{k-p}}{[1+\sigma (\left[ p-m\right] _{q}-1)]\theta (p,m)-\sum \limits _{k=j+p}^{\infty }[1+\sigma (\left[ k-m \right] _{q}-1)]\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k}\ z^{k-p}} \right| \\&\le \frac{\sum \limits _{k=j+p}^{\infty }(\left[ k-m\right] _{q}-\left[ p-m \right] _{q})[1+\sigma (\left[ k-m\right] _{q}-1)]\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k\ }\left| z\right| ^{k-p}}{[1+\sigma ( \left[ p-m\right] _{q}-1)]\theta (p,m)-\sum \limits _{k=j+p}^{\infty }[1+\sigma (\left[ k-m\right] _{q}-1)]\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k}\ \left| z\right| ^{k-p}}\\&\le \tfrac{\sum \limits _{k=j+p}^{\infty }(\left[ k-m\right] _{q}-\left[ p-m \right] _{q})[1+\sigma (\left[ k-m\right] _{q}-1)]\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k\ }}{[1+\sigma (\left[ p-m\right] _{q}-1)]\theta (p,m)-\sum \limits _{k=j+p}^{\infty }[1+\sigma (\left[ k-m\right] _{q}-1)]\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k}\ }=\beta \left| b\right| . \end{aligned}

So, we have $$f(z)\in F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )\;$$by applying the maximum modulus theorem, which completes the proof of Lemma 1. $$\square$$

### Remark 1

Letting $$q\rightarrow 1^{-}$$ and $$n=m=0$$ in Lemma 1, we obtain the result obtained by Aouf and Mostafa , Lemma 1, with $$b_{k}=1$$].

The following lemma can be established similarly.

### Lemma 2

Let $$f\in T(p,j)$$ is given by (1.2). Then$$f\in G_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta ),$$

if and only if

\begin{aligned} \sum \limits _{k=j+p}^{\infty }{[}\left[ p-m\right] _{q}{+\sigma (}\left[ k-m\right] _{q}{-}\left[ p-m\right] _{q}{)]\theta (k,m)\Psi }_{q,p}^{n,m}{(k,\lambda )a}_{k}{\le \beta } \left| b\right| \left[ p-m\right] _{q}{.} \end{aligned}
(2.4)

## 3- Inclusion results

In this part, we showed inclusion relations for each of the classes $$F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )\;$$and $$G_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )\;$$including $$(q,j,\delta ,m)-\;$$ neighborhood were defined by (1.15) and (1.16).

### Theorem 1

Suppose$$f\in T(p,j)\;$$includes in $$F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta ),\;$$ then

\begin{aligned} F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta ){\subset N}_{j,\eta ,m}^{p,q}{(h),} \end{aligned}
(3.1)

since h(z) is defined by (1.13) and $$\eta$$is given by

\begin{aligned} {\eta =}\tfrac{\left[ j+p-m\right] _{q}\beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)]{\theta (p,m)}}{(\left[ j+p-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )}{\ (}\left[ p-m\right] _{q}{>}\left| b\right| {).} \end{aligned}
(3.2)

### Proof

Let $$f\in F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )$$, then by using (2.1) of Lemma 1, we obtain

\begin{aligned}&(\left[ j+p-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )\sum \limits _{k=j+p}^{\infty }a_{k}\\&\le \sum \limits _{k=j+p}^{\infty }(\left[ k-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ k-m\right] _{q}-1)] {\theta (k,m)}\Psi _{q,p}^{n,m}(k,\lambda )a_{k} \end{aligned}
\begin{aligned} \le \beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)] {\theta (p,m)}, \end{aligned}
(3.3)

which quickly gives

\begin{aligned} \sum \limits _{k=j+p}^{\infty }a_{k}\le \tfrac{\beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)]{\theta (p,m)}}{(\left[ j+p-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )}. \end{aligned}
(3.4)

Making use of (2.1) with (3.4), we obtain

\begin{aligned}&[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )\sum \limits _{k=j+p}^{\infty }\left[ k-m\right] _{q}a_{k}\le \\&\le \beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)] {\theta (p,m)}+ \\&\quad (\left[ p-m\right] _{q}-\beta \left| b\right| )[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )\sum \limits _{k=j+p}^{\infty }a_{k}\\&\le \beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)] {\theta (p,m)}+\frac{(\left[ p-m\right] _{q}-\beta \left| b\right| )\beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)]{\theta (p,m)}}{(\left[ j+p-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})}\\&=\frac{\left[ j+p-m\right] _{q}\beta \left| b\right| [1+\sigma ( \left[ p-m\right] _{q}-1)]{\theta (p,m)}}{\left[ j+p-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q}}. \end{aligned}

Hence

\begin{aligned} \sum \limits _{k=j+p}^{\infty }\left[ k-m\right] _{q}a_{k}&\le \tfrac{\left[ j+p-m\right] _{q}\beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)]{\theta (p,m)}}{(\left[ j+p-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)] {\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )}\nonumber \\&=\eta \ (\left[ p-m \right] _{q}>\left| b\right| ), \end{aligned}
(3.5)

by means of (1.14), we obtained (3.1) which asserted by Theorem 1. $$\square$$

### Remark 2

Letting $$q\rightarrow 1^{-}$$ and $$n=m=0\;$$in Theorem 1, we obtain the result obtained by Aouf and Mostafa , Theorem 2, with $$b_{k}=1$$].

In a similar manner, we proved the following inclusion relationship by using (2.4) of Lemma 2 recompensed (2.1) of Lemma 1 on functions in $$G_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta ).$$

### Theorem 2

Assume $$f\in T(p,j)\;$$includes in $$G_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta ),$$ then

\begin{aligned} G_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta ){\subset }N_{j,\delta ,m}^{p,q}{(h),} \end{aligned}
(3.6)

such that$$h(z)\;$$is defined by (1.13) and $$\delta \;$$is introduced by

\begin{aligned} {\delta =}\frac{\left[ j+p-m\right] _{q}\beta \left| b\right| \left[ p-m\right] _{q}}{[\left[ p-m\right] _{q}+\sigma (\left[ j+p-m \right] _{q}-\left[ p-m\right] _{q})]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )}={.} \end{aligned}
(3.7)

## 4- Neighborhoods properties

In this section, we determine the neighborhood for each of the classes $$F_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta )$$and $$G_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta )$$. If there exists a function $$\rho (z)\in F_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta ),$$ satisfies (4.1), then $$f(z)\in T(p,j)$$is said to be in the class $$F_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta )$$

\begin{aligned} \left| \frac{f(z)}{\rho (z)}-1\right|<\left[ p-m\right] _{q}-\gamma \ \ (z\in {\mathbb {U}};\ 0\le \gamma <\left[ p-m\right] _{q}). \end{aligned}
(4.1)

Analogously,if we find a function $$\rho (z)\in G_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta )\;$$which the inequality (4.1) achieve, then we can say for $$f(z)\in T(p,j),\ f(z)\in G_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta ).$$

### Theorem 3

Let $$f\in T(p,j)$$includes in $$F_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta )$$and

\begin{aligned} \gamma&=\left[ p-m\right] _{q}\nonumber \\&\quad -\tfrac{\eta (\left[ j+p-m \right] _{q}+\beta \vert b \vert -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )}{\left[ j+p-m\right] _{q}\left\{ (\left[ j+p-m \right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )-\beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)]\right\} {\theta (p,m)}}{,} \end{aligned}
(4.2)

then

\begin{aligned} {N}_{j,\eta ,m}^{p,q}{(h)\subset }F_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta ){,} \end{aligned}
(4.3)

where

\begin{aligned} {\eta \le }\left[ p-m\right] _{q}\left[ j+p-m\right] _{q}\left\{ 1- \tfrac{\beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)] {\theta (p,m)}}{(\left[ j+p-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]{ \theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )}\right\} {.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \end{aligned}

### Proof

Assume $$f\in N_{j,\eta ,m}^{p,q}(h).$$ From (1.15) we find that

\begin{aligned} \sum \limits _{k=j+p}^{\infty }\left[ k-m\right] _{q}\left| a_{k}-b_{k}\right| \le \eta , \end{aligned}
(4.4)

\begin{aligned} \sum \limits _{k=j+p}^{\infty }\left| a_{k}-b_{k}\right| \le \frac{ \eta }{\left[ j+p-m\right] _{q}}\ \ (p,\ j\in {\mathbb {N}}). \end{aligned}
(4.5)

Next, since $$\rho (z)\in F_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta ),\;$$by using (3.4), we have

\begin{aligned} \sum \limits _{k=j+p}^{\infty }b_{k}\le \tfrac{\beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)]{\theta (p,m)}}{(\left[ j+p-m \right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )}, \end{aligned}
(4.6)

so that

\begin{aligned} \left| \frac{f(z)}{\rho (z)}-1\right|&\le \frac{\sum \limits _{k=j+p}^{\infty }\left| a_{k}-b_{k}\right| }{ 1-\sum \limits _{k=j+p}^{\infty }b_{k}}\\&\le \tfrac{\eta (\left[ j+p-m\right] _{q}+\beta \left| b\right| - \left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )}{\left[ j+p-m\right] _{q}\left\{ ( \left[ j+p-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]\Psi _{q,p}^{n,m}(j+p,\lambda ) {\theta (j+p,m)}-\beta \left| b\right| [1+\sigma (\left[ p-m \right] _{q}-1)]{\theta (p,m)}\right\} }\\&=\left[ p-m\right] _{q}-\gamma , \end{aligned}

provided that $$\gamma$$ is given by (4.2) and by the above definition, $$f\in F_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta )$$, so the proof of Theorem 3 is finished. $$\square$$

The proof of Theorem 4 below is similar to the proof of Theorem 3, we omit the details involved.

### Theorem 4

Let $$f\in T(p,j)$$includes in $$G_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta )$$and

\begin{aligned}&\gamma =\left[ p-m\right] _{q}\nonumber \\&\quad -\tfrac{\delta [\left[ p-m\right] _{q}+\sigma (\left[ j+p-m\right] _{q}-\left[ p-m\right] _{q})]\Psi _{q,p}^{n,m}(j+p,\lambda ){\theta (j+p,m)}}{\left[ j+p-m \right] _{q}\left\{ [\left[ p-m\right] _{q}+\sigma (\left[ j+p-m\right] _{q}- \left[ p-m\right] _{q})]\Psi _{q,p}^{n,m}(j+p,\lambda ){\theta (j+p,m) }-\beta \left[ p-m\right] _{q}\left| b\right| \right\} }{,} \end{aligned}
(4.7)

then

\begin{aligned} N_{j,\delta ,m}^{p,q}{(h)\subset }G_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta ){,} \end{aligned}
(4.8)

where

\begin{aligned} {\delta \le }\left[ p-m\right] _{q}\left[ j+p-m\right] _{q}\left\{ 1- \tfrac{\beta \left[ p-m\right] _{q}\left| b\right| }{(\left[ p-m \right] _{q}+\sigma (\left[ j+p-m\right] _{q}-\left[ p-m\right] _{q})]\Psi _{q,p}^{n,m}(j+p,\lambda ){\theta (j+p,m)}}\right\} {.} \end{aligned}
(4.9)

### Remarks

(1) Taking $$m=0\;$$in our outcomes, we obtain the outcomes obtained by Aouf and Madian , with $$\varrho =0$$]; (2) Taking $$q\rightarrow 1^{-}\;$$ and $$\lambda =1\;$$in our outcomes, we obtain the outcomes obtained by Aouf et al. ; (3) Taking $$q\rightarrow 1^{-}\;$$and $$n=0\;$$in our outcomes, we obtain the outcomes obtained by El- El-Ashwah et al. , with $$b_{k}=1$$ and $$m=0$$]; (4) Taking $$q\rightarrow 1^{-}$$ in Theorems 1,2,3 and 4, respectively, we obtain new outcomes for the classes $$F_{p}^{n,m}(j,\lambda ,\sigma ,b,\beta ),\ G_{p}^{n,m}(j,\lambda ,\sigma ,b,\beta ),$$ $$F_{p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta )\;$$and $$G_{p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta ),\;$$ respectively;  (5) Taking (a) $$\lambda =1$$, (b) $$\lambda =1\;$$and $$m=0,\;$$(c) Letting $$q\rightarrow 1^{-}$$ and $$\sigma =0,\;$$(d)$$\ \lambda =1\;$$and $$\sigma =0,\;$$(E) $$\sigma =0,\;$$(F) Letting $$q\rightarrow 1^{-}$$ and $$\sigma =1,\;$$(G)$$\ \lambda =1\;$$and $$\sigma =1,\;$$(H) $$\sigma =1,\;$$in Lemma 1, Theorems 1 and 3, respectively, we get to new outcomes for the classes $$F_{q,p}^{n,m}(j,\sigma ,b,\beta ),\ F_{q,p}^{n}(j,\sigma ,b,\beta ),SF_{p}^{n,m}(j,\lambda ,\sigma ,b,\beta ),SF_{q,p}^{n,m}(j,b,\beta ),S_{q,p}^{n,m}(j,\lambda ,b,\beta ),$$ $$KF_{p}^{n,m}(j,\lambda ,b,\beta ),\ KF_{q,p}^{n,m}(j,b,\beta )\;$$and$$\ K_{q,p}^{n,m}(j,\lambda ,b,\beta ),\;$$ respectively;  (6) Taking (a) $$\lambda =1$$, (b) $$\lambda =1\;$$and $$m=0,\;$$(c) Letting $$q\rightarrow 1^{-}$$ and $$\sigma =1,\;$$(d)$$\ \lambda =1\;$$and $$\sigma =1,\;$$(E) $$\lambda =\sigma =1\;$$and $$m=0,\;$$(F) Letting $$q\rightarrow 1^{-}$$ and $$\sigma =0,\;$$(G)$$\ \lambda =1\;$$and $$\sigma =0,\;$$ (H) $$\lambda =1\;$$and $$m=\sigma =0,$$(I) $$\sigma =1\;$$(J) $$\sigma =0\;$$in Lemma 2, Theorems 2 and 4, respectively, we get to new outcomes for the classes $$G_{q,p}^{n,m}(j,\sigma ,b,\beta ),\ G_{q,p}^{n}(j,\sigma ,b,\beta ),L_{p}^{n,m}(j,\lambda ,b,\beta ),M_{q,p}^{n,m}(j,b,\beta ),G_{q,p}^{n}(j,b,\beta ),O_{p}^{n,m}(j,\lambda ,b,\beta ),$$

$$R_{q,p}^{n,m}(j,b,\beta ),\ P_{q,p}^{n}(j,b,\beta ),G_{q,p}^{n,m}(j,\lambda ,b,\beta )\;$$and $$GL_{p}^{n,m}(j,\lambda ,b,\beta ),$$respectively.

## Conclusions

Throughout the paper, we defined new subclasses of complex order $$F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )$$and $$G_{q,p}^{n,m}(j, \lambda ,\sigma ,b,\beta )\;$$by using $$D_{\lambda ,q,p}^{n}f^{(m)}(z)\;$$ operator. Also, we introduced coefficients estimates theorems and neighborhoods properties for this classes. This paper generalized many results for different authors. There was connection between q-analysis and (p,q)-analysis see Srivastava . Srivastava , page 340] applied some obvious parametric, argument variations and considered $$0<q<p\le 1,$$ also translated the classical q-number $$\left[ n\right] _{q}$$ to $$\left[ n\right] _{p,q}\;$$as follows:

\begin{aligned} \left[ n\right] _{p,q}=\left\{ \begin{array}{ccc} \frac{p^{n}-q^{n}}{p-q} &{} if &{} n\in {\mathbb {N}} \\ 0 &{} if &{} n=0 \end{array} \right. =p^{n-1}\left[ n\right] _{\frac{q}{p}}. \end{aligned}

## Availability of data and material

During the current study the data sets are derived arithmetically.

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