# Differential subordination applications to a class of meromorphic multivalent functions associated with Mittag-Leffler function

## Abstract

In this paper, using the principal of differential subordination, we obtain some properties of certain class of p-valent meromorphic functions, which are defined by Mittag-Leffler function.

## Introduction

Denote by Σp,m the class of analytic meromorphic multivalent functions of the form:

$$f(z)=\frac{1}{z^{p}}+\sum\limits_{k=m}^{\infty }a_{k}z^{k}\ \left(p\in \mathbb{N}=\{1,2,\ldots\};m>-p\right),$$
(1)

where $$\mathbb {U}^{\ast }=\{z\in \mathbb {C}$$ and $$0<|z|<1\}=\mathbb {U}\backslash \{0\}.\$$We note that Sigmap,1−p=Σp.

For two functions f(z) and g(z), analytic in $$\mathbb {U}, f(z)$$ is subordinate to g(z)(f(z)g(z)) in $$\mathbb {U}$$, if there exists a function ω(z), analytic in $$\mathbb {U}$$ with ω(0)=0 and $$\left \vert \omega (z)\right \vert <1, f(z)=g(\omega (z)) (z\in \mathbb {U})$$ and if g(z) is univalent in $$\mathbb {U}$$, then (see for details  and also )

$$f(z)\prec g(z)\Longleftrightarrow f(0)=g(0)\text{ and }f(\mathbb{U})\subset g(\mathbb{U}).$$

The Hadamard product of f(z) and g(z) given by

$$g(z)=\frac{1}{z^{p}}+\sum\limits_{k=m}^{\infty }b_{k}z^{k}\$$

is defined by

$$\left(f\ast g\right) \left(z\right) =\frac{1}{z^{p}}+\sum\limits_{k=m}^{ \infty }a_{k}b_{k}z^{k}=\left(g\ast f\right) \left(z\right).$$
(2)

The Mittag-Leffler function Eα(z) (z$$\in \mathbb {U}^{\ast }$$) ( and ) see also ([5, 6] and ) is defined by

$$E_{\alpha }(z)=\sum\limits_{k=0}^{\infty }\frac{1}{\Gamma (k\alpha +1)}z^{k},\alpha \in \mathbb{C},\Re(\alpha)>0.$$

For $$\alpha,\beta,\gamma \in \mathbb {C}$$, (α)>0, max { 0, (c)−1} and (c)>0, Srivastava and Tomovski  generalized Mittag-Leffler function by the function

$$E_{\alpha,\beta }^{\gamma,c}(z)=\sum\limits_{k=0}^{\infty }\frac{(\gamma)_{kc}}{ \Gamma (k\alpha +\beta)k!}z^{nk}$$
(3)

and proved that it is an entire function in the complex z-plane, where

$$(\gamma)_{\theta }=\frac{\Gamma (\gamma +\theta)}{\Gamma (\gamma)}\left\{ \begin{array}{c} 1,\ \ \ \ \ \ \ \ \theta =0 \\ \gamma (\gamma +1)\ldots(\gamma +\theta -1),\ \theta \neq 0 \end{array} \right..$$

Mostafa and Aouf  (see also ) used the function $$E_{\alpha,\beta }^{\gamma,c}(z)$$ and defined the meromorphic function

$$\begin{array}{@{}rcl@{}} \mathcal{M}_{p,\alpha,\beta }^{\gamma,c}(z) &=&z^{-p}\Gamma (\beta)E_{\alpha,\beta }^{\gamma,c}(z) \\ &=&z^{-p}+\sum\limits_{k=m}^{\infty }\frac{\Gamma (\beta)\Gamma \lbrack \gamma +(k+p)c]}{\Gamma (\gamma)\Gamma \lbrack \beta +(k+p)\alpha ](k+p)!} z^{k}, \\ &&\left(\Re \left(\alpha \right) =0\ \text{when\ }\Re \left(c\right) =1\ \text{with}\ \beta \neq 0\right), \end{array}$$
(4)

and for f(z)Σp,m, they defined the operator

$$\begin{array}{@{}rcl@{}} \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z) &=&\mathcal{M}_{p,\alpha,\beta }^{\gamma,c}(z)\ast f(z) \\ &=&z^{-p}+\sum\limits_{k=m}^{\infty }\frac{\Gamma (\beta)\Gamma \lbrack \gamma +(k+p)c]}{\Gamma (\gamma)\Gamma \lbrack \beta +(k+p)\alpha ](k+p)!} a_{k}z^{k}. \end{array}$$
(5)

From (5) it is easy to have

$$cz(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z))^{\prime }=\gamma \mathcal{ H}_{p,\alpha,\beta }^{\gamma +1,c}f(z)-(\gamma +pc)\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z) \left(c>0\right)$$
(6)

and

$$\alpha z\left(\mathcal{H}_{p,\alpha,\beta +1}^{\gamma,c}f\left(z\right) \right)^{^{\prime }}=\beta \ \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f\left(z\right) -\left(p\alpha +\beta \right) \ \mathcal{H}_{p,\alpha,\beta +1}^{\gamma,c}f\left(z\right),\alpha \neq 0.$$
(7)

We note that:

(i) $$\mathcal {H}_{p,0,\beta }^{1,1}f(z)=f(z);$$

(ii) $$\mathcal {H}_{p,0,\beta }^{2,1}f(z)=\left (p+1\right) f(z)+zf^{^{\prime }}(z);$$

(iii) $$\mathcal {H}_{1,0,\beta }^{2,1}f(z)=2f(z)+zf^{^{\prime }}(z).$$

Using the operator $$\mathcal {H}_{p,\alpha,\beta }^{\gamma,c}f(z)$$, we have the following definition.

### Definition 1.

For fixed A and B (−1≤B<A≤1), we say that a function fΣp,m is in the class $$\Sigma _{p,m}^{\gamma,c}\left (\alpha,\beta ;A,B\right)$$ if it satisfies

$$-\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{p}\prec \frac{1+Az}{1+Bz}.$$
(8)

In view of the definition of differential subordination, (8) is equivalent to

$$\left\vert \frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}+p}{Bz^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}+pA}\right\vert <1.$$
(9)

We note that:

(i)

$$\begin{array}{@{}rcl@{}} \Sigma_{p,1}^{1,1}\left(0,1;A,B\right) &=&\Sigma_{p}\left(A,B\right) \left(-1\leq B<A\leq 1;\mathbb{U}^{\ast }\right) \\ &=&\left\{ f\in \Sigma_{p}:-\frac{z^{p+1}f^{^{\prime }}(z)}{p}\prec \frac{ 1+Az}{1+Bz}\right\}, \end{array}$$

the class Σp(A,B) was introduced and studied by Mogra .

(ii)

$$\begin{array}{@{}rcl@{}} \Sigma_{p,m}^{\gamma,c}\left(\alpha,\beta ;1-\frac{2\eta }{p},-1\right) &=&\Sigma_{p,m}^{\gamma,c}\left(\alpha,\beta,\eta \right) \left(0\leq \eta < p\right) \\ &=&\left\{ f\in \Sigma_{p,m}:\Re\{-z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}\}>\eta \right\}. \\ && \end{array}$$
(10)

## Preliminary results

The following lemmas will be required in our investigation.

### Lemma 1

. Let h be a convex (univalent) function in $$\mathbb {U}$$ with h(0)=1. Also let

$$\phi (z)=1+d_{p+m}z^{p+m}+d_{p+m+1}z^{p+m+1}+\ldots\text{,}$$
(11)

be analytic in $$\mathbb {U}.$$ If

$$\phi (z)+\frac{z\phi^{^{\prime }}(z)}{\tau }\prec h(z)\ \left(\Re \left(\tau \right) \geq 0;\ \tau \neq 0;\ z\in \mathbb{U}\right),$$
(12)

then

$$\phi (z)\prec \Psi (z)=\frac{\tau }{p+m}z^{-\frac{\tau }{p+m} }\int\limits_{0}^{z}t^{\frac{\tau }{p+m}-1}h(t)dt.$$
(13)

### Lemma 2

. Let μbe a positive measure on the unit interval [0,1]. Let g(z,t)be a complex valued function defined on $$\mathbb {U}\times \left [ 0,1\right ]$$ such that g(.,t) is analytic in $$\mathbb {U}$$ for each t[0,1] and such that g(z,.) is μ integrable on [0,1] for all $$z\in \mathbb {U}.$$ In addition, suppose that {g(z,t)}>0,g(−r,t)is real and

$$\Re \left\{ \frac{1}{g(z,t)}\right\} \geq \frac{1}{g(-r,t)} \left(\left\vert z\right\vert \leq r<1;t\in \left[ 0,1\right] \right).$$

If the function G is defined by

$$G(z)=\int\limits_{0}^{1}g(z,t)d\mu (t),$$

then

$$\Re \left\{ \frac{1}{G(z)}\right\} \geq \frac{1}{G(-r)} \left(\left\vert z\right\vert \leq r<1\right).$$

Each of the identities (asserted by Lemma 2) is fairly well known (cf., e.g., [, ch. 14]).

### Lemma 3

. For real or complex numbers a, b, and c (c≠0,−1,−2,…)

$$\int\limits_{0}^{1}t^{b-1}\left(1-t\right)^{c-b-1}\left(1-tz\right)^{-a}dt=\frac{\Gamma \left(b\right) \Gamma \left(c-b\right) }{\Gamma \left(c\right) }_{2}F_{1}\left(a,b;c;z\right) \ \left(\Re \left(c\right) <\Re \left(b\right) >0\right) ;$$
$$_{2}F_{1}\left(a,b;c;z\right) =\left(1-z\right)_{\text{ \ \ \ \ } 2}^{-a}F_{1}\left(a,c-b;c;\frac{z}{z-1}\right) \left(z\neq 1\right)$$
(14)

and

$$_{2}F_{1}\left(a,b;c;z\right) =_{2}F_{1}\left(b,a;c;z\right).$$
(15)

### Lemma 4

. Let Φbe analytic in $$\mathbb {U}$$ with

$$\Phi (0)=1\text{ and }\Re \{\Phi (z)\}>\frac{1}{2}.$$

Then, for any function F, analytic in $$\mathbb {U}, \left (\Phi \ast F\right) \left (\mathbb {U}\right)$$ is contained in the convex hull of $$F\left (\mathbb {U}\right).$$

We used the technique used by ( and ).

## Main inclusion relationships

Unless otherwise mentioned, we assume throughout this paper that $$-1\leq B<A\leq 1,\alpha,\beta,\gamma \in \mathbb {C}, \Re (\alpha)>0$$, max { 0, (c)−1}, (c)>0,δ>0,f(z) given by (1) and $$z\in \mathbb {U}^{\ast }.$$

### Theorem 1

Let γ≠0 and f(z) satisfy:

$$\mathcal{-}\frac{\left(1-\delta \right) z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}+\delta z^{p+1}\left(\mathcal{H }_{p,\alpha,\beta }^{\gamma +1,c}f(z)\right)^{^{\prime }}}{p}\prec \frac{ 1+Az}{1+Bz},$$
(16)

then

$$-\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{p}\prec \Psi (z)\prec \frac{1+Az}{1+Bz},$$
(17)

where

$$\Psi (z)=\left\{ \begin{array}{cc} \frac{A}{B}+\left(1-\frac{A}{B}\right) \left(1+Bz\right)_{\text{ \ \ \ } 2}^{-1}F_{1}\left(1,1;\frac{\gamma }{\delta c\left(p+m\right) }+1;\frac{Bz }{1+Bz}\right), & B\neq 0 \\ 1+\frac{\gamma }{\gamma +\delta c\left(p+m\right) }Az, & B=0. \end{array} \right.$$
(18)

is the best dominant of (17). Furthermore,

$$\Re \left\{ -\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{p}\right\} >\rho \ \left(0\leq \rho <1\right),$$
(19)

where

$$\rho =\left\{ \begin{array}{cc} \frac{A}{B}+\left(1-\frac{A}{B}\right) \left(1-B\right)_{\text{ \ \ \ } 2}^{-1}F_{1}\left(1,1;\frac{\gamma }{\delta c\left(p+m\right) }+1;\frac{B}{ B-1}\right), & B\neq 0, \\ 1-\frac{\gamma }{\gamma +\delta c\left(p+m\right) }A, & B=0. \end{array} \right.$$
(20)

The result is the best possible.

### Proof

Let

$$\phi (z)=-\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{p},$$
(21)

where ϕ is given by (11). Differentiating (21) and using (6), we get

$$\mathcal{-}\frac{\left(1-\delta \right) z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}+\delta z^{p+1}\left(\mathcal{H }_{p,\alpha,\beta }^{\gamma +1,c}f(z)\right)^{^{\prime }}}{p}=\phi (z)+ \frac{\delta cz\phi^{^{\prime }}(z)}{\gamma }\prec \frac{1+Az}{1+Bz}.$$

Now, by using Lemma 1 for $$\tau =\frac {\gamma }{\delta c},$$ we get

$$\begin{array}{@{}rcl@{}} -\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{p} &\prec &\Psi (z)=\frac{\gamma }{\delta c\left(p+m\right) } z^{-\frac{\gamma }{\delta c\left(p+m\right) }}\int\limits_{0}^{z}t^{\frac{ \gamma }{\delta c\left(p+m\right) }-1}\left(\frac{1+At}{1+Bt}\right) dt \\ &=&\left\{ \begin{array}{cc} \frac{A}{B}+\left(1-\frac{A}{B}\right) \left(1+Bz\right)_{\text{ \ \ \ } 2}^{-1}F_{1}\left(1,1;\tfrac{\gamma }{\delta c\left(p+m\right) }+1;\tfrac{ Bz}{1+Bz}\right), & B\neq 0 \\ 1+\frac{\gamma }{\gamma +\delta c\left(p+m\right) }Az, & B=0. \end{array} \right. \end{array}$$

This proves (17) of Theorem 1. In order to prove (20), we need to show that

$$\inf_{\left\vert z\right\vert <1}\left\{ \Re(\Psi(z))\right\} =\Psi (-1).$$
(22)

We have

$$\Re\left\{ \frac{1+Az}{1+Bz}\right\} \geq \frac{1-Ar}{1-Br} \left(\left\vert z\right\vert \leq r<1\right).$$

Putting

$$G\left(z,\zeta \right) =\frac{1+A\zeta z}{1+B\zeta z}\text{ and }dv(\zeta)= \frac{\gamma }{\delta c\left(p+m\right) }\zeta^{\frac{\gamma }{\delta c\left(p+m\right) }-1}d\zeta \left(0\leq \zeta \leq 1\right),$$

which is a positive measure on [0,1], we obtain

$$\Psi (z)=\int\limits_{0}^{1}G\left(z,\zeta \right) dv(\zeta).$$

Then

$$\Re(\Psi (z))\geq \int\limits_{0}^{1}\frac{1-A\zeta r}{1-B\zeta r}dv(\zeta)=\Psi (-r) \left(\left\vert z\right\vert \leq r<1\right).$$

Assuming r→1 in the above inequality, we obtain (22). The result in (19) is the best possible and Ψ is the best dominant of (17). This completes the proof of Theorem 1. □

### Theorem 2

Let $$f(z) \in \Sigma _{p,m}^{\gamma,c}\left (\alpha,\beta,\eta \right) \left (0\leq \eta < p\right),$$ then

$$\Re \left\{ -z^{p+1}\left[ \left(1-\delta \right) \left(\mathcal{H} _{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}+\delta \left(\mathcal{H}_{p,\alpha,\beta }^{\gamma +1,c}f(z)\right)^{^{\prime }}\right] \right\} >\eta \left(\left\vert z\right\vert <R\right),$$
(23)

where

$$R=\left\{ \frac{\sqrt{c^{2}\delta^{2}\left(p+m\right)^{2}+\gamma^{2}} -c\delta \left(p+m\right) }{\gamma }\right\}^{\frac{1}{p+m}}.$$
(24)

The result is the best possible.

### Proof

Since $$f(z)\in \Sigma _{p,m}^{\gamma,c}\left (\alpha,\beta,\eta \right),$$ let

$$-z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}=\eta +\left(p-\eta \right) u(z),$$
(25)

where u(z) in the form (11) and {u(z)}>0. Differentiating (25) and using (6), we get

$$-\frac{z^{p+1}\left[ \left(1-\delta \right) \left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}+\delta \left(\mathcal{H} _{p,\alpha,\beta }^{\gamma +1,c}f(z)\right)^{^{\prime }}\right] +\eta }{ p-\eta }=u(z)+\frac{c\delta zu^{^{\prime }}(z)}{\gamma }.$$
(26)

Applying the following estimate ,

$$\frac{\left\vert zu^{^{\prime }}(z)\right\vert }{\Re \left\{ u(z)\right\} } \leq \frac{2(p+m)r^{p+m}}{1-r^{2(p+m)}} \left(\left\vert z\right\vert =r<1\right) ;$$

in (26), we get

$$\begin{array}{@{}rcl@{}} &&\Re \left\{ -\frac{z^{p+1}\left[ \left(1-\delta \right) \left(\mathcal{H} _{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}+\delta \left(\mathcal{H}_{p,\alpha,\beta }^{\gamma +1,c}f(z)\right)^{^{\prime }}\right] +\eta }{p-\eta }\right\} \\ &\geq &\Re \left(u(z)\right) \left(1-\frac{2c\delta (p+m)r^{p+m}}{\gamma \left(1-r^{2(p+m)}\right) }\right). \end{array}$$
(27)

It is easily seen that the right-hand side of (27) is positive, if r<R, where R is given by (24). In order to show that the bound R is the best possible, we consider the function fΣp,m defined by

$$-z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}=\eta +\left(p-\eta \right) \frac{1+z^{p+m}}{1-z^{p+m}}.$$

Noting that

$$\begin{array}{@{}rcl@{}} &&-\frac{z^{p+1}\left[ \left(1-\delta \right) \left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}+\delta \left(\mathcal{H} _{p,\alpha,\beta }^{\gamma +1,c}f(z)\right)^{^{\prime }}\right] +\eta }{ p-\eta} \\ &=&\frac{\gamma \left(1-z^{2\left(p+m\right) }\right) +2c\delta (p+m)z^{p+m}}{\gamma \left(1-z^{p+m}\right)^{2}}=0, \end{array}$$

for

$$z=R\text{ }\Re \left(\frac{i\pi }{p+m}\right).$$

This completes the proof of Theorem 2. □

Putting δ=1 in Theorem 2, we obtain the following result.

### Corollary 1

If $$f(z) \in \Sigma _{p,m}^{\gamma,c}\left (\alpha,\beta,\eta \right) \left (0\leq \eta < p\right),$$ then $$f(z) \in \Sigma _{p,m}^{\gamma +1,c}\left (\alpha,\beta,\eta \right)$$ for |z|<R, where

$$R^{\ast }=\left\{ \frac{\sqrt{c^{2}\left(p+m\right)^{2}+\gamma^{2}} -c\left(p+m\right) }{\gamma }\right\}^{\frac{1}{p+m}}.$$

The result is the best possible.

### Theorem 3

If the function f(z)Σp,m satisfies

$$z^{p}\left[ \left(1-\delta \right) \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)+\delta \mathcal{H}_{p,\alpha,\beta }^{\gamma +1,c}f(z)\right] \prec \frac{1+Az}{1+Bz},$$

then

$$z^{p}\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\prec \Psi_{1}(z)\prec \frac{1+Az}{1+Bz}$$

and

$$\Re \left\{ z^{p}\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right\} >\rho,$$

where Ψ1(z) is in the form (18) and ρ given by (20). The result is the best possible.

### Proof

The proof follows by taking the same lines as in the proof of Theorem 1 and taking $$\phi (z)=z^{p}\mathcal {H}_{p,\alpha,\beta }^{\gamma,c}f(z)$$ in (21). □

For the function f(z) in the class Σp,m, Kumar and Shukla  defined the integral operator гμ,p:Σp,mΣp,m as follows:

$$\begin{array}{@{}rcl@{}} \digamma_{\mu,p}(f)(z) &=&\frac{\mu }{z^{\mu +p}}\int\limits_{0}^{z}t^{\mu +p-1}f(t)dt \\ &=&z^{-p}+\sum\limits_{k=m}^{\infty }\frac{\mu }{k+p+\mu }a_{k}z^{k}\ \left(\mu >0\right). \end{array}$$
(28)

From (28), we get

$$z\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z)\right)^{^{\prime }}=\mu \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)-\left(\mu +p\right) \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z).$$
(29)

### Theorem 4

Let the function f(z)given by (1) be in the class $$\Sigma _{p,m}^{\gamma,c}\left (\alpha,\beta ;A,B\right)$$ and гμ,p(f)(z) defined by (28). Then

$$-\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z)\right)^{^{\prime }}}{p}\prec \Theta (z)\prec \frac{1+Az}{ 1+Bz},$$
(30)

where

$$\Theta (z)=\left\{ \begin{array}{cc} \frac{A}{B}+\left(1-\frac{A}{B}\right) \left(1+Bz\right)_{\text{ \ \ \ } 2}^{-1}F_{1}\left(1,1;\frac{\mu }{\left(p+m\right) }+1;\frac{Bz}{1+Bz} \right), & B\neq 0 \\ 1+\frac{\mu }{\mu +p+m}Az, & B=0. \end{array} \right.$$
(31)

is the best dominant of (31). Furthermore,

$$\Re\left\{ -\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z)\right)^{^{\prime }}}{p}\right\} >\sigma \text{ } \left(0\leq \sigma <1\right),$$
(32)

where

$$\sigma =\left\{ \begin{array}{cc} \frac{A}{B}+\left(1-\frac{A}{B}\right) \left(1-B\right)_{\text{ \ \ \ } 2}^{-1}F_{1}\left(1,1;\frac{\mu }{\left(p+m\right) }+1;\frac{B}{B-1} \right), & B\neq 0 \\ 1-\frac{\mu }{\mu +p+m}A, & B=0. \end{array} \right.$$
(33)

The result is the best possible.

### Proof

Let

$$L(z)=-\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z)\right)^{^{\prime }}}{p},$$
(34)

where L in the form (11). Differentiating (34) and using (29), we get

$$-\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{p}=L(z)+\frac{z}{\mu }L^{^{\prime }}(z)\prec \frac{1+Az}{1+Bz}.$$

Now the remaining part of Theorem 4 follows by using the technique used in proving Theorem 1. □

### Theorem 5

Let the function гμ,p(f)(z) defined by (28) satisfy:

$$z^{p}\left[ \left(1-\delta \right) \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z)+\delta \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right] \prec \frac{1+Az}{1+Bz},$$
(35)

then

$$\Re\left\{ z^{p}\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z)\right\} >\theta,$$
(36)

where

$$\theta =\left\{ \begin{array}{cc} \frac{A}{B}+\left(1-\frac{A}{B}\right) \left(1-B\right)_{\text{ \ \ \ } 2}^{-1}F_{1}\left(1,1;\frac{\mu }{\delta \left(p+m\right) }+1;\frac{B}{B-1} \right), & B\neq 0 \\ 1-\frac{\mu }{\mu +\delta \left(p+m\right) }A, & B=0. \end{array} \right.$$

The result is the best possible.

### Proof

Let

$$K(z)=z^{p}\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z),$$
(37)

where K in the form (11). Differentiating (37) and using (29) and (35), we get

$$K(z)+\frac{\delta z}{\mu }K^{^{\prime }}(z)\prec \frac{1+Az}{1+Bz}.$$

Now the remaining part of Theorem 5 follows by using the technique used in proving Theorem 1. □

### Theorem 6

Let the function f(z)Σp,m satisfy:

$$\mathcal{-}\frac{z^{p+1}\left[ \left(1-\delta \right) \left(\mathcal{H} _{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z)\right)^{^{\prime }}+\delta \left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}\right] }{p}\prec \frac{1+Az}{1+Bz},$$

then

$$\Re\left\{ -\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z)\right)^{^{\prime }}}{p}\right\} >\theta,$$

where гμ,p(f)(z) is given by (28) and θ is given as in Theorem 5. The result is the best possible.

### Proof

The proof follows by taking the same lines as in Theorem 5. □

### Theorem 7

Let f(z)be in the class Σp,m. Also, let g(z)Σp,m satisfy:

$$\Re\left\{ z^{p}\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}g(z)\right\} >0.$$

If

$$\left\vert \frac{\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)}{\mathcal{H} _{p,\alpha,\beta }^{\gamma,c}g(z)}-1\right\vert <1,$$

then

$$\Re \left\{ -\frac{z\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)} \right\} >0\text{ }\left(\left\vert z\right\vert <R_{0}\right),$$
(38)

where

$$R_{0}=\frac{\sqrt{9(p+m)^{2}+4p(2p+m)}-3(p+m)}{2(2p+m)}.$$
(39)

### Proof

Let

$$\phi (z)=\frac{\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)}{\mathcal{H} _{p,\alpha,\beta }^{\gamma,c}g(z)}-1=e_{p+m}z^{p+m}+e_{p+m+1}z^{p+m+1}+\ldots \text{,}$$
(40)

we note that ϕ is analytic in $$\mathbb {U}$$, with ϕ(0)=0 and |ϕ(z)|≤|z|p+m. Then, by applying the familiar Schwarz Lemma , we have ϕ(z)=zp+mΨ(z) is analytic in $$\mathbb {U}$$ and $$\left \vert \Psi (z)\right \vert \leq 1 \left (z\in \mathbb {U}\right).$$ Therefore, (40) leads to

$$\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)=\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}g(z)\left(z^{p+m}\Psi (z)+1\right).$$
(41)

Differentiating (41), we obtain

$$\frac{z\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)}=\frac{z\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}g(z)\right)^{^{\prime }}}{ \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}g(z)}+\frac{z^{p+m}\left[ \left(p+m\right) \Psi (z)+z\Psi^{^{\prime }}(z)\right] }{1+z^{p+m}\Psi (z)}.$$
(42)

Letting $$\chi (z)=z^{p}\mathcal {H}_{p,\alpha,\beta }^{\gamma,c}g(z),$$ we see that the function χ is of the form (11) and is analytic in $$\mathbb {U}$$, {χ(z)}>0 and

$$\frac{z\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}g(z)\right)^{^{\prime }}}{\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}g(z)}=\frac{z\chi^{^{\prime }}(z)}{\chi (z)}-p,$$

so, we find from (42) that

$$\Re \left\{ -\tfrac{z\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)} \right\} \geq p-\left\vert \frac{z\chi^{^{\prime }}(z)}{\chi (z)} \right\vert -\left\vert \tfrac{z^{p+m}\left[ \left(p+m\right) \Psi (z)+z\Psi^{^{\prime }}(z)\right] }{1+z^{p+m}\Psi (z)}\right\vert.$$
(43)

$$\left\vert \frac{\chi^{^{\prime }}(z)}{\chi (z)}\right\vert \leq \frac{ 2(p+m)r^{p+m-1}}{1-r^{2(p+m)}}\text{ and }\left\vert \frac{\left(p+m\right) \Psi (z)+z\Psi^{^{\prime }}(z)}{1+z^{p+m}\Psi (z)}\right\vert \leq \frac{p+m }{1-r^{p+m}}\left(\left\vert z\right\vert =r<1\right),$$

in (43), we have

$$\Re \left\{ -\frac{z\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)} \right\} \geq \frac{p-3(p+m)r^{p+m}-(2p+m)r^{2(p+m)}}{1-r^{2(p+m)}},$$

which is certainly positive, provided that r<R0,R0 given by (39). □

### Theorem 8

Let the function f(z)Σp,m satisfy:

$$z^{p}\left[ \left(1-\delta \right) \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)+\delta \mathcal{H}_{p,\alpha,\beta }^{\gamma +1,c}f(z)\right] \prec \frac{1+Az}{1+Bz},$$

then

$$\Re\left\{ \left(z^{p}\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{\frac{1}{q}}\right\} >\epsilon^{\frac{1}{q}}\text{ }\left(q\in \mathbb{N} \right),$$

where ε in the form (20). The result is the best possible.

### Proof

Let

$$\phi (z)=z^{p}\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z),$$
(44)

where ϕ in the form (11). Differentiating (44) and using (6), we have

$$z^{p}\left[ \left(1-\delta \right) \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)+\delta \mathcal{H}_{p,\alpha,\beta }^{\gamma +1,c}f(z)\right] =\phi (z)+\frac{\delta cz\phi^{^{\prime }}(z)}{\gamma }\prec \frac{1+Az}{1+Bz}.$$

Now the remaining part of Theorem 8 follows by using the technique used in proving Theorem 1, and using the inequality:

$$\Re (w^{\frac{1}{q}})\geq \left(\Re (w)\right)^{\frac{1}{q}}\text{ }\left(\Re (w)>0;q\in \mathbb{N}\right),$$

we have the result asserted by Theorem 8. □

### Theorem 9

Let the function $$f(z)\in \Sigma _{p,m}^{\gamma,c}\left (\alpha, \beta ;A,B\right)$$ and let g(z)Σp,m satisfy:

$$\Re\left(z^{p}g(z)\right)>\frac{1}{2}.$$

Then,

$$\left(f\ast g\right) (z)\in \Sigma_{p,m}^{\gamma,c}\left(\alpha,\beta ;A,B\right).$$

### Proof

We have

$$-\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\left(f\ast g\right) (z)\right)^{^{\prime }}}{p}=-\frac{z^{p+1}\left(\mathcal{H} _{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{p}\ast z^{p}g(z).$$

Since

$$\Re \left(z^{p}g(z)\right) >\frac{1}{2}$$

and $$\frac {1+Az}{1+Bz}$$ is convex in $$\mathbb {U}$$, it follows from (8) and Lemma 4 that $$\left (f\ast g\right) (z)\in \Sigma _{p,m}^{\gamma,c}\left (\alpha,\beta ;A,B\right),$$ which completes the proof of Theorem 9. □

### Remark 1

For different value of γ, c, α, β, and p in the above results, we obtain results corresponding to the functions given in the introduction.

## Availability of data and materials

All data generated or analyzed during this study are included in this published article.

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