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# Some properties for meromorphic functions associated with integral operators

## Abstract

In the present paper we aim at proving some subordinations properties for meromorphic functions analytic in the punctured unit disc $$\Delta ^*=\{z:0<|z|<1\}$$ with a simple pole at the origin. The functions under investigation are associated with two integral operators $$\mathcal {P}_\sigma ^\gamma$$ and $$\mathcal {Q}_\sigma ^\gamma$$ (see Lashin in Comput Math Appl 59:524–531, 2010, https://doi.org/10.1016/j.camwa.2009.06.015). Several other results and numerical examples are also obtained.

## Introduction and Preliminaries

Let $$\Sigma$$ denote the class of functions of the form

\begin{aligned} f(z)=1/z+\sum _{\kappa =1}^\infty c_\kappa z^\kappa \end{aligned}
(1)

which are analytic in the punctured unit disc $$\Delta ^*=\{z:0<|z|<1\}$$ with a simple pole at the origin.

### Definition 1

For $$f(z)\in \Sigma$$, given by (1) and $$h(z)\in \Sigma$$ defined by

\begin{aligned} h(z)=1/z+\sum _{\kappa =1}^\infty h_\kappa z^\kappa , \end{aligned}

Hadamard product (or convolution) of f(z) and h(z) is given by

\begin{aligned} (f*h)(z)=1/z+\sum _{\kappa =1}^\infty c_\kappa h_\kappa z^\kappa =(h*f)(z). \end{aligned}

### Definition 2

[2] For two functions f and g, analytic in $$\Delta =\{z:|z|<1\}$$, we say that the function f is subordinate to g in $$\Delta$$, written $$f\prec g$$, if there exists a Schwarz function $$\omega (z)$$ which is analytic in $$\Delta$$, satisfying the following conditions:

\begin{aligned} \omega (0)=0\quad \quad \quad \text {and}\quad \quad |\omega (z)|<1;\quad (z\in \Delta ), \end{aligned}

such that

\begin{aligned} f(z)=g(\omega (z));\quad \quad (z\in \Delta ). \end{aligned}

In particular, if the function g is univalent in $$\Delta$$, we have the following equivalence (see also [3, 4]):

\begin{aligned} f(z)\prec g(z)\quad (z\in \Delta )\quad \Longleftrightarrow \quad f(0)=g(0)\quad \text {and}\quad f(\Delta )\subset g(\Delta ). \end{aligned}

In 2010, Lashin [1] defined the following integral operators $$\mathcal {P}_\sigma ^\gamma ,\,\mathcal {Q}_\sigma ^\gamma ,\,\mathcal {J}_\sigma :\Sigma \rightarrow \Sigma$$ as follows:

\begin{aligned}{} & {} \mathcal {P}_\sigma ^\gamma f(z)= \frac{\sigma ^\gamma }{z^{\sigma +1}\Gamma (\gamma )}\int _0^z s^{\sigma } \left( log\left( \frac{z}{s} \right) \right) ^{\gamma -1}f(s)ds\qquad (\gamma ,\sigma>0,\,z\in \Delta ^*),\\{} & {} \mathcal {Q}_\sigma ^\gamma f(z)= \frac{\Gamma (\sigma +\gamma )}{z^{\sigma +1}\Gamma (\sigma )\Gamma (\gamma )}\int _0^z s^{\sigma } \left( 1-\frac{s}{z} \right) ^{\gamma -1}f(s)ds\qquad (\gamma ,\sigma >0,\,z\in \Delta ^*), \end{aligned}

and

\begin{aligned} \mathcal {J}_\sigma f(z)= \frac{\sigma }{z^{\sigma +1}}\int _0^z s^{\sigma } f(s)ds\qquad (\sigma >0,\,z\in \Delta ^*), \end{aligned}

where $$\Gamma (\gamma )$$ is the familiar Gamma function.

Using the integral representation of the Gamma functions, it can be shown that

### Definition 3

Let $$f(z)\in \Sigma$$ be given by (1), then

\begin{aligned} \mathcal {P}_\sigma ^\gamma f(z)= & {} 1/z+\sum _{\kappa =1}^\infty \left( \frac{\sigma }{\kappa +\sigma +1} \right) ^\gamma c_\kappa z^\kappa \qquad \qquad \qquad (\gamma ,\sigma >0,\,z\in \Delta ^*), \end{aligned}
(2)
\begin{aligned} \mathcal {Q}_\sigma ^\gamma f(z)= & {} 1/z+\frac{\Gamma (\sigma +\gamma )}{\Gamma (\sigma )}\sum _{\kappa =1}^\infty \frac{\Gamma (\kappa +\sigma +1)}{\Gamma (\kappa +\sigma +\gamma +1)}c_\kappa z^\kappa \qquad (\gamma ,\sigma >0,\,z\in \Delta ^*), \end{aligned}
(3)

and

\begin{aligned} \mathcal {J}_\sigma f(z)= 1/z+\sum _{\kappa =1}^\infty \frac{\sigma }{\kappa +\sigma +1} c_\kappa z^\kappa \qquad \qquad \qquad \qquad \qquad (\sigma >0,\,z\in \Delta ^*). \end{aligned}
(4)

By (3) and (4), we can easily obtain that

\begin{aligned} z(\mathcal {P}_{\sigma }^{\gamma } f(z))' &= {} \sigma \mathcal {P}_{\sigma }^{\gamma -1} f(z) -(\sigma +1) \mathcal {P}_{\sigma }^{\gamma } f(z) \qquad (\sigma>0,\,\gamma > 1) \end{aligned}
(5)

and

\begin{aligned} z(\mathcal {Q}_{\sigma }^{\gamma } f(z))' = & {} (\sigma +\gamma -1) \mathcal {Q}_{\sigma }^{\gamma -1} f(z) -(\sigma +\gamma ) \mathcal {Q}_{\sigma }^{\gamma } f(z) \qquad (\sigma>0,\,\gamma > 1). \end{aligned}
(6)

### Remark 1

1. (i)

Putting $$\gamma =1$$ in the integral operators $$\mathcal {P}_{\sigma }^{\gamma }$$ and $$\mathcal {Q}_{\sigma }^{\gamma }$$, we obtain $$\mathcal {Q}_{\sigma }^{1}=\mathcal {P}_{\sigma }^{1}=\mathcal {J}_\sigma$$;

2. (ii)

The operator $$\mathcal {J}_\sigma$$ was defined and studied by Kumar and Shukla [5, Theorem 4.2, with $$p=1$$].

In order to prove our main results, we recall the following lemmas:

### Lemma 1

[6, 7] If g(z) is a convex function in $$\Delta$$ with $$g(0)=1$$, $$q(z)=1+q_1(z)+q_2(z)+...$$ is analytic in $$\Delta$$ and $$\delta \in \mathbb {C},$$ $$\text {Re}(\delta )>0,$$ then

\begin{aligned} q(z)+\frac{zq'(z)}{\delta }\prec g(z) \end{aligned}

implies

\begin{aligned} q(z)\prec \delta z^{-\delta } \int _0^z s^{\delta -1} g(s) ds=\hbar (z)\prec g(z) \end{aligned}

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

### Definition 4

Let ab and c be complex and real numbers with $$c\ne 0,-1,-2,...$$. The Gaussian hypergeometric function is defined by the power series

\begin{aligned} \begin{aligned} _2F_1(a,b;c;z)&=1+\frac{ab}{c}z+\frac{a(a+1)b(b+1)}{2! c(c+1)}z^2+...\\ {}&=\sum _{\kappa =0}^\infty \frac{(a)_\kappa (b)_\kappa }{(c)_\kappa (1)_\kappa } z^\kappa , \end{aligned} \end{aligned}

where $$(a)_\kappa =\Gamma (a+\kappa )/\Gamma (a)$$ is the Pochhammer symbol.

### Lemma 2

[8, 9] For abc real numbers other than $$0,-1,-2,...$$ and $$c>b>0,$$ we have

\begin{aligned} \int _0^1 s^{b-1} (1-s)^{c-b-1} (1-sz)^{-a} ds= & {} \frac{\Gamma (b)\Gamma (c-b)}{\Gamma (c)}\, {_2}F_1(a,b;c;z) \end{aligned}
(7)
\begin{aligned} {_2}F_1(a,b;c;z)= & {} (1-z)^{-a} {_2}F_1(a,c-b;c;z/(z-1)) \end{aligned}
(8)
\begin{aligned} {_2}F_1(1,1;2;z)= & {} -(1/z)ln(1-z) \end{aligned}
(9)
\begin{aligned}{} & {} c(c-1)(z-1) {_2}\,F_1(a,b;c-1;z)+ c[c-1-(2c-a-b-1)z]\, {_2}F_1(a,b;c;z) +(c-a)(c-b)z\, {_2}F_1(a,b;c+1;z)=0. \end{aligned}
(10)

### Lemma 3

For any real number $$\alpha \ne 0,$$ we have

\begin{aligned} _2F_1(1,1;2;\alpha z/(\alpha z+1))= & {} \frac{(1+\alpha z) ln(1+\alpha z)}{\alpha z}, \end{aligned}
(11)
\begin{aligned} _2F_1(1,1;3;\alpha z/(\alpha z+1))= & {} \frac{2(1+\alpha z)}{\alpha z}\left[ 1-\frac{ln(1+\alpha z)}{\alpha z} \right] , \end{aligned}
(12)
\begin{aligned} _2F_1(1,1;4;\alpha z/(\alpha z+1))= & {} \frac{3(1+\alpha z)}{2(\alpha z)^3}[2 ln(1+\alpha z)-\alpha z(2-\alpha z)], \end{aligned}
(13)
\begin{aligned} _2F_1(1,1;5;\alpha z/(\alpha z+1))= & {} \frac{2(1+\alpha z)}{(\alpha z)^3}\left[ \frac{2(\alpha z)^2-3\alpha z+6}{3}-\frac{2 ln(1+\alpha z)}{\alpha z} \right] . \end{aligned}
(14)

### Proof

To prove the relation (11), substituting for $$z=\alpha z/(\alpha z+1)$$ in (9), we obtain

\begin{aligned} \begin{aligned} _2F_1(1,1;2;\alpha z/(\alpha z+1))&=-\left( \frac{1+\alpha z}{\alpha z}\right) \ln \left( 1-\frac{\alpha z}{\alpha z+1} \right) \\&=-\left( \frac{1+\alpha z}{\alpha z}\right) \ln \left( \frac{1}{\alpha z+1} \right) \\&= \frac{(1+\alpha z) ln(1+\alpha z)}{\alpha z}. \end{aligned} \end{aligned}

Now, we prove the relation (12), by replacing $$a=b=1$$ and $$c=2$$ in (10), we obtain

\begin{aligned} 2(z-1) _2F_1(1,1;1;z)+2(1-z) \,_2F_1(1,1;2;z)+z \,_2F_1(1,1;3,z)=0 \end{aligned}

then

\begin{aligned} \begin{aligned} \,_2F_1(1,1;3,z)&= \frac{1}{z}[-2(z-1) _2F_1(1,1;1;z)-2(1-z) \,_2F_1(1,1;2;z)]\\&= \frac{2}{z}\left[ 1+\frac{1-z}{z}ln(1-z)\right] . \end{aligned} \end{aligned}

By replacing $$z=\alpha z/(\alpha z+1)$$ in the above equation, we obtain

\begin{aligned} \begin{aligned} _2F_1(1,1;3,\alpha z/(\alpha z+1))&= \frac{2(1+\alpha z)}{\alpha z}\left[ 1+\frac{1}{\alpha z}ln \left( 1-\frac{\alpha z}{1+\alpha z}\right) \right] \\&=\frac{2(1+\alpha z)}{\alpha z}\left[ 1-\frac{ln (1+\alpha z)}{\alpha z}\right] . \end{aligned} \end{aligned}

Similarly, we can obtain the relations (13) and (14). $$\square$$

The purpose of this paper is to prove some subordinations properties for meromorphic functions analytic in the punctured unit disc $$\Delta ^*=\{z:0<|z|<1\}.$$ The functions under investigation are associated with two integral operators $$\mathcal {P}_\sigma ^\gamma$$ and $$\mathcal {Q}_\sigma ^\gamma$$. Several other results and numerical examples are also obtained.

## Main results

Unless otherwise mentioned, we assume throughout this paper that $$\gamma>2,\,\sigma >0$$ and $$-1\le B<A\le 1.$$

### Theorem 1

Let

\begin{aligned} \frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_\sigma ^{\gamma -2}f(z)}{\mathcal {P}_\sigma ^{\gamma -1} f(z)}- \frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma }f(z)} \right) \prec \frac{1+A z}{1+B z}\qquad (\sigma>0,\,\gamma >2) \end{aligned}
(15)

then we have

\begin{aligned} \frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma } f(z)}\prec \hbar (z) \prec \frac{1+A z}{1+B z}\qquad (z\in \Delta ) \end{aligned}

where

\begin{aligned} \hbar (z)=(1+Bz)^{-1}\left[ \,{_2}F_1\bigg (1,1;1+\sigma ;Bz/(1+Bz)\bigg ) +\frac{\sigma A z}{\sigma +1 } \,{_2}F_1\bigg (1,1;2+\sigma ;Bz/(1+Bz)\bigg ) \right] \end{aligned}
(16)

and $$\hbar (z)$$ is the best dominant. Furthermore,

\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma } f(z)} \right) >\eta \end{aligned}

where

\begin{aligned} \eta =(1-B)^{-1}\left[ \,{_2}F_1\bigg (1,1;1+\sigma ;B/(B-1)\bigg ) -\frac{\sigma A}{\sigma +1}\, {_2}F_1\bigg (1,1;2+\sigma ;B/(B-1)\bigg ) \right] . \end{aligned}
(17)

### Proof

Suppose that

\begin{aligned} q(z)= \frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma } f(z)}, \end{aligned}
(18)

then $$q(z)=1+a_1 z+a_2 z^2+...$$ is analytic in $$\Delta$$ with $$q(0)=1$$. Using the logarithmic differentiation of the both sides of (18) with respect to z and with the aid of the identity (5), we get

\begin{aligned} \left( \frac{1}{\sigma } \right) \frac{ zq'(z)}{ q(z)}= \frac{\mathcal {P}_\sigma ^{\gamma -2}f(z)}{\mathcal {P}_\sigma ^{\gamma -1} f(z)}-\frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma } f(z)} \end{aligned}
(19)

By using (18) and (19), we obtain

\begin{aligned} \frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_\sigma ^{\gamma -2}f(z)}{\mathcal {P}_\sigma ^{\gamma -1} f(z)}- \frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma }f(z)} \right) =q(z)+\left( \frac{1}{\sigma }\right) zq'(z). \end{aligned}

Thus, by using Lemma 1, for $$\delta =\sigma$$, we have

\begin{aligned} \frac{\mathcal {P}_\sigma ^{\gamma -1}f(z)}{\mathcal {P}_\sigma ^{\gamma } f(z)}\prec \sigma z^{-\sigma } \int _0^z s^{\sigma -1}\frac{1+A s}{1+Bs}ds =\hbar (z). \end{aligned}

Using the identities (7) and (8), we can written $$\hbar (z)$$ as following:

\begin{aligned} \hbar (z)= & {} \sigma \int _0^1 t^{\sigma -1}\frac{1+A tz}{1+Btz}dt \\= & {} \sigma \bigg [ \int _0^1 t^{\sigma -1}(1+Btz)^{-1}dt+A z \int _0^1 t^{\sigma }(1+Btz)^{-1}dt\bigg ]\\= & {} (1+Bz)^{-1}\bigg [\, _2F_1\bigg (1,1;1+\sigma , Bz/(1+Bz) \bigg )+\frac{\sigma A z}{\sigma +1}\, _2F_1\bigg (1,1;2+\sigma , Bz/(1+Bz)\bigg ) \bigg ]. \end{aligned}

This completes the proof of (16) of Theorem 1. Now, to prove the assertion (17) of Theorem 1, it suffices to show that

\begin{aligned} \inf _{|z|<1}\{\hbar (z)\}=\hbar (-1). \end{aligned}
(20)

Indeed, for $$|z|\le r<1,$$

\begin{aligned} \text {Re}\bigg (\frac{1+Az}{1+Bz}\bigg )\ge \frac{1-A r}{1-Br}. \end{aligned}

Upon setting

\begin{aligned} K(t,z)=\frac{1+Atz}{1+Btz} \text { and } d\nu (t)=\sigma t^{\sigma -1} dt\qquad \qquad (0\le t\le 1,\, z\in \Delta ), \end{aligned}

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

\begin{aligned} \hbar (z)=\int _0^1 K(t,z) d\nu (t), \end{aligned}

so that

\begin{aligned} \text {Re}(\hbar (z))\ge \int _0^1 \bigg (\frac{1-Atr}{1-Btr}\bigg ) d\nu (t)=\hbar (-r)\qquad \qquad (|z|\le r<1). \end{aligned}

Letting $$r\rightarrow 1^-$$ in the above inequality, we obtain the assertion (20). The result in (17) is the best possible as the function $$\hbar (z)$$ is the best dominant of (16). The proof of Theorem 1 is completed. $$\square$$

Letting $$\sigma =1$$ in Theorem 1 and using the identities (11) and (12), we have

### Corollary 1

Let

\begin{aligned} \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}^{\gamma -2}f(z)}{\mathcal {P}^{\gamma -1} f(z)}- \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma }f(z)} \right) \prec \frac{1+A z}{1+B z}\qquad (\gamma >2) \end{aligned}

then we have

\begin{aligned} \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)}\prec \hbar (z) \prec \frac{1+A z}{1+B z}\qquad (z\in \Delta ) \end{aligned}

where

\begin{aligned} \hbar (z)=\left\{ \begin{array}{lll} \bigg (1-\frac{A}{B} \bigg )\frac{\ln (1+Bz)}{Bz}+\frac{A}{B} &{} \quad &{} \quad B\ne 0 \\ &{} \quad &{} \\ 1+\frac{A}{2}z &{} \quad &{} \quad B=0, \end{array} \right. \end{aligned}
(21)

and $$\hbar (z)$$ is the best dominant. Furthermore,

\begin{aligned} \text {Re}\left( \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)} \right) >\eta \end{aligned}

where

\begin{aligned} \eta =\left\{ \begin{array}{lll} \bigg (\frac{A}{B}-1 \bigg )\frac{\ln (1-B)}{B}+\frac{A}{B} &{} \quad &{} \quad B\ne 0 \\ &{} \quad &{} \\ 1- \frac{A}{2} &{} \quad &{} B=0. \end{array} \right. \end{aligned}
(22)

Letting $$B\ne 0$$ in Corollary 1, we have

### Corollary 2

If

\begin{aligned} \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}^{\gamma -2}f(z)}{\mathcal {P}^{\gamma -1} f(z)}- \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma }f(z)} \right) \prec \frac{1+\frac{B\ln (1-B)}{B+\ln (1-B)} z}{1+B z}\qquad (\gamma >2) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)} \right) >0\qquad \qquad (z\in \Delta ). \end{aligned}

Letting $$B=-1$$ in Corollary 2, we obtain the following special case

### Example 1

If

\begin{aligned} \text {Re}\bigg ( \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}^{\gamma -2}f(z)}{\mathcal {P}^{\gamma -1} f(z)}- \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma }f(z)} \right) \bigg )> \frac{2\ln 2-1}{2\ln 2-2}\thickapprox -0.61\qquad (\gamma >2) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)} \right) >0\qquad \qquad (z\in \Delta ). \end{aligned}

Letting $$A=1-2\lambda ,\,(0\le \lambda <1)$$ and $$B=-1$$ in Corollary 1, we have

### Corollary 3

If

\begin{aligned} \text {Re}\bigg (\frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}^{\gamma -2}f(z)}{\mathcal {P}^{\gamma -1} f(z)}- \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma }f(z)} \right) \bigg )>\lambda \qquad (\gamma >2) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {P}^{\gamma -1}f(z)}{\mathcal {P}^{\gamma } f(z)} \right) >(2\lambda -1)+2(1-\lambda )\ln 2. \end{aligned}

Letting $$\sigma =2$$ in Theorem 1 and using the identities (12) and (13), we have

### Corollary 4

Let

\begin{aligned} \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_2^{\gamma -2}f(z)}{\mathcal {P}_2^{\gamma -1} f(z)}- \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma }f(z)} \right) \prec \frac{1+A z}{1+B z}\qquad (\gamma >2) \end{aligned}

then we have

\begin{aligned} \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)}\prec \hbar (z) \prec \frac{1+A z}{1+B z}\qquad (z\in \Delta ) \end{aligned}

where

\begin{aligned} \hbar (z)=\left\{ \begin{array}{lll} \bigg (\frac{A}{B}-1 \bigg )\bigg (\frac{2[\ln (1+Bz)-Bz]}{B^2z^2}\bigg )+\frac{A}{B} &{} \quad &{} \quad B\ne 0 \\ &{} \quad &{} \\ \frac{2}{3}Az+1 &{} \quad &{} B=0, \end{array} \right. \end{aligned}
(23)

and $$\hbar (z)$$ is the best dominant. Furthermore,

\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)} \right) >\eta \end{aligned}

where

\begin{aligned} \eta =\left\{ \begin{array}{lll} \bigg (\frac{A}{B}-1 \bigg )\frac{2[\ln (1-B)+B]}{B^2}+\frac{A}{B} &{} \quad &{} \quad B\ne 0 \\ &{} \quad &{} \\ 1-\frac{2}{3}A &{} \quad &{} B=0, \end{array} \right. \end{aligned}
(24)

Letting $$B\ne 0$$ in Corollary 4, we have

### Corollary 5

If

\begin{aligned} \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_2^{\gamma -2}f(z)}{\mathcal {P}_2^{\gamma -1} f(z)}- \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma }f(z)} \right) \prec \frac{1+\frac{B\ln (1-B)}{B+\ln (1-B)} z}{1+B z}\qquad (\gamma >2) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)} \right) >0 \qquad \qquad (z\in \Delta ). \end{aligned}

Letting $$B=-1$$ in Corollary 5, we obtain the following special case

### Example 2

If

\begin{aligned} \text {Re}\bigg ( \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_2^{\gamma -2}f(z)}{\mathcal {P}_2^{\gamma -1} f(z)}- \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma }f(z)} \right) \bigg )> \frac{4\ln 2-3}{4 \ln 2-2}\thickapprox -0.29 \qquad (\gamma >2) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)} \right) >0 \qquad \qquad (z\in \Delta ). \end{aligned}

Letting $$A=1-2\lambda ,\,(0\le \lambda <1)$$ and $$B=-1$$ in Corollary 4, we have

### Corollary 6

If

\begin{aligned} \text {Re}\bigg ( \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_2^{\gamma -2}f(z)}{\mathcal {P}_2^{\gamma -1} f(z)}- \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma }f(z)} \right) \bigg )>\lambda \qquad (\gamma >2) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_2^{\gamma -1}f(z)}{\mathcal {P}_2^{\gamma } f(z)} \right) >(2\lambda -1)-4(1-\lambda )(\ln 2-1). \end{aligned}

Letting $$\sigma =3$$ in Theorem 1 and using the identities (13) and (14), we have

### Corollary 7

Let

\begin{aligned} \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_3^{\gamma -2}f(z)}{\mathcal {P}_3^{\gamma -1} f(z)}- \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma }f(z)} \right) \prec \frac{1+A z}{1+B z}\qquad (\gamma >2) \end{aligned}

then we have

\begin{aligned} \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)}\prec \hbar (z) \prec \frac{1+A z}{1+B z}\qquad (z\in \Delta ) \end{aligned}

where

\begin{aligned} \hbar (z)=\left\{ \begin{array}{lll} \bigg (1-\frac{A}{B} \bigg )\frac{3}{B^3z^3}\bigg [\ln (1+Bz)-Bz+\frac{B^2 z^2}{2}\bigg ]+\frac{A}{B} &{} \quad &{} \quad B\ne 0 \\ &{} \quad &{} \\ 1+ \frac{3}{4}Az &{} \quad &{} B=0, \end{array} \right. \end{aligned}
(25)

and $$\hbar (z)$$ is the best dominant. Furthermore,

\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)} \right) >\eta \end{aligned}

where

\begin{aligned} \eta =\left\{ \begin{array}{lll} \bigg (\frac{A}{B}-1 \bigg )\frac{3}{B^3}\bigg [\ln (1-B)+B+\frac{B^2}{2}\bigg ]+\frac{A}{B} &{} \quad &{} \quad B\ne 0 \\ &{} \quad &{} \\ 1- \frac{3}{4}A &{} \quad &{} B=0. \end{array} \right. \end{aligned}
(26)

Letting $$B\ne 0$$ in Corollary 7, we have

### Corollary 8

If

\begin{aligned} \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_3^{\gamma -2}f(z)}{\mathcal {P}_3^{\gamma -1} f(z)}- \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma }f(z)} \right) \prec \frac{1+\frac{B\ln (1-B)}{B+\ln (1-B)} z}{1+B z}\qquad (\gamma >2) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)} \right) >0\qquad \qquad (z\in \Delta ). \end{aligned}

Letting $$B=-1$$ in Corollary 8, we obtain the following special case

### Example 3

If

\begin{aligned} \text {Re}\bigg (\frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_3^{\gamma -2}f(z)}{\mathcal {P}_3^{\gamma -1} f(z)}- \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma }f(z)} \right) \bigg )> \frac{12 \ln 2-19}{12\ln 2 -20}\thickapprox 0.91\qquad (\gamma >2) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)} \right) >0\qquad \qquad (z\in \Delta ). \end{aligned}

Letting $$A=1-2\lambda ,\,(0\le \lambda <1)$$ and $$B=-1$$ in Corollary 7, we have

### Corollary 9

If

\begin{aligned} \text {Re}\bigg ( \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)}\left( 1+ \frac{\mathcal {P}_3^{\gamma -2}f(z)}{\mathcal {P}_3^{\gamma -1} f(z)}- \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma }f(z)} \right) \bigg )>\lambda \qquad (\gamma >2) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {P}_3^{\gamma -1}f(z)}{\mathcal {P}_3^{\gamma } f(z)} \right) > (2\lambda -1)+3(1-\lambda )(2\ln 2-3). \end{aligned}

### Theorem 2

Let

\begin{aligned} \frac{\mathcal {Q}_\sigma ^{\gamma -1}f(z)}{\mathcal {Q}_\sigma ^{\gamma } f(z)}\left( \frac{\mathcal {Q}_\sigma ^{\gamma -2}f(z)}{\mathcal {Q}_\sigma ^{\gamma -1} f(z)}-\frac{\sigma +\gamma -1}{\sigma +\gamma -2} \frac{\mathcal {Q}_\sigma ^{\gamma -1}f(z)}{\mathcal {Q}_\sigma ^{\gamma }f(z)}+\frac{\sigma +\gamma -1}{\sigma +\gamma -2} \right) \prec \frac{1+A z}{1+B z}\qquad (\sigma>0,\,\gamma >2) \end{aligned}

then we have

\begin{aligned} \frac{\mathcal {Q}_\sigma ^{\gamma -1}f(z)}{\mathcal {Q}_\sigma ^{\gamma } f(z)}\prec \hbar (z) \prec \frac{1+A z}{1+B z}\qquad (z\in \Delta ) \end{aligned}

where

\begin{aligned} \hbar (z)=(1+Bz)^{-1}\left[ \,_2F_1\bigg (1,1;\sigma +\gamma -1;Bz/(1+Bz) \bigg )+\frac{(\sigma +\gamma -2) A z}{\sigma +\gamma -1} \,_2F_1\bigg (1,1;\sigma +\gamma ;Bz/(1+Bz)\bigg ) \right] \end{aligned}
(27)

and $$\hbar (z)$$ is the best dominant. Furthermore,

\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{\gamma -1}f(z)}{\mathcal {Q}_\sigma ^{\gamma } f(z)} \right) >\eta \end{aligned}

where

\begin{aligned} \eta =(1-B)^{-1}\left[ \,_2F_1\bigg (1,1;\sigma +\gamma -1;B/(B-1)\bigg ) -\frac{(\sigma +\gamma -2) A }{\sigma +\gamma -1} \,_2F_1\bigg (1,1;\sigma +\gamma ;B/(B-1)\bigg ) \right] . \end{aligned}
(28)

### Proof

Suppose that

\begin{aligned} q(z)= \frac{\mathcal {Q}_\sigma ^{\gamma -1}f(z)}{\mathcal {Q}_\sigma ^{\gamma } f(z)}, \end{aligned}
(29)

then $$q(z)=1+a_1 z+a_2 z^2+...$$ is analytic in $$\Delta$$ with $$q(0)=1$$. Using the logarithmic differentiation of the both sides of (29) with respect to z, and with the aid of the identities (6), we get

\begin{aligned} \left( \frac{1}{\sigma +\gamma -2} \right) \frac{z q'(z)}{ q(z)}= \frac{\mathcal {Q}_\sigma ^{\gamma -2}f(z)}{\mathcal {Q}_\sigma ^{\gamma -1} f(z)}-\frac{\sigma +\gamma -1}{\sigma +\gamma -2}\frac{\mathcal {Q}_\sigma ^{\gamma -1}f(z)}{\mathcal {Q}_\sigma ^{\gamma } f(z)}+\frac{1}{\sigma +\gamma -2} \end{aligned}
(30)

By using (29) and (30), we obtain

\begin{aligned} \frac{\mathcal {Q}_\sigma ^{\gamma -1}f(z)}{\mathcal {Q}_\sigma ^{\gamma } f(z)}\left( \frac{\mathcal {Q}_\sigma ^{\gamma -2}f(z)}{\mathcal {Q}_\sigma ^{\gamma -1} f(z)}-\frac{\sigma +\gamma -1}{\sigma +\gamma -2} \frac{\mathcal {Q}_\sigma ^{\gamma -1}f(z)}{\mathcal {Q}_\sigma ^{\gamma }f(z)}+\frac{\sigma +\gamma -1}{\sigma +\gamma -2} \right) =q(z)+\left( \frac{1}{\sigma +\gamma -2}\right) zq'(z). \end{aligned}

Thus, by using Lemma 1, for $$\delta =\sigma +\gamma -2$$, the estimates (27) and (28) can be proved on the same lines as that of (16) and (17). This completes the proof of Theorem 2. $$\square$$

Letting $$\gamma =3-\sigma$$ in Theorem 2 and using the identities (11) and (12), we have

### Corollary 10

Let

\begin{aligned} \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}\left( 2+ \frac{\mathcal {Q}_\sigma ^{1-\sigma }f(z)}{\mathcal {Q}_\sigma ^{2-\sigma } f(z)}-2 \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma }f(z)} \right) \prec \frac{1+A z}{1+B z}\qquad (0<\sigma <1) \end{aligned}

then we have

\begin{aligned} \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}\prec \hbar (z) \prec \frac{1+A z}{1+B z}\qquad (z\in \Delta ) \end{aligned}

where $$\hbar (z)$$ given as (21) and $$\hbar (z)$$ is the best dominant. Furthermore,

\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)} \right) >\eta \end{aligned}

where $$\eta$$ given as (22).

Letting $$B\ne 0$$ in Corollary 10, we have

### Corollary 11

If

\begin{aligned} \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}\left( 2+ \frac{\mathcal {Q}_\sigma ^{1-\sigma }f(z)}{\mathcal {Q}_\sigma ^{2-\sigma } f(z)}-2 \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma }f(z)} \right) \prec \frac{1+\frac{B\ln (1-B)}{B+\ln (1-B)} z}{1+B z}\qquad (0<\sigma <1) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)} \right) >0\qquad \qquad (z\in \Delta ). \end{aligned}

Letting $$B=-1$$ in Corollary 11, we obtain the following special case

### Example 4

If

\begin{aligned} \text {Re}\bigg ( \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}\left( 2+ \frac{\mathcal {Q}_\sigma ^{1-\sigma }f(z)}{\mathcal {Q}_\sigma ^{2-\sigma } f(z)}-2 \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma }f(z)} \right) \bigg )> \frac{2\ln 2-1}{2\ln 2-2}\thickapprox -0.61\qquad (0<\sigma <1) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)} \right) >0\qquad \qquad (z\in \Delta ). \end{aligned}

Letting $$A=1-2\lambda ,\,(0\le \lambda <1)$$ and $$B=-1$$ in Corollary 10, we have

### Corollary 12

If

\begin{aligned} \text {Re}\bigg ( \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}\left( 2+ \frac{\mathcal {Q}_\sigma ^{1-\sigma }f(z)}{\mathcal {Q}_\sigma ^{2-\sigma } f(z)}-2 \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma }f(z)} \right) \bigg )>\lambda \qquad (0<\sigma <1) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)} \right) >(2\lambda -1)+2(1-\lambda )\ln 2. \end{aligned}

Letting $$\gamma =4-\sigma$$ in Theorem 2 and using the identities (12) and (13), we have

### Corollary 13

Let

\begin{aligned} \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}\left( \frac{3}{2}+ \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}-\frac{3}{2} \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma }f(z)} \right) \prec \frac{1+A z}{1+B z}\qquad (0<\sigma <2) \end{aligned}

then we have

\begin{aligned} \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}\prec \hbar (z) \prec \frac{1+A z}{1+B z}\qquad (z\in \Delta ) \end{aligned}

where $$\hbar (z)$$ given as (23) and $$\hbar (z)$$ is the best dominant. Furthermore,

\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)} \right) >\eta \end{aligned}

where $$\eta$$ given as (24).

Letting $$B\ne 0$$ in Corollary 13, we have

### Corollary 14

If

\begin{aligned} \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}\left( \frac{3}{2}+ \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}-\frac{3}{2} \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma }f(z)} \right) \prec \frac{1+\frac{B\ln (1-B)}{B+\ln (1-B)} z}{1+B z}\qquad (0<\sigma <2) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)} \right) >0 \qquad \qquad (z\in \Delta ). \end{aligned}

Letting $$B=-1$$ in Corollary 15, we obtain the following special case

### Example 5

If

\begin{aligned} \text {Re}\bigg ( \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}\left( \frac{3}{2}+ \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}-\frac{3}{2} \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma }f(z)} \right) \bigg )> \frac{4\ln 2-3}{4 \ln 2-2}\thickapprox -0.29 \qquad (0<\sigma <2) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)} \right) >0 \qquad \qquad (z\in \Delta ). \end{aligned}

Letting $$A=1-2\lambda ,\,(0\le \lambda <1)$$ and $$B=-1$$ in Corollary 13, we have

### Corollary 15

If

\begin{aligned} \text {Re}\bigg ( \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}\left( \frac{3}{2}+ \frac{\mathcal {Q}_\sigma ^{2-\sigma }f(z)}{\mathcal {Q}_\sigma ^{3-\sigma } f(z)}-\frac{3}{2} \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma }f(z)} \right) \bigg )>\lambda \qquad (0<\sigma <2) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)} \right) >(2\lambda -1)-4(1-\lambda )(\ln 2-1). \end{aligned}

Letting $$\gamma =5-\sigma$$ in Theorem 2 and using the identities (13) and (14), we have

### Corollary 16

Let

\begin{aligned} \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)}\left( \frac{4}{3}+ \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}-\frac{4}{3} \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma }f(z)} \right) \prec \frac{1+A z}{1+B z}\qquad (0<\sigma <3) \end{aligned}

then we have

\begin{aligned} \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)}\prec \hbar (z) \prec \frac{1+A z}{1+B z}\qquad (z\in \Delta ) \end{aligned}

where $$\hbar (z)$$ given as (25) and $$\hbar (z)$$ is the best dominant. Furthermore,

\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)} \right) >\eta \end{aligned}

where given as (26).

Letting $$B\ne 0$$ in Corollary 16, we have

### Corollary 17

If

\begin{aligned} \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)}\left( \frac{4}{3}+ \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}-\frac{4}{3} \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma }f(z)} \right) \prec \frac{1+\frac{B\ln (1-B)}{B+\ln (1-B)} z}{1+B z}\qquad (0<\sigma <3) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)} \right) >0\qquad \qquad (z\in \Delta ). \end{aligned}

Letting $$B=-1$$ in Corollary 17, we obtain the following special case

### Example 6

If

\begin{aligned} \text {Re} \bigg (\frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)}\left( \frac{4}{3}+ \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}-\frac{4}{3} \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma }f(z)} \right) \bigg )> \frac{12 \ln 2-19}{12\ln 2 -20}\thickapprox 0.91\qquad (0<\sigma <3) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)} \right) >0\qquad \qquad (z\in \Delta ). \end{aligned}

Letting $$A=1-2\lambda ,\,(0\le \lambda <1)$$ and $$B=-1$$ in Corollary 16, we have

### Corollary 18

If

\begin{aligned} \text {Re} \bigg (\frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)}\left( \frac{4}{3}+ \frac{\mathcal {Q}_\sigma ^{3-\sigma }f(z)}{\mathcal {Q}_\sigma ^{4-\sigma } f(z)}-\frac{4}{3} \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma }f(z)} \right) \bigg )>\lambda \qquad (0<\sigma <3) \end{aligned}

then

\begin{aligned} \text {Re}\left( \frac{\mathcal {Q}_\sigma ^{4-\sigma }f(z)}{\mathcal {Q}_\sigma ^{5-\sigma } f(z)} \right) > (2\lambda -1)+3(1-\lambda )(2\ln 2-3). \end{aligned}

## Conclusion

The purpose of the current work is to demonstrate various subordination characteristics for analytical meromorphic functions in $$\Delta ^*=\{z:0<|z|<1\}$$. Lashin [1] defined and studied the integral operators $$\mathcal {P}_\sigma ^\gamma$$ and $$\mathcal {Q}_\sigma ^\gamma$$ on the meromorphic functions. We use these operators to study and demonstrate some subordination characteristics for meromorphic functions. In addition, we obtain some particular cases and numerical examples of our main results.

## Availability of data and materials

During the current study the data sets are derived arithmetically.

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3. Bulboacâ, T.: Differential Subordinations and Superordinations: Recent Results. House of Scientific Book Publishing, Cluj-Napoca (2005)

4. Miller, S.S., Mocanu, P.T.: Differential Subordinations: Theory and Applications. Marcel Dekker Incorporated, New York (2000)

5. Kumar, V., Shukla, S.L.: Certain integrals for classes of p-valent meromorphic functions. Bull. Austral. Math. Soc. 25, 35–97 (1982). https://doi.org/10.1017/S0004972700005062

6. Hallenbeck, D.J., Ruscheweyh, S.T.: Subordination by convex functions. Proc. Am. Math. Soc. 52, 191–195 (1975). https://doi.org/10.1090/S0002-9939-1975-0374403-3

7. Miller, S.S., Mocanu, P.T.: Differential subordination and univalent functions. Michigan Math. J. 28, 157–171 (1981). https://doi.org/10.1307/mmj/1029002507

8. Abramowitz, M., Stegun, I.A.: Hand Book of Mathematical functions. Dover Publications, Inc., New York (1971)

9. Whittaker, E.T., Watson, G.N.: A Course on Modern Analysis: An Introduction to the General Theory of Infinite Processes and of Analytic Functions; With An Account of the Principal Transcendental Functions, 4th edn. Cambridge University Press, Cambridge (1927)

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The authors would like to express their sincere thanks to the referees for their valuable comments, which helped improve the manuscript.

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Kota, W.Y., El-Ashwah, R.M. Some properties for meromorphic functions associated with integral operators. J Egypt Math Soc 31, 5 (2023). https://doi.org/10.1186/s42787-023-00163-4