# Certain Subclasses ofp-Valent Meromorphic

Authors: A. O. Mostafa; M. K. Aouf; A. Shamandy; E. A. Adwan
Journal: Journal of Complex Analysis
 (2013)

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## Abstract

We introduce two classes ofp-valent meromorphic functions associated
with a new operator and derive several interesting results for these classes.


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## Body

## 1. Introduction

LetΣp denote the class of functions of the form
(1)f(z)=z-p+∑n=1∞‍an-pzn-p(p∈ℕ={1,2,3,…}),
which are analytic and p-valent in the punctured unit disc U*={z∈ℂ:0<|z|<1}=U∖{0}. Let Pk(β) be the class of functions ϕ(z) analytic in U satisfying the properties h(0)=1 and(2)∫02π‍|ℜ{h(z)}-β1-β|dθ≤kπ,
where z=reiθ, k≥2, and 0≤β<1. The class Pk(β)was introduced by Padmanabhan and Parvatham [1]. For β=0, the class Pk(0)=Pk  was introduced by Pinchuk [2]. Also we note that P2(β)=P(β), where P(β) is the class of functions with positive real part greater than β and  P2(0)=P, where P is the class of functions with positive real part. From (2), we have h(z)∈Pk(β) if and only if there exists h1,h2∈P(β) such  that
(3)h(z)=(k4+12)h1(z)-(k4-12)h2(z)(z∈U).
It is known that the class Pk(β) is a convex set (see [3]).For functionsf(z)∈Σp given by (1) and g(z)∈Σp given by
(4)g(z)=z-p+∑n=1∞‍an-pzn-p(p∈ℕ),
the Hadamard product (or convolution) of f(z) and g(z) is defined by
(5)(f*g)(z)=z-p+∑n=1∞‍an-pbn-pzn-p=(g*f)(z).Aqlan et al. [4] defined the operator Qβ,pα:Σp→Σp by
(6)Qβ,pαf(z)={z-p+Γ(α+β)Γ(β)∑n=1∞‍Γ(n+β)Γ(n+β+α)an-pzn-p(α>0;β>-1;p∈ℕ;f∈Σp)f(z)(α=0;β>-1;p∈ℕ;f∈Σp).Mostafa [5] used Aqlan et al. operator and defined the following linear operator Hp,β,μα:Σp→Σp as follows.First put(7)Gβ,pα(z)=z-p+Γ(α+β)Γ(β)∑n=1∞‍Γ(n+β)Γ(n+β+α)zn-p(p∈ℕ)
and let Gβ,p,μα* be defined by
(8)Gβ,pα(z)*Gβ,p,μα*(z)=1zp(1-z)μ(μ>0;p∈ℕ).
Then
(9)Hp,β,μαf(z)=Gβ,pα*(z)*f(z)(f∈Σp).
Using (7) and (9), we have
(10)Hp,β,μαf(z)=z-p+Γ(β)Γ(α+β)∑n=1∞‍Γ(n+β+α)(μ)nΓ(n+β)(1)nan-pzn-p,
where (ν)n denotes the Pochhammer symbol given by
(11)(ν)n=Γ(ν+n)Γ(ν)={1(n=0)ν(ν+1)⋯(ν+n-1)(n∈ℕ).
It is readily verified from (10) that (see [5])
(12)z(Hp,β,μαf(z))′=(α+β)Hp,β,μα+1f(z)-(α+β+p)Hp,β,μαf(z),(13)z(Hp,β,μαf(z))′=μHp,β,μ+1αf(z)-(μ+p)Hp,β,μαf(z).It is noticed that by puttingμ=1 in (10), we obtain the operator
(14)Hp,β,1αf(z)=Hp,βαf(z)=z-p+Γ(β)Γ(α+β)∑n=1∞‍Γ(n+α+β)Γ(n+β)an-pzn-p.Now, by using the linear operatorHp,β,μαf(z), we introduce classes of p-valent Bazilevic functions of Σp as follows.Definition 1.
A functionf(z)∈Σp is said to be in the class Kp(α,β,μ,δ,ρ,γ) if it satisfies the following condition:
(15)[(Hp,β,μ+1αf(z)f(z)Hp,β,μαf(z))(1-γ)(zpHp,β,μαf(z))δ+γ(Hp,β,μα+1f(z)f(z)Hp,β,μαf(z))(zpHp,β,μαf(z))δ]∈𝒫k(ρ)(α,μ>0;β>-1;k≥2;γ≥0,δ>0;0≤ρ<1;p∈ℕ;z∈U).Definition 2.
A functionf(z)∈Σp is said to be in the class Mp(α,β,μ,δ,ρ,γ) if it satisfies the following condition:
(16)[(Hp,β,μ+1αf(z)f(z)Hp,β,μαf(z))(1-γ)(zpHp,β,μαf(z))δ+γ(Hp,β,μ+1αf(z)f(z)Hp,β,μαf(z))(zpHp,β,μαf(z))δ]∈𝒫k(ρ)(α,μ>0;β>-1;k≥2;γ≥0,δ>0;0≤ρ<1;p∈ℕ;z∈U).In this paper, we investigate several properties of the classesKp(α,β,μ,δ,ρ,γ) and Mp(α,β,μ,δ,ρ,γ).


## 2. Main Results

Unless otherwise mentioned, we assume throughout this paper thatk≥2, α,δ,γ,μ>0, β>-1, α+β>0, 0≤ρ<1, p∈ℕ and all powers are understood as principle values.To prove our results we need the following lemma.Lemma 3 (see [6]).
Letu=u1+iu2, v=v1+iv2 and Φ(u,v) be a complex-valued function satisfying the conditions:(i)
Φ(u,v) is continuous in a domain D∈ℂ2.(ii)
(0,1)∈D and Φ(1,0)>0.(iii)
ℜ{Φ(iu2,v1)}>0 whenever (iu2,v1)∈D and v1≤-(1/2)(1+u22).
Ifh(z)=1+cnzn+cn+1zn+1+⋯ is analytic in U such that (h(z),zh′(z))∈D and ℜ{Φ(h(z),zh′(z))}>0 for z∈U, then ℜ{h(z)}>0 in U.Employing the technique used by Noor and Muhammad [7] and Aouf and Seoudy [8] for multivalent functions, we prove the following theorems.Theorem 4.
Iff(z)∈Kp(α,β,μ,δ,ρ,γ), then
(17)(zpHp,β,μαf(z))δ∈𝒫k(η),
where η is given by
(18)η=2δρ(α+β)+γ2δ(α+β)+γ.Proof.
Setting(19)(zpHp,β,μαf(z))δ=H(z)=(1-η)h(z)+η=(k4+12){(1-η)h1(z)+η}-(k4-12){(1-η)h2(z)+η},
where hi(z)(i=1,2) are analytic in U with hi(0)=1(i=1,2), and h(z) is given by (3). Differentiating both sides of (19) with respect to z and using (12) in  the resulting equation, we obtain
(20)[(Hp,β,μ+1αf(z)f(z)Hp,β,μαf(z))(1-γ)(zpHp,β,μαf(z))δ+γ(Hp,β,μα+1f(z)f(z)Hp,β,μαf(z))(zpHp,β,μαf(z))δ]={(Hp,β,μ+1αf(z)f(z)Hp,β,μαf(z))γ(1-η)zh′(z)δ(α+β)(1-η)h(z)+η+γ(1-η)zh′(z)δ(α+β)}∈Pk(ρ)(z∈U),
which implies that
(21)11-ρ{γ(1-η)zhi′(z)δ(α+β)(Hp,β,μ+1αf(z)f(z)Hp,β,μαf(z))η-ρ+(1-η)hi(z)+γ(1-η)zhi′(z)δ(α+β)}∈P(z∈U;i=1,2).
We form the functional Φ(u,v) by choosing u=hi(z), v=zhi′(z), that is,
(22)Φ(u,v)=η-ρ+(1-η)u+γ(1-η)vδ(α+β).
Clearly, the first two conditions of Lemma 3 are satisfied. Now, we verify the condition (iii) as follows:
(23)ℜ{Φ(iu2,v1)}=η-ρ+ℜ{γ(1-η)v1δ(α+β)}≤η-ρ-γ(1-η)(1+u22)2δ(α+β)=A+Bu222C,
where
(24)A=2δ(α+β)(η-ρ)-γ(1-η),B=-γ(1-η),C=2δ(α+β).
We note that ℜ{Φ(iu2,v1)}<0 if and only if A=0, B<0. From η given by (18), we have 0≤η<1, A=0, and B<0.  Therefore, applying Lemma 3, we have hi(z)∈P(i=1,2) and consequently H(z)∈Pk(η)for z∈U. This completes the proof of Theorem 4.Similarly, we can prove the following theorem for the classMp(α,β,μ,δ,ρ,γ).Theorem 5.
Iff(z)∈Mp(α,β,μ,δ,ρ,γ), then
(25)(zpHp,β,μαf(z))δ∈𝒫k(χ),
where χ is given by
(26)χ=2δμρ+γ2δμ+γ.Theorem 6.
Iff(z)∈Kp(α,β,μ,δ,ρ,γ), then
(27)(zpHp,β,μαf(z))δ/2∈𝒫k(ζ),
where ζ is given by
(28)ζ=γ+γ2+4ρδ(α+β)[δ(α+β)+γ]2[δ(α+β)+γ].Proof.
Letf(z)∈Kp(α,β,μ,δ,ρ,γ) and
(29)(zpHp,β,μαf(z))δ=M(z)=[(1-ζ)h(z)+ζ]2=(k4+12)[(1-ζ)h1(z)+ζ]2-(k4-12)[(1-ζ)h2(z)+ζ]2,
where hi(z)(i=1,2) are analytic in U with hi(0)=1(i=1,2) and h(z) is given by (3). Differentiating both sides of (29) with respect to z and using (12) in  the resulting equation, we obtain
(30)[(Hp,β,μα+1f(z)f(z)Hp,β,μαf(z))(1-γ)(zpHp,β,μαf(z))δ+γ(Hp,β,μα+1f(z)f(z)Hp,β,μαf(z))(zpHp,β,μαf(z))δ]={2γ(1-ζ)zh′(z)δ(α+β)[(1-ζ)h(z)+ζ]2+[(1-ζ)h(z)+ζ]×2γ(1-ζ)zh′(z)δ(α+β)}∈Pk(ρ)(z∈U),
which implies that
(31)11-ρ{(Hp,β,μα+1f(z)f(z)Hp,β,μαf(z))2γ(1-ζ)zh′(z)δ(α+β)[(1-ζ)h(z)+ζ]2+[(1-ζ)h(z)+ζ]×2γ(1-ζ)zh′(z)δ(α+β)-ρ}∈P(i=1,2).
We form the functional Φ(u,v) by choosing u=hi(z), v=zhi′(z), that is,
(32)Φ(u,v)=[(1-ζ)u+ζ]2+[(1-ζ)u+ζ]2γ(1-ζ)vδ(α+β)-ρ.
Clearly, the conditions (i) and (ii) of Lemma 3 are satisfied. Now, we verify the condition (iii) as follows:
(33)ℜ{Φ(iu2,v1)}=ζ2-(1-ζ)2u22+2γ(1-ζ)v1δ(α+β)-ρ≤ζ2-ρ-(1-ζ)2u22-γζ(1-ζ)(1+u22)δ(α+β)=A+Bu222C,
where
(34)A=ζ2δ(α+β)-γζ(1-ζ)-ρδ(α+β),B=-[(1-ζ)2+γζ(1-ζ)],C=δ(α+β)2.
We note that ℜ{Φ(iu2,v1)}<0 if and only if A=0, B<0. From ζ as given by (28), we have 0≤ζ<1, A=0, and B<0. Therefore, applying Lemma 3, we have hi(z)∈P(i=1,2) and consequently M(z)∈Pk(ζ)for z∈U. This completes the proof of Theorem 6.Similarly, we can prove the following theorem for the classMp(α,β,μ,δ,ρ,γ).Theorem 7.
Iff(z)∈Mp(α,β,μ,δ,ρ,γ), then
(35)(zpHp,β,μαf(z))δ/2∈𝒫k(λ),
where λ is given by
(36)λ=γ+γ2+ρδμ(2γ+δμ)(2γ+δμ).Remark 8.
Puttingμ=1, in Theorems 4 and 6, we obtain the corresponding results for the operator Hp,βα defined in (14).



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*Source: 851318-2013-04-03.xml*