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

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

• (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 a, b 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 a, b, c 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.