Formula for meromorphic symmetric differences operator with application in the theory of subordination inequality

Abstract

This investigation deals with a new symmetric formula for a class of meromorphic analytic functions in the puncher open unit disk. Accordingly, the symmetric formula is employed to define a convolution linear operator associated with a special function type the incomplete hypergeometric function. By making utilize the recommended operators, a new sub-classes of meromorphic functions is presented discussing some subordination results. As exceptional examples of our results, certain more acknowledged outcomes are also provided.

1. Introduction

Suppose that $X_a^{+}$ is the class of meromorphic functions in $\overline{U}=\{z\in \mathbb{C},0<|z|<1\}$, denoting by $H(z)$ and taking the form

$$H(z-q)=\frac{1}{z-q}+\sum _{n=1}^{\infty }{b}_n{(z-q)}^n,z\in U,$$

And assume that $X_q^{-}$ is the negative class of meromorphic functions in $\overline{U}$, such that

$$H(z-q)=\frac{1}{z-q}-\sum _{n=1}^{\infty }{b}_n{(z-q)}^n,z-q\in U,b_n\ge 0,$$

Remembering the attitude of subordination between analytic functions can be defined as follows: assume that the functions h and p are analytic in the unit disk, then the function h is subordinate to p if there occurs an analytic function $m(z)$ in U achieving

$$h(z)=p(m(z)),|z|<1.$$

This definition is represented by

$$h(z)\prec p(z),|z|<1.$$

Constructing usage of the definition of subordination in the space of holomorphic functions, Miller and Mocanu [1] and other researchers presented a lot of motivating concerning the differential subordination keeping the behaviors of well-known integral and differential operators (see [2], [3], [4], [5], [6], [7]). Additionally, Miller and Mocanu [1] considered the dual concept of differential subordinations, which is called the superordination (see also [6]). It should be commented that in current years, numerous authors found various remarkable outcomes concerning different integral operators connected with differential subordination and superordination. This concept is used to define classes of analytic functions, which are called Ma-Minda type classes [8]. In 2000, Polatoglu and Bolcal [9] furnished the boundaries of the radius of α-convexity for definite relations of analytic functions in U utilizing subordination concept and the authors in [10] achieved adequate conditions for functions being starlike of order α. Similarly, Baner [7] considered the geometric relations in the superior half plane and showing that they have parametric illustration utilizing subordination idea. Recent studies on the class of meromorphic functions are indicated in the efforts [11], [12], [13].

We have the following classes:

Definition 1.1

Let ϒ be analytic function in a range including $H(\overline{U})$ and $\mathrm{\Upsilon }\ne 0\in U$. Let $\mathrm{\Psi }(z)$ be a specified analytic function in U. The function $H\in X_q^{+}(X_q^{-})$ is known ϒ-like with respect to Ψ if

$$-\left(\frac{(z-q)H^{\prime }(z-q)}{\mathrm{\Upsilon }(H(z-q))}\right)\prec \mathrm{\Psi }(z),z-q\in \overline{U},$$

where $\mathrm{\Upsilon }(0)=0,\Re ({\mathrm{\Upsilon }}^{\prime })(0)>0$ (the necessary conditions). As a special case, when $\mathrm{\Upsilon }(ϵ)=ϵ$, we have

$$-\left(\frac{(z-q)H^{\prime }(z-q)}{H(z-q)}\right)\prec \mathrm{\Psi }(z),z-q\in \overline{U}.$$

Definition 1.2

Let $\mathrm{\Upsilon },\mathrm{\Omega }\in X_q^{+}(X_q^{-})$, where

$$Y(z)=\frac{1}{z-q}+\sum _{n=1}^{\infty }{\gamma }_n{(z-q)}^n$$

and

$$\mathrm{\Omega }(z)=\frac{1}{z-q}+\sum _{n=1}^{\infty }{\eta }_n{(z-q)}^n,$$

under the conditions ${\gamma }_n\ge 0,{\eta }_n\ge 0$. Then the Hadamard product is defined by

$$(Y⁎\mathrm{\Omega })(z-q)=\frac{1}{z-q}+\sum _{n=1}^{\infty }\left({\gamma }_n{\eta }_n\right){(z-q)}^n\in X_q^{+}$$

and

$$(Y⁎\mathrm{\Omega })(z-q)=\frac{1}{z-q}-\sum _{n=1}^{\infty }\left({\gamma }_n{\eta }_n\right){(z-q)}^n\in X_q^{-}.$$

In this effort, we shall define a new difference-differential symmetric operator for functions in the class $X_q^{+}$ and similarly in $X_q^{-}$. We study the operator via different classes of analytic functions, especially the starlike sub-classes. Some consequences results are given.

2. Symmetric meromorphic operator (SMO)

The formula for the Symmetric Meromorphic Operator (SMO) of one dimensional parameter $(\alpha \in [0,1])$, which takes the class described in the preceding part, is covered in this section. The key step in discussing the operator geometrically is to demonstrate that the new symmetric operator formula belongs to the same class. In order to ensure that the coefficients have non-negative values, the constraints on α are computed. SMO's semi-group attribute is examined. The primary class of analytic functions known as Y-like involving SMO is given last.

Now we define a new meromorphic difference-differential operator of a complex variable by utilizing the class of meromorphic functions $X_q^{+}$ as follows. Let $H\in X_q^{+}(X_q^{-})$ then

$$\begin{gathered} {\mathrm{\Delta }}_{\alpha }^{0}H(z-q)=H(z-q)=\frac{1}{z-q}+\sum _{n=1}^{\infty }{b}_n{(z-a)}^n \\ {\mathrm{\Delta }}_{\alpha }^{1}H(z-q)=-\alpha (z-q)H^{\prime }(z-q)+(1-\alpha )(z-q)H^{\prime }(-(z-q))=\frac{1}{z-q}+\sum _{n=1}^{\infty }n(-\alpha +(1-\alpha ){(-1)}^n)b_n{(z-q)}^n \\ {\mathrm{\Delta }}_{\alpha }^{2}H(z-q)={\mathrm{\Delta }}_{\alpha }({\mathrm{\Delta }}_{\alpha }^{1}H(z-q))=\frac{1}{z-q}+\sum _{n=1}^{\infty }{(n(-\alpha +(1-\alpha ){(-1)}^n))}^2{b}_n{(z-q)}^n⋮{\mathrm{\Delta }}_{\alpha }^{k}H(z-q)={\mathrm{\Delta }}_{\alpha }({\mathrm{\Delta }}_{\alpha }^{k-1}H(z-q))=\frac{1}{z-q}+\sum _{n=1}^{\infty }{(n(-\alpha +(1-\alpha ){(-1)}^n))}^k{b}_n{(z-q)}^n. \end{gathered}$$ (2.1)

$$(z-q\in U,b_n\ge 0,\alpha \in [0,1],k\in \mathbb{N},{(-1)}^n=\cos (n\pi )+i\sin (n\pi ))$$

Note that the term

$${♭}_n:=(n(-\alpha +(1-\alpha ){(-1)}^n))$$

may take a complex value. Therefore, the operator (2.1), ${\mathrm{\Delta }}_{\alpha }^{k}H(z-q)$ indicates complex coefficients. To make sure that it is in the class $X_q^{+}(X_q^{-})$ we have a set of conditions on ${♭}_n$.

We have the following remark:

Remark 2.1

• •Clearly, when $H\in X_q^{+}(X_q^{-})$ then ${\mathrm{\Delta }}_{\alpha }^{k}H\in X_q^{+}(X_q^{-})$ respectively, under some restriction on $\alpha \in [0,1]$.
• •
  By the definition of ${\mathrm{\Delta }}^{k}H$, we confirm that

  $${\mathrm{\Delta }}_{\alpha }^{k}H(z-q)=\frac{1}{z-q}+\sum _{n=1}^{\infty }{(n(-\alpha +(1-\alpha ){(-1)}^n))}^k{b}_n{(z-a)}^n=(\frac{1}{z-q}+\sum _{n=1}^{\infty }{(n(-\alpha +(1-\alpha ){(-1)}^n))}^k{(z-q)}^n)⁎(\frac{1}{z-a}+\sum _{n=1}^{\infty }{b}_n{(z-q)}^n)=(\frac{1}{z-q}+\sum _{n=1}^{\infty }{\left({♭}_n\right)}^k{(z-q)}^n)⁎(\frac{1}{z-q}+\sum _{n=1}^{\infty }{b}_n{(z-q)}^n):={\mathcal{B}}_{\alpha }(z-q)⁎H(z-q)$$

  obligates to the Hadamard product.
• •
  It is clear that ${♭}_n=0$ if and only if one of the following cases is achieved

  • ▹
    $0<\alpha <1,\Re ({(-1)}^n)=0,\Re ({(-1)}^{n+1})=1,\Im ({(-1)}^n)=0\mathrm{\&}\Im ({(-1)}^{n+1})=0$;
  • ▹
    $\alpha =\left|\frac{{(-1)}^n}{{(-1)}^n+1}\right|,\Im ({(-1)}^{n+1})<0,0<\Im ({(-1)}^n)<-\Im ({(-1)}^{n+1}),\mathrm{\&}\Re ({(-1)}^n)=\frac{\Im ({(-1)}^n)(\Re ({(-1)}^{n+1})-1)}{\Im ({(-1)}^{n+1})}$;
  • ▹
    $$\alpha =\left|\frac{{(-1)}^n}{{(-1)}^n+1}\right|,\Im ({(-1)}^{n+1})=0,\Im ({(-1)}^n)=0,\Re ({(-1)}^{n+1})<1,\mathrm{\&}$$

    $$0<\Re ({(-1)}^n)<1-\Re ({(-1)}^{n+1});$$
  • ▹
    $$\alpha =\left|\frac{{(-1)}^n}{{(-1)}^n+1}\right|,\Im ({(-1)}^{n+1})>0,-\Im ({(-1)}^{n+1})<\Im ({(-1)}^n)<0,\mathrm{\&}$$

    $$\Re ({(-1)}^n)=\frac{\Im ({(-1)}^n)(\Re ({(-1)}^{n+1})-1)}{Im({(-1)}^{n+1})};$$
  • ▹
    Over the reals number, we have $\alpha \le \frac{{(-1)}^n}{{(-1)}^n+1}$.
• •
  For ♭ > 0, we have the following restrictions:

  • ▹
    $\alpha \in (0,1),{(-1)}^n<-1,{(-1)}^{n+1}\le 0$;
  • ▹
    $\frac{{(-1)}^n}{{(-1)}^n+1}<\alpha <1,{(-1)}^n<-1,0<{(-1)}^{n+1}<{(-1)}^{n+1}-1$;
  • ▹
    $\alpha \in (0,1),{(-1)}^n=-1,{(-1)}^{n+1}<0$;
  • ▹
    $\frac{{(-1)}^n}{{(-1)}^n+1}>\alpha >1,{(-1)}^n>-1,0>{(-1)}^{n+1}>{(-1)}^{n+1}-1$.

In the sequel, we assume that ${♭}_n$ satisfies one of the above conditions to be sure that ${♭}_n\ge 0$. That is

$$\alpha \le \frac{{(-1)}^n}{{(-1)}^n+1},{(-1)}^n\ne 0,n>1.$$

Recently, there is attention towards the symmetric differential operators. Ibrahim and Darus presented a symmetric differential and integral operators for the class of the normalized analytic functions in the open unit disk [14]. Later, the researchers formulated different types of generalized symmetric differential operators [14], [15], [16].

Proposition 2.2

(Semi-group property) The SMO satisfies the semi-group property, as follows:

$${\mathrm{\Delta }}_{\alpha }^k(c_{1}H(z-q)+c_{2}G(z-q))=c_1{\mathrm{\Delta }}_{\alpha }^{k}H(z-q)+c_2{\mathrm{\Delta }}_{\alpha }^{k}G(z-q),$$

where $H(z-a),G(z-q)\in X_q^{+}(X_q^{-})$ .

Proof

Let $H(z-q),G(z-q)\in X_q^{+}(X_q^{-})$. Then for $k=1$, we have

$${\mathrm{\Delta }}_{\alpha }(c_{1}H(z-q)+c_{2}G(z-q))=-\alpha (z-q){(c_{1}H(z-q)+c_{2}G(z-q))}^{\prime }+(1-\alpha )(z-q){(c_{1}H(-(z-q))+c_{2}G(-(z-q)))}^{\prime })=-\alpha (z-q){(c_{1}H(z-q))}^{\prime }+(1-\alpha )(z-q){(c_{1}H(-(z-q)))}^{\prime })+-\alpha (z-q){(c_{2}G(z-q))}^{\prime }+(1-\alpha )(z-q){(c_{2}G(-(z-q)))}^{\prime })=c_1{\mathrm{\Delta }}_{\alpha }^{k}H(z-q)+c_2{\mathrm{\Delta }}_{\alpha }^{k}G(z-q).$$

By indication, we have the desire assertion. □

Our proofs are based on the following outcomes.

Lemma 2.3

(see[2]) Let$\omega (z)$be convex univalent in the unit disk$U:=\{z\in \mathbb{C}:|z|<1\}$and κ and$\nu \in \mathbb{C}\setminus \{0\}$with

$$\Re \{1+\frac{z{\omega }^{″}(z)}{{\omega }^{\prime }(z)}+\frac{\kappa }{\nu }\}>0.$$

If$\zeta (z)$is holomorphic in Π and

$$\kappa \zeta (z)+\nu z{\zeta }^{\prime }(z)\prec \kappa \omega (z)+\nu z{\omega }^{\prime }(z)$$

then

$$\zeta (z)\prec \omega (z)$$

and ω is the best dominant.

Lemma 2.4

(see[1]) Let$\omega (z)$be univalent in U and ϑ and ϱ be holomorphic in a domain Γ containing $\omega (U)$ with $\vartheta (\iota )\ne 0$ when $\iota \in \omega (U)$. Set

$$\mathrm{\Theta }(z)=z{\omega }^{\prime }(z)\varrho (\omega (z)),\rho (z)=\vartheta (\omega (z))+\mathrm{\Theta }(z).$$

Suppose that $\mathrm{\Theta }(z)$ is starlike univalent in Π, and

$$\Re \frac{z{\rho }^{\prime }(z)}{\mathrm{\Theta }(z)}>0$$

for $z\in U$. If

$$\vartheta (\zeta (z))+z{\zeta }^{\prime }(z)\varrho (\zeta (z))\prec \vartheta (\omega (z))+z{\omega }^{\prime }(z)\varrho (\omega (z))$$

then

$$\zeta (z)\prec \omega (z)$$

and $\omega (z)$ is the best dominant.

In this effort, we deal with the generalized class of meromorphic functions

Definition 2.5

Let ϒ be analytic function in a domain containing $H(\overline{U})$ and $\mathrm{\Upsilon }\ne 0\in U$. Let $\omega (z)$ be a specified holomorphic function in U. The function $H\in X_q^{+}(X_q^{-})$ is called ϒ-like with respect to ω if

$$-\left(\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}{\mathrm{\Upsilon }({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}\right)\prec \omega (z),z-q\in U,$$ (2.2)

where $\mathrm{\Upsilon }(0)=0,\Re ({\mathrm{\Upsilon }}^{\prime })(0)>0$ (the necessary conditions). As a special case of (2.2), when $\mathrm{\Upsilon }(ϵ)=ϵ$, we have

$$-\left(\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}{{\mathrm{\Delta }}_{\alpha }^{k}H(z-q)}\right)\prec \omega (z),z-q\in U.$$

Our aim is to illustrate a set of conditions that makes H satisfies (2.2).

3. Subordination inequalities

In this section, we obtain several adequate assumptions for subordination of holomorphic functions in the classes $X_q^{+}$ and $X_q^{-}$. Through various classes of analytic functions, particularly the starlike subclasses, we investigate the operator. Results of some ramifications are provided. We deal with the subordination inequality in Definition 2.5. We shall illustrate a set of conditions that implies the relation when ω is starlike then SMO is also starlike. The consequences are discovered for special values of the parameters. Sharp results and the best dominated function are also presented.

Theorem 3.1

Assume that $\omega (z)$ is convex univalent function in U with ${\omega }^{\prime }(z)\ne 0$ and

$$\Re \{1+\frac{z{\omega }^{″}(z)}{{\omega }^{\prime }(z)}+\frac{\kappa }{\nu }\}>0,\nu \ne 0.$$

Suppose that $-\frac{(z-q){({\mathrm{\Delta }}^{k}H(z-q))}^{\prime }}{\mathrm{\Upsilon }({\mathrm{\Delta }}^{k}H(z-q))}$ is holomorphic in U. If $H\in X_q^{+}$ fulfills the subordination

$$-\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}{\mathrm{\Upsilon }({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}\{\kappa +\nu [1+\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{″}}{{({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}]-\frac{(z-q){\mathrm{\Upsilon }}^{\prime }({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}{\mathrm{\Upsilon }({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}\}\prec \kappa \omega (z)+\nu z{\omega }^{\prime }(z).$$

Then

$$-\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}{\mathrm{\Upsilon }({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}\prec \omega (z),z\in U$$

and $\omega (z)$ is the best dominant.

Proof

Suppose that $\omega (z)$ is given by

$$\omega (z)=-\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}{\mathrm{\Upsilon }({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))},z\in U.$$

Hence, we have

$$\kappa \zeta (z)+\nu z{\zeta }^{\prime }(z)=-\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}{\mathrm{\Upsilon }({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}\{\kappa +\nu [1+\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{″}}{{({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}]-\frac{(z-q){\mathrm{\Upsilon }}^{\prime }({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}{\mathrm{\Upsilon }({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}\}\prec \kappa \omega (z)+\nu z{\omega }^{\prime }(z).$$

At that point, by the supposition of this result, we attain that the declaration of the result indicates an application of Lemma 2.3. □

For $\mathrm{\Upsilon }(ϵ)=ϵ$, in Theorem 3.1, we state the following corollary.

Corollary 3.2

Let the following subordination fulfilled

$$-\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}{({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}\{\kappa +\nu [1+\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{″}}{{({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}]-\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}{({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}\}\prec \kappa \omega (z)+\nu z{\omega }^{\prime }(z)$$

then

$$-\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}{({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}\prec \omega (z).$$

Again, let $\mathrm{\Upsilon }(ϵ)=ϵ,k=0$ in Theorem 3.1, we state the next corollary.

Corollary 3.3

Suppose that the inequality

$$-\frac{(z-q){(H(z-q))}^{\prime }}{(H(z-q))}\{\kappa +\nu [1+\frac{(z-q){(H(z-q))}^{″}}{{(H(z-q))}^{\prime }}]-\frac{(z-q){(H(z-q))}^{\prime }}{(H(z-q))}\}\prec \kappa \omega (z)+\nu z{\omega }^{\prime }(z)$$

is satisfied, then

$$-\frac{(z-q){(H(z-q))}^{\prime }}{(H(z-q))}\prec \omega (z).$$

Theorem 3.4

Consider the function $\omega (z)$ is univalent in U such that $\omega (z)\ne 0,z\in U$ , and

$$\frac{z{\omega }^{\prime }(z)}{\omega (z)}\in {\mathcal{S}}^{⁎}$$

is starlike univalent in U and

$$\Re \{\frac{c}{d}\omega (z)+[1+\frac{z{\omega }^{″}(z)}{{\omega }^{\prime }(z)}-\frac{z{\omega }^{\prime }(z)}{\omega (z)}]\}>0.$$

$$(c\in \mathbb{C},d\ne 0,{\omega }^{\prime }(z)\ne 0,z\in U)$$

If $F\in X_q^{-}$ satisfies

$$c[-\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}{\mathrm{\Upsilon }({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}]+d[1-\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{″}}{{({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}-\frac{(z-a)({\mathrm{\Upsilon }}^{\prime }({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}{\mathrm{\Upsilon }({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}]\prec c\omega (z)+d\frac{z{\omega }^{\prime }(z)}{\omega (z)}$$

then

$$-\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}{{\mathrm{\Delta }}^{k}H(z-q)}\prec \omega (z)$$

and $\omega (z)$ is the best dominant.

Proof

Suppose that $\zeta (z)$ is given by

$$\zeta (z)=-\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}{\mathrm{\Upsilon }({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))},z\in U.$$

By setting $\theta (m)=cm$ and $\varphi (m)=\frac{d}{m},d\ne 0$, it can clearly indicated that $\theta (m)$ is analytic in $\mathbb{C}$, is analytic in $\varrho (m)﹨\{0\}$. Also, we obtain

$$\mathrm{\Theta }(z)=z{\omega }^{\prime }(z)\varrho (\omega (z))=d\frac{{\omega }^{\prime }(z)}{\omega (z)}$$

and

$$g(z)=\theta (\omega (z))+\mathrm{\Theta }(z)=c\omega (z)+d\frac{z{\omega }^{\prime }(z)}{\omega (z)}.$$

It is clear that $\mathrm{\Theta }(z)$ is starlike univalent in U and

$$\Re \{\frac{z{g}^{\prime }(z)}{\mathrm{\Theta }(z)}\}=Re\{\frac{c}{d}\omega (z)+[1+\frac{z{\omega }^{″}(z)}{{\omega }^{\prime }(z)}-\frac{z{\omega }^{\prime }(z)}{\omega (z)}]\}>0.$$

$$c\zeta (z)+d\frac{z{\zeta }^{\prime }(z)}{\zeta (z)}=c[-\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{\prime }}{\mathrm{\Upsilon }({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}+\frac{(z-q){({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}^{″}}{{({\mathrm{\Delta }}^{k}H(z-q))}^{\prime }}-\frac{(z-q){\mathrm{\Upsilon }}^{\prime }({\mathrm{\Delta }}_{\alpha }^{k}H(z-q)))}{\mathrm{\Upsilon }({\mathrm{\Delta }}_{\alpha }^{k}H(z-q))}]\prec c\omega (z)+d\left(\frac{z{\omega }^{\prime }(z)}{\omega (z)}\right).$$

 □

Taking $\mathrm{\Upsilon }(ϵ)=ϵ,k=0$ in Theorem 3.4, we have the following outcome.

Corollary 3.5

If the subordination

$$c[-\frac{(z-q)H^{\prime }(z-q)}{H(z-q)}]+d[1+\frac{(z-q)H^{″}(z-q))}{H^{\prime }(z-q)}-\frac{(z-q)H^{\prime }(z-q)}{H(z-q)}]\prec c\omega (z)+d\left(\frac{z{\omega }^{\prime }(z)}{\omega (z)}\right)$$

satisfies then

$$-\frac{(z-q)H^{\prime }(z-q)}{H(z-q)}\prec \omega (z).$$

We proceed to discuss multi-convoluted formula of the suggested operator taking the structure (3.1)

$$\vartheta \left(\frac{{\mathrm{\Delta }}_{\alpha }^k(H(z-q)⁎G(z-q))}{{\mathrm{\Delta }}_{\alpha }^k(H(z-q)⁎W(z-q))}\right)\prec \vartheta -\left(\frac{(\mu -\nu )(z-q)}{1+\nu (z-q)}\right),$$ (3.1)

$$(\nu \in [-1,\mu ),\mu \in (\nu ,1],k\ge 0,\vartheta >0)$$

where

$$G(z-q)=\frac{1}{z-q}-\sum _{n=1}^{\infty }{g}_n{(z-q)}^n,z-q\in U,g_n\ge 0$$

and

$$W(z-q)=\frac{1}{z-q}-\sum _{n=1}^{\infty }{w}_n{(z-q)}^n,z-q\in U,w_n\ge 0.$$

We have the following result

Theorem 3.6

The functions$H,G$and W have the structure(3.1)if they satisfy

$$\sum _{n=1}^{\infty }{♭}_n^k{b}_n(\vartheta g_n(1+\nu ))-w_n(\vartheta (1+\nu )+(\mu -\nu )))\le \mu -\nu .$$

The function$H_n$provided by achieves equality

$$H_n(z-q)=\frac{1}{z-q}+\frac{\mu -\nu }{{♭}_n^k(\vartheta g_n(1+\nu ))-w_n(\vartheta (1+\nu )+(\mu -\nu )))}{(z-q)}^n.$$

Proof

From the inequality (3.1), we have

$$\vartheta \left(\frac{{\mathrm{\Delta }}_{\alpha }^k(H(z-q)⁎G(z-q))}{{\mathrm{\Delta }}_{\alpha }^k(H(z-q)⁎W(z-q))}\right)=\vartheta -\left(\frac{(\mu -\nu )(m(z)-q)}{1+\nu (m(z)-q)}\right),$$

where $m(z)<1,m(0)=0$ is a Schwarz function. Now by assuming $m(z)=z,z\in U$ we have

$$\vartheta \left(\frac{{\mathrm{\Delta }}_{\alpha }^k(H(z-q)⁎G(z-q))}{{\mathrm{\Delta }}_{\alpha }^k(H(z-q)⁎W(z-q))}\right)=\vartheta -\left(\frac{(\mu -\nu )(z-q)}{1+\nu (z-q)}\right).$$

A computation yields

$$\left|\frac{\vartheta {\sum }_{n=1}^{\infty }{♭}_n^k{b}_n(g_n-w_n){(z-q)}^{n+1}}{(\mu -\nu )-{\sum }_{n=1}^{\infty }{♭}_n^k{b}_n(\vartheta g_n\nu +((\mu -\nu )-\vartheta \nu )w_n)){(z-q)}^{n+1}}\right|\le \frac{\vartheta {\sum }_{n=1}^{\infty }{♭}_n^k{b}_n|g_n-w_n||{(z-q)}^{n+1}|}{(\mu -\nu )-{\sum }_{n=1}^{\infty }{♭}_n^k{b}_n(\vartheta g_n\nu +((\mu -\nu )-\vartheta \nu )w_n))|{(z-q)}^{n+1}|}\le 1.$$

Considering that $z-q\to 1$, we obtain the assertion of this theorem. □

4. Linear operator associated with special functions

In this section, we introduce hybrid linear operators involving the SMO and special function terms. We shall introduce two types of hybrid linear operators associating with hypergeometric function and Mittag-Leffler function respectively. To make the connection, we request to make a change in the usual formula to fit the expression of SMO. Therefore, we present a modify expression to be sure that the new formula belongs to the main class of the analytic functions. The next step is convoluted SMO with the meromorphic hypergeometric function and Mittag-Leffler function. Proceeding with these convoluted operators, we formulate them in new classes of analytic functions of Y-type using the subordination inequality. To study them geometrically, we illustrate a set of conditions to get the starlikeness. Examples with graphs are presented in the sequel. Sharp and the best dominant results are also considered.

4.1. Hyper-geometric function

Define a special function analog to hypergeometric functions

$$\mathrm{\Phi }(\rho ,\varrho ,z-q)=\frac{1}{z-q}+\sum _{n=1}^{\infty }\frac{{(\rho )}_n}{{(\varrho )}_n}{(z-q)}^n,$$

where $\rho \in \mathbb{C}\setminus \{0\},\varrho \ne 0,-1,-2,...$ and ${(\complement )}_n$ is the Pochhammer symbol. Corresponding to $\mathrm{\Phi }(\rho ,\varrho ,z-q)$, we define a linear operator utilizing the Hadamard product as follows:

$${\ell }_{\alpha }^k(\rho ,\varrho )(z-q)=\mathrm{\Phi }(\rho ,\varrho ,z-q)⁎{\mathrm{\Delta }}_{\alpha }^{k}H(z-q)=\frac{1}{z-q}+\sum _{n=1}^{\infty }\left(\frac{{(\rho )}_n}{{(\varrho )}_n}\right)\left({♭}_n^k{b}_n\right){(z-q)}^n.$$ (4.1)

Note that, when $k=0$ in (4.1), we get the Carlson-Sheffer operator of a meromorphic function [17]. Moreover, it is clear that ${\ell }_{\alpha }^k(\rho ,\varrho )(z-q)\in X_q^{+}(X_q^{-})$ when $H(z-q)\in X_q^{+}(X_q^{-})$, and that $\left(\frac{{(\rho )}_n}{{(\varrho )}_n}\right)\ge 0,\forall n\ge 1$.

We define the following class

$$-\left(\frac{(z-q){({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{\prime }}{\mathrm{\Upsilon }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}\right)\prec \omega (z),z-q\in U,$$ (4.2)

where $\mathrm{\Upsilon }(0)=0,\Re ({\mathrm{\Upsilon }}^{\prime })(0)>0$ (the necessary conditions). As a special case of (4.2), when $\mathrm{\Upsilon }(ϵ)=ϵ$, we have

$$-\left(\frac{(z-q){({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{\prime }}{{\ell }_{\alpha }^k(\rho ,\varrho )(z-q)}\right)\prec \omega (z),z-q\in U.$$

Theorem 4.1

Suppose that the function$\omega (z)$is convex univalent in U with${\omega }^{\prime }(z)\ne 0$and

$$\Re \{1+\frac{z{\omega }^{″}(z)}{{\omega }^{\prime }(z)}+\frac{\kappa }{\nu }\}>0,\nu \ne 0.$$

Also, assume that$-\frac{(z-q){({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{\prime }}{\mathrm{\Upsilon }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}$is analytic in U. If$H\in X_q^{+}$achieves the inequality

$$-\frac{(z-q){({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{\prime }}{\mathrm{\Upsilon }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}\{\kappa +\nu [1+\frac{(z-q){({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{″}}{{({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{\prime }}]-\frac{(z-q){\mathrm{\Upsilon }}^{\prime }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}{\mathrm{\Upsilon }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}\}\prec \kappa \omega (z)+\nu z{\omega }^{\prime }(z).$$

Then

$$-\frac{(z-q){({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{\prime }}{\mathrm{\Upsilon }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}\prec \omega (z),z\in U$$

and$\omega (z)$is the best dominant.

Proof

Suppose that $\omega (z)$ is given by

$$\omega (z)=-\frac{(z-q){({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{\prime }}{\mathrm{\Upsilon }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))},z\in U.$$

Hence, we have

$$\kappa \zeta (z)+\nu z{\zeta }^{\prime }(z)=-\frac{(z-q){({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{\prime }}{\mathrm{\Upsilon }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}\{\kappa +\nu [1+\frac{(z-q){({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{″}}{{({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{\prime }}]-\frac{(z-q){\mathrm{\Upsilon }}^{\prime }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}{\mathrm{\Upsilon }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}\}\prec \kappa \omega (z)+\nu z{\omega }^{\prime }(z).$$

Then by the conditions of the theorem we get that the desired affirmation using Lemma 2.3. □

Theorem 4.2

Consume that the function $\omega (z)$ is univalent in U with $\omega (z)\ne 0,z\in U,\frac{z{\omega }^{\prime }(z)}{\omega (z)}$ is starlike univalent in U and

$$\Re \{\frac{c}{d}\omega (z)+[1+\frac{z{\omega }^{″}(z)}{{\omega }^{\prime }(z)}-\frac{z{\omega }^{\prime }(z)}{\omega (z)}]\}>0.$$

$$(c\in \mathbb{C},d\ne 0,{\omega }^{\prime }(z)\ne 0,z\in U)$$

If $F\in X_q^{-}$ fulfills

$$c[-\frac{(z-q){({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{\prime }}{\mathrm{\Upsilon }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}]+d[1-\frac{(z-q){({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{″}}{{({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{\prime }}-\frac{(z-q)({\mathrm{\Upsilon }}^{\prime }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}{\mathrm{\Upsilon }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}]\prec c\omega (z)+d\frac{z{\omega }^{\prime }(z)}{\omega (z)}$$

then

$$-\frac{(z-q){({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{\prime }}{{\ell }_{\alpha }^k(\rho ,\varrho )(z-q)}\prec \omega (z)$$

and $\omega (z)$ is the best dominant.

Proof

Consider $\zeta (z)$ is given by

$$\zeta (z)=-\frac{(z-q){({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{\prime }}{\mathrm{\Upsilon }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))},z\in U.$$

By setting $\theta (m)=cm$ and $\varphi (m)=\frac{d}{m},d\ne 0$, it can easily observed that $\theta (m)$ is analytic in $\mathbb{C}$, is analytic in $\varrho (m)﹨\{0\}$. Also we obtain

$$\mathrm{\Theta }(z)=z{\omega }^{\prime }(z)\varrho (\omega (z))=d\frac{{\omega }^{\prime }(z)}{\omega (z)}$$

and

$$g(z)=\theta (\omega (z))+\mathrm{\Theta }(z)=c\omega (z)+d\frac{z{\omega }^{\prime }(z)}{\omega (z)}.$$

Obviously, $\mathrm{\Theta }(z)$ is starlike with a positive real structured as follows:

$$\Re \{\frac{z{g}^{\prime }(z)}{\mathrm{\Theta }(z)}\}=\Re \{\frac{c}{d}\omega (z)+[1+\frac{z{\omega }^{″}(z)}{{\omega }^{\prime }(z)}-\frac{z{\omega }^{\prime }(z)}{\omega (z)}]\}>0.$$

$$c\zeta (z)+d\frac{z{\zeta }^{\prime }(z)}{\zeta (z)}=c[-\frac{(z-q){({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{\prime }}{\mathrm{\Upsilon }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}+\frac{(z-q){({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{″}}{{({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}^{\prime }}-\frac{(z-q){\mathrm{\Upsilon }}^{\prime }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q)))}{\mathrm{\Upsilon }({\ell }_{\alpha }^k(\rho ,\varrho )(z-q))}\prec c\omega (z)+d\left(\frac{z{\omega }^{\prime }(z)}{\omega (z)}\right).$$

 □

Example 4.3

Let $H(z-q)\in X_q^{+}$ such that

$$H(z-q)=\frac{1}{z-q}+\frac{(z-q)}{1-(z-q)}=\frac{1}{z-q}+\sum _{n=1}^{\infty }{(z-q)}^n.$$

Then,

$${\mathrm{\Delta }}_{1}H(z-q)=-\alpha (z-q)H^{\prime }(z-q)+(1-\alpha )(z-q)H^{\prime }(-(z-q))=-(z-q)H^{\prime }(z-q);$$

and for $\alpha =0.5$, we have

$${\mathrm{\Delta }}_{0.5}H(z-q)=0.5(z-q)(\frac{1}{{(-q+z+1)}^2}-\frac{1}{{(q-z)}^2})-\frac{0.5(-2q+2z-1)}{{(q-z+1)}^2(z-q)};$$

and when $\alpha =0.25$, we get

$${\mathrm{\Delta }}_{0.25}H(z-q)=0.75(z-q)(\frac{1}{{(-q+z+1)}^2}-\frac{1}{{(q-z)}^2})-\frac{0.25(-2q+2z-1)}{{(q-z+1)}^2(z-q)}$$

Fig. 1 indicates the behavior of H under ${\mathrm{\Delta }}_{\alpha }$ for different values of $\alpha \in [0,1]$, while Fig. 2 presents its derivative.

Figure 1.

The 3D and cantor plot of the function H(z − q) and its behavior under the suggested operator ${\mathrm{\Delta }}_{\alpha }^k$, where (A1-A2) indicated the graph of H(z − q), (B1-B2) presented the graph of Δ₁ H(z − q), (C1-C2) plotted the graph of Δ_0.5 H(z − q) and (D1-D2) provided the plot of Δ_0.25 H(z − q) respectively.

Figure 2.

The 3D and cantor plot of H′(z − q) and its behavior under the suggested operator ${\mathrm{\Delta }}_{\alpha }^k$. From the left: (A1-A2) indicated the plot of H′(z − q), and (B1-B2) presented the graph of ${({\mathrm{\Delta }}_{1}H(z-q))}^{\prime }$ respectively.

Theorem 4.4

The functions $H,\mathrm{\Phi }({\rho }_1,{\varrho }_1,z-q)$ and $\mathrm{\Phi }({\rho }_2,{\varrho }_2,z-q)$ have the structure (3.1) if they satisfy

$$\sum _{n=1}^{\infty }{♭}_n^k{b}_n(\vartheta (\frac{{({\rho }_1)}_n}{{({\varrho }_1)}_n})(1+\nu ))-(\frac{{({\rho }_2)}_n}{{({\varrho }_2)}_n})(\vartheta (1+\nu )+(\mu -\nu )))\le \mu -\nu .$$

The function $H_n$ provided by achieves equality

$$H_n(z-q)=\frac{1}{z-q}+\frac{\mu -\nu }{{♭}_n^k(\vartheta (\frac{{({\rho }_1)}_n}{{({\varrho }_1)}_n})(1+\nu ))-(\frac{{({\rho }_2)}_n}{{({\varrho }_2)}_n})(\vartheta (1+\nu )+(\mu -\nu )))}{(z-q)}^n.$$

Proof

From the inequality (3.1), we have

$$\vartheta \left(\frac{{\mathrm{\Delta }}_{\alpha }^k(H(z-q)⁎\mathrm{\Phi }({\rho }_1,{\varrho }_1,z-q))}{{\mathrm{\Delta }}_{\alpha }^k(H(z-q)⁎\mathrm{\Phi }({\rho }_2,{\varrho }_2,z-q))}\right)=\vartheta -\left(\frac{(\mu -\nu )(m(z)-q)}{1+\nu (m(z)-q)}\right),$$

where $m(z)<1,m(0)=0$ is a Schwarz function. Now by assuming $m(z)=z,z\in U$ we have

$$\vartheta \left(\frac{{\mathrm{\Delta }}_{\alpha }^k(H(z-q)⁎\mathrm{\Phi }({\rho }_1,{\varrho }_1,z-q))}{{\mathrm{\Delta }}_{\alpha }^k(H(z-q)⁎\mathrm{\Phi }({\rho }_2,{\varrho }_2,z-q))}\right)=\vartheta -\left(\frac{(\mu -\nu )(z-q)}{1+\nu (z-q)}\right).$$

A computation yields

$$\left|\frac{\vartheta {\sum }_{n=1}^{\infty }{♭}_n^k{b}_n((\frac{{({\rho }_1)}_n}{{({\varrho }_1)}_n})-(\frac{{({\rho }_2)}_n}{{({\varrho }_2)}_n})){(z-q)}^{n+1}}{(\mu -\nu )-{\sum }_{n=1}^{\infty }{♭}_n^k{b}_n(\vartheta (\frac{{({\rho }_1)}_n}{{({\varrho }_1)}_n})\nu +((\mu -\nu )-\vartheta \nu )(\frac{{({\rho }_2)}_n}{{({\varrho }_2)}_n}))){(z-q)}^{n+1}}\right|\le \frac{\vartheta {\sum }_{n=1}^{\infty }{♭}_n^k{b}_n|(\frac{{({\rho }_1)}_n}{{({\varrho }_1)}_n})-(\frac{{({\rho }_2)}_n}{{({\varrho }_2)}_n})||{(z-q)}^{n+1}|}{(\mu -\nu )-{\sum }_{n=1}^{\infty }{♭}_n^k{b}_n(\vartheta (\frac{{({\rho }_1)}_n}{{({\varrho }_1)}_n})\nu +((\mu -\nu )-\vartheta \nu )(\frac{{({\rho }_2)}_n}{{({\varrho }_2)}_n})))|{(z-q)}^{n+1}|}\le 1.$$

Considering that $z-q\to 1$, we obtain the assertion of this theorem. □

4.2. Mittag-Leffler function MLF

The following series indicate the MLF

$${\mathrm{\Xi }}_{\alpha ,\beta }(z)=\sum _{n=0}^{\infty }\frac{z^n}{\mathrm{\Gamma }(an+b)}.$$

In the same manner of the above instruction, we obtain

$${\mathrm{\Xi }}_{\alpha ,\beta }(z-q)=\frac{1}{z-q}+\sum _{n=1}^{\infty }\frac{{(z-q)}^n}{\mathrm{\Gamma }(an+b)}.$$

Corresponding to ${\mathrm{\Xi }}_{\alpha ,\beta }(z-q)$, we define a linear operator utilizing the Hadamard product as follows:

$${\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q)={\mathrm{\Xi }}_{\alpha ,\beta }(z-q)⁎{\mathrm{\Delta }}_{\alpha }^{k}H(z-q)=\frac{1}{z-q}+\sum _{n=1}^{\infty }\left(\frac{z^k}{\mathrm{\Gamma }(an+b)}\right)\left({♭}_n^k{b}_n\right){(z-q)}^n.$$

Note that, when $k=0$, we get the Carlson-Sheffer operator of a meromorphic function [17]. Moreover, it is clear that ${\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q)\in X_q^{+}(X_q^{-})$ when $H(z-q)\in X_q^{+}(X_q^{-})$.

We define the following class

$$-\left(\frac{(z-q){({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}^{\prime }}{\mathrm{\Upsilon }({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}\right)\prec \omega (z),z-q\in U,$$

where $\mathrm{\Upsilon }(0)=0,\Re ({\mathrm{\Upsilon }}^{\prime })(0)>0$ (the necessary conditions). As a special case, when $\mathrm{\Upsilon }(ϵ)=ϵ$, we have

$$-\left(\frac{(z-q){({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}^{\prime }}{{\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q)}\right)\prec \omega (z),z-q\in U.$$

The following outcomes are similar to the above subsection, therefore, we ignore the proofs.

Theorem 4.5

Suppose that the function$\omega (z)$is convex univalent in U with${\omega }^{\prime }(z)\ne 0$and

$$\Re \{1+\frac{z{\omega }^{″}(z)}{{\omega }^{\prime }(z)}+\frac{\kappa }{\nu }\}>0,\nu \ne 0.$$

Also, assume that$-\frac{(z-q){({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}^{\prime }}{\mathrm{\Upsilon }({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}$is analytic in U. If$H\in X_q^{+}$achieves the inequality

$$-\frac{(z-q){({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}^{\prime }}{\mathrm{\Upsilon }({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}\{\kappa +\nu [1+\frac{(z-q){({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}^{″}}{{({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}^{\prime }}]-\frac{(z-q){\mathrm{\Upsilon }}^{\prime }({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}{\mathrm{\Upsilon }({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}\}\prec \kappa \omega (z)+\nu z{\omega }^{\prime }(z).$$

Then

$$-\frac{(z-q){({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}^{\prime }}{\mathrm{\Upsilon }({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}\prec \omega (z),z\in U$$

and$\omega (z)$is the best dominant.

Theorem 4.6

Consume that the function$\omega (z)$is univalent in U with$\omega (z)\ne 0,z\in U,\frac{z{\omega }^{\prime }(z)}{\omega (z)}$is starlike univalent in U and

$$\Re \{\frac{c}{d}\omega (z)+[1+\frac{z{\omega }^{″}(z)}{{\omega }^{\prime }(z)}-\frac{z{\omega }^{\prime }(z)}{\omega (z)}]\}>0.$$

$$(c\in \mathbb{C},d\ne 0,{\omega }^{\prime }(z)\ne 0,z\in U)$$

If$F\in X_q^{-}$fulfills

$$c[-\frac{(z-q){({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}^{\prime }}{\mathrm{\Upsilon }({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}]+d[1-\frac{(z-q){({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}^{″}}{{({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}^{\prime }}-\frac{(z-q)({\mathrm{\Upsilon }}^{\prime }({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}{\mathrm{\Upsilon }({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}]\prec c\omega (z)+d\frac{z{\omega }^{\prime }(z)}{\omega (z)}$$

then

$$-\frac{(z-q){({\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q))}^{\prime }}{{\mathrm{\Sigma }}_{\alpha }^k(a,b)(z-q)}\prec \omega (z)$$

and$\omega (z)$is the best dominant.

Theorem 4.7

The functions$H,{\mathrm{\Xi }}_{\alpha ,\beta }(z-q)$and${\mathrm{\Xi }}_{\alpha ,\beta }(z-q)$have the structure(3.1)if they satisfy

$$\sum _{n=1}^{\infty }{♭}_n^k{b}_n(\vartheta (\frac{1}{\mathrm{\Gamma }(a_{1}n+b_1)})(1+\nu ))-(\frac{1}{\mathrm{\Gamma }(a_{2}n+b_2)})(\vartheta (1+\nu )+(\mu -\nu )))\le \mu -\nu .$$

The function$H_n$provided by achieves equality

$$H_n(z-q)=\frac{1}{z-q}+\frac{(\mu -\nu ){(z-q)}^n}{{♭}_n^k(\vartheta (\frac{1}{\mathrm{\Gamma }(a_{1}n+b_1)})(1+\nu ))-(\frac{1}{\mathrm{\Gamma }(a_{2}n+b_2)})(\vartheta (1+\nu )+(\mu -\nu )))}.$$

5. Conclusion

The above study indicated several facts in the theory of the meromorphic analytic functions. We formulated new symmetric operators. The first one is based on a special class of meromorphic functions and the second one is a linear operator given by a special function type incomplete hypergeometric function. We formulated the sufficient conditions, in view of the subordination results, to prove that the operators are in ϒ-like class.

Our new method can be employed to describe the meromorphic solution of classes of algebraic differential equations of complex variables geometrically. This class of differential equations are very important in optics and other mathematical physics applications. From above investigation, we studied a class of first order differential inequality. Therefore, the meromorphic solution is indicated as an upper solution corresponding to its differential equation. For future works, may a researcher formulate the operator in terms of second order differential inequality using the second order differential subordination.

Funding statement

This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.

CRediT authorship contribution statement

Ibtisam Aldawish: Conceived and designed the experiments; Performed the experiments; Analyzed and interpreted the data; Contributed reagents, materials, analysis tools or data; Wrote the paper.

Declaration of Competing Interest

The authors declare no conflict of interest.

Data availability

Data included in article/supp. material/referenced in article.