# On a Class of Meromorphic Functions Related to Cho-Kwon-Srivastava Operator

Authors: F. Ghanim; M. Darus
Journal: ISRN Applied Mathematics
 (2012)

---

## Abstract

We introduce new classes,Σn,λ*,μ,k(α,β,ρ) and Σn,λμ,k(α,β,ρ),
of meromorphic functions defined by means of the Hadamard product of Cho-Kwon-Srivastava operator, and we define here a similar transformation by means of an operator
given by Ghanim and Darus, in U={z:0<|z|<1}, and investigate a number of inclusion
relationships of these classes. We also derive some interesting properties of these
classes.


---

## Body

## 1. Introduction

LetΣ denote the class of meromorphic functions f(z) normalized by(1.1)f(z)=1z+∑n=1∞anzn,
which are analytic in the punctured unit disk U={z:0<|z|<1}. For 0≤β, we denote by S*(β) and k(β), the subclasses of Σ consisting of all meromorphic functions, which are, respectively, starlike of order β and convex of order β in U.For functionsfj(z)(j=1;2) defined by(1.2)fj(z)=1z+∑n=1∞an,jzn,
we denote the Hadamard product (or convolution) of f1(z) and f2(z) by(1.3)(f1*f2)=1z+∑n=1∞an,1an,2zn.LetPn(ρ) be the class of functions p(z) analytic in U satisfying the properties p(0)=1 and(1.4)∫02π|Rp(z)-ρ1-ρ|dθ≤nπ,
where z=reiθ,n≥2, and 0≤ρ<1. This class has been introduced in [1]. We note that Pn(0)=Pn and P2(ρ)=P(ρ), see ([2, 3]), the class of analytic functions with positive real part greater than ρ and P2(0)=P, the class of functions with positive real part. From (1.6), we can write p∈Pn(ρ) as(1.5)p(z)=(n4+12)p1(z)+(12-n4)p2(z),
where pi(z)∈P(ρ),i=1,2 and z∈U.Let us define the functionϕ̃(α,β;z) by(1.6)ϕ̃(α,β;z)=1z+∑n=0∞|(α)n+1(β)n+1|zn,
for β≠0,-1,-2,…, and α∈ℂ/{0}, where (λ)n=λ(λ+1)n+1 is the Pochhammer symbol. We note that(1.7)ϕ̃(α,β;z)=1z2F1(1,α,β;z),
where(1.8)2F1(b,α,β;z)=∑n=0∞(b)n(α)n(β)nznn!
is the well-known Gaussian hypergeometric function.Let us put(1.9)qλ,μ(z)=1z+∑n=1∞(λn+1+λ)μzn,(λ>0,μ≥0).
Corresponding to the functions ϕ̃(α,β;z) and qλ,μ(z), and using the Hadamard product for f(z)∈Σ, we define a new linear operator L(α,β,λ,μ) on Σ by(1.10)L(α,β,λ,μ)f(z)=(f(z)*ϕ̃(α,β;z)*qλ,μ(z))=1z+∑n=1∞|(α)n+1(β)n+1|(λn+1+λ)μanzn.
The meromorphic functions with the generalized hypergeometric functions were considered recently by Dziok and Srivastava [4, 5], Liu [6], Liu and Srivastava [7–9], and Cho and Kim [10].For a functionf∈L(α,β,λ,μ)f(z), we define(1.11)Iα,β,λμ,0=L(α,β,λ,μ)f(z)
and for k=1,2,3,…,(1.12)Iα,β,λμ,kf(z)=z(Ik-1L(α,β,λ,μ)f(z))′+2z=1z+∑n=1∞nk|(α)n+1(β)n+1|(λn+1+λ)μanzn.
Note that if n=β,k=0, the operator Iα,n,λμ,0 has been introduced by Cho et al. [11] for μ∈ℕ0=ℕ∪0. It was known that the definition of the operator Iα,n,λμ,0 was motivated essentially by the Choi-Saigo-Srivastava operator [12] for analytic functions, which includes a simpler integral operator studied earlier by Noor [13] and others (cf. [14–16]). Note also the operator Iα,β0,k has been recently introduced and studied by Ghanim and Darus, [17], [18] and [19], respectively. To our best knowledge, the recent work regarding operator Iα,n,λμ,0 was charmingly studied by Piejko and Sokół [20]. Moreover, the operator Iα,β,λμ,k was defined and studied by Ghanim and Darus [21]. In the same direction, we will study the operator Iα,β,λμ,k given in (1.12).Now, it follows from (1.10) and (1.12) that(1.13)z(Iα,β,λμ,kf(z))′=αIα+1,β,λμ,kf(z)-(α+1)Iα,β,λμ,kf(z).In the present paper, we will use of the operatorIα,β,λμ,kf(z) and introduce some new classes of meromorphic functions.Definition 1.1.
Letf∈Σ.Then f∈Σn,λ*,μ,k(α,β,ρ) if and only if
(1.14)-(1-α)z2(Iα,β,λμ,kf(z))′-αz2(Iα,β,λμ,kf(z))′∈Pn(ρ),
where α>0,n≥2,0≤ρ<1, and z∈U.Definition 1.2.
Letf∈Σ.Then f∈Σn,λμ,k(α,β,ρ) if and only if
(1.15)(1-α)z(Iα,β,λμ,kf(z))′-αz(Iα,β,λμ,kf(z))′∈Pn(ρ),
where α>0,n≥2,0≤ρ<1, and z∈U.


## 2. Preliminary Results

Lemma 2.1 (see [22]).
Ifp(z) is analytic in  U with  p(0)=1, and if  α is a complex number satisfying  ℜ(α)≥0, then
(2.1)R{p(z)+αzp′(z)}>β(0≤β<1)
implies
(2.2)Rp(z)>β+(1-β)(2γ-1),
where γ is given by
(2.3)γ=γ(α)=∫01(1+tRα)-1dt,
which is an increasing function of ℜ(α) and 1/2≤γ<1. The estimate is sharp in the sense that the bound cannot be improved.Lemma 2.2 (see [23]).
Ifp(z) is analytic in  U,  p(0)=1, and  ℜp(z)>1/2,z∈U, then for any function  F analytic in  U, the function  p*F takes values in the convex hull of the image of  U under  F.Lemma 2.3 (see [24]).
Letp(z)=1+b1z+b2z2+⋯∈P(ρ). Then
(2.4)Rp(z)≥2ρ-1+2(1-ρ)1+|z|.


## 3. Main Results

Theorem 3.1.
Letf∈Σn,λ*,μ,k(α,β,ρ). Then
(3.1)-z2(Iα,β,λμ,kf(z))′∈Pn(ρ1),
where ρ1 is given by
(3.2)ρ1=ρ+(1-ρ)(2γ-1),(3.3)γ=∫01(1+tRα)-1dt.Proof.
Let(3.4)-z2(Iα,β,λμ,kf(z))′=p(z)=(n4+12)p1(z)+(12-n4)p2(z).
Then p(z) is analytic in U with p(0)=1. Applying the identity (1.13) in (3.4) and differentiating the resulting equation with respect to z, we have
(3.5)-(1-α)z2(Iα,β,λμ,kf(z))′-αz2(Iα,β,λμ,kf(z))′=[p(z)+αzp′(z)].
Since f∈Σn,λ*,μ,k(α,β,ρ), so {p(z)+αzp′(z)}∈Pn(ρ) for z∈U. This implies that
(3.6)R{pi(z)+αzpi′(z)}>ρ,i=1,2.
Using Lemma 2.1, we see that ℜ{pi(z)}>ρ1, where ρ1 is given by (3.2).
Consequently,p∈Pn(ρ1) for z∈U, and the proof is complete.Theorem 3.2.
Letf∈Σn,λ*,μ,k(0,β,ρ) for  z∈U. Then  f∈Σn,λ*,μ,k(α,β,ρ) for  |z|<ℜ(α), where
(3.7)R(α)=1α+1+α2.Proof.
Set(3.8)-z2(Iα,β,λμ,kf(z))′=(1-ρ)h(z)+ρ,h∈Pn.
Now proceeding as in Theorem 3.1, we have
(3.9)-(1-α)z2(Iα,β,λμ,kf(z))′-αz2(Iα,β,λμ,kf(z))′-ρ=(1-ρ){h(z)+αzh′(z)}=(1-ρ)[(n4+12){h1(z)+αzh1′(z)}+(12-n4){h2(z)+αzh2′(z)}],
where we have used (1.6) and h1,h2∈P,z∈U. Using the following well-known estimate [25]
(3.10)|zhi′(z)|≤2r1-r2R{hi(z)}(|z|=r<1,i=1,2),
we have
(3.11)R{hi(z)+αzhi′(z)}≥R{hi(z)+α|zhi′(z)|}≥Rhi(z){1-2αr1-r2}.
The right-hand side of this inequality is positive if r<R(α), where ℜ(α) is given by (3.7). Consequently, it follows from (3.9) that f∈Σn,λ*,μ,k(α,β,ρ) for |z|<R(α). Sharpness of this result follows by taking hi(z)=(1+z)/(1-z) in (3.9), i=1,2.Theorem 3.3.
Letf∈Σn,λ*,μ,k(0,β,ρ), and let
(3.12)Fδ(f(z))=δzδ+1∫0ztδf(t)dt(δ>0,z∈U).
Then
(3.13)-z2(Iα,β,λμ,kF(f(z)))′∈Pn(ρ2),
where ρ2 is given by
(3.14)ρ2=ρ+(1-ρ)(2γ1-1),(3.15)γ1=∫01(1+tR(1/δ))-1dt.Proof.
Setting(3.16)-z2(Iα,β,λμ,kF(f(z)))′=p(z)=(n4+12)p1(z)+(12-n4)p2(z),
then p(z) is analytic in U with p(0)=1. Using the following operator identity:
(3.17)z(Iα,β,λμ,kF(f(z)))′=δ(Iα,β,λμ,kF(f(z)))-(δ+1)(Iα,β,λμ,kF(f(z))),
in (3.16), and differentiating the resulting equation with respect to z, we find that
(3.18)-z2(Iα,β,λμ,kf(z))′={p(z)+1δzp′(z)}∈Pn(ρ).
Using Lemma 2.1, we see that -z2(Iα,β,λμ,kF(f(z)))′∈Pn(ρ2) for z∈U, where ρ2 is given by (3.14).
Hence, the proof is complete.Theorem 3.4.
Letϕ(z)∈Σ satisfy the following inequality:
(3.19)R(zϕ(z))>12,z∈U.
If f∈Σn,λμ,k(α,β,ρ), then ϕ*f∈Σn,λμ,k(α,β,ρ).
Proof.
LetG=ϕ*f. Then
(3.20)(1-α)z(Iα,β,λμ,kG(z))+αz(Iα,β,λμ,kG(z))=(1-α)z(Iα,β,λμ,k(ϕ*f)(z))+αz(Iα,β,λμ,k(ϕ*f)(z))=zϕ(z)*h(z),h∈Pn(ρ)=(n4+12)[(1-ρ){zϕ(z)*h1(z)+ρ}]+(12-n4)[(1-ρ){zϕ(z)*h2(z)+ρ}],h1,h2∈P.
Since ℜ(zϕ(z))>1/2 and by using Lemma 2.2, we can conclude that G=ϕ*f∈Σn,λμ,k(α,β,ρ).Theorem 3.5.
Letϕ(z)∈Σ satisfy the inequality (3.19) and  f∈Σn,λ*,μ,k(0,β,ρ). Then  ϕ*f∈Σn,λ*,μ,k(0,β,ρ).Proof.
We have(3.21)-z2(Iα,β,λμ,k(ϕ*f)(z))′=-z2(Iα,β,λμ,kf(z))′*zϕ(z),z∈U.
Now the remaining part of Theorem 3.5 follows by employing the techniques that we used in proving Theorem 3.4 above.Theorem 3.6.
Letf∈Σn,λμ,k(α,β,ρ3) and  g∈Σn,λμ,k(α,β,ρ4), and let  F=f*g. Then  F∈Σn(α,β,k,ρ5), where
(3.22)ρ5=1-4(1-ρ3)(1-ρ4)[1-1α∫01u(1/(1-α))-11+udu].
This result is sharp.Proof.
Sincef∈Σn,λμ,k(α,β,ρ3) and g∈Σn,λμ,k(α,β,ρ4), it follows that
(3.23)S(z)=(1-α)z(Iα,β,λμ,kf(z))+αz(Iα,β,λμ,kf(z))∈Pn(ρ3),S*(z)=(1-α)z(Iα,β,λμ,kg(z))+αz(Iα,β,λμ,kg(z))∈Pn(ρ4),
and so using identity (1.13) in the above equation, we have
(3.24)Iα,β,λμ,kf(z)=1αz-1-1/α∫0zt1/α-1S(t)dt,Iα,β,λμ,kg(z)=1αz-1-1/α∫0zt1/α-1S*(t)dt.
Using (3.24), we have
(3.25)Iα,β,λμ,kf(z)=1αz-1-1/α∫0zt1/α-1Q(t)dt,
where
(3.26)Q(z)=(n4+12)q1(z)+(12-n4)q2(z)=1αz-1-1/α∫0zt1/α-1(S*S*)(t)dt.
Now
(3.27)S(z)=(n4+12)s1(z)+(12-n4)s2(z),(3.28)S*(z)=(n4+12)s1*(z)+(12-n4)s2*(z),
where si∈P(ρ3) and si*∈P(ρ4),i=1,2. Since
(3.29)Pi*(z)=si*(z)-ρ42(1-ρ4)+12∈P(12),i=1,2,
we obtain that (si*pi*)(z)∈P(ρ3), by using the Herglots formula. Thus
(3.30)(si*si*)(z)∈P(ρ5)
with
(3.31)ρ5=1-2(1-ρ3)(1-ρ4).
Using (3.25), (3.26), (3.28), (3.31), and Lemma 2.3, we have
(3.32)Rqi(z)=1α∫01u1/α-1R{(si*si*)(uz)}du≥1α∫01u1/α-1(2ρ5-1+2(1-ρ5)1+u|z|)du≥1α∫01u1/α-1(2ρ5-1+2(1-ρ5)1+u)du=1-4(1-ρ3)(1-ρ4)[1-1α∫01u(1/(1-α))-11+udu].
From this, we conclude that F∈Σn,λμ,k(α,β,ρ5), where ρ5 is given by (3.22). We discuss the sharpness as follows: we take
(3.33)S(z)=(n4+12)1+(1-ρ3)z1-z+(12-n4)1-(1-ρ3)z1+z,S*(z)=(n4+12)1+(1-ρ4)z1-z+(12-n4)1-(1-ρ4)z1+z.
Since
(3.34)(1+(1-ρ3)z1-z)*(1+(1-ρ4)z1-z)=1-4(1-ρ3)(1-ρ4)+4(1-ρ3)(1-ρ4)1-z,
it follows from (3.26) that
(3.35)qi(z)=1α∫01u1/α-1{1-4(1-ρ3)(1-ρ4)+4(1-ρ3)(1-ρ4)1-z}du⟶1-4(1-ρ3)(1-ρ4){1-1α∫01u(1/(1-α))-11+udu}
as z→-1. This completes the proof.Note 1.
Some other works related to Cho-Kwon-Srivastava operator can also be referred to [26–29].



---

*Source: 272678-2012-05-06.xml*