# On Certain Sufficiency Criteria forp-Valent Meromorphic Spiralike Functions

Authors: Muhammad Arif
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We consider some subclasses of meromorphic multivalent functions and obtain certain simple sufficiency criteria for the functions belonging to these classes. We also study the mapping properties of these classes under an integral operator.


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## Body

## 1. Introduction

Let∑‍(p,n) denote the class of functions f(z) of the form
(1.1)f(z)=z-p+∑k=n∞‍akzk-p+1(p∈ℕ),
which are analytic and p-valent in the punctured unit disk 𝕌={z:0<|z|<1}. Also let ∑λ*‍(p,n,α) and ∑cλ‍(p,n,α) denote the subclasses of ∑‍(p,n) consisting of all functions f(z) which are defined, respectively, by
(1.2)Re(-eiλzf′(z)f(z))>αcosλ(z∈𝕌),Re(-eiλ(zf′(z))′f′(z))>αcosλ(z∈𝕌).We note that forλ=0 and n=1, the above classes reduce to the well-known subclasses of ∑‍(p) consisting of meromorphic multivalent functions which are, respectively, starlike and convex of order α (0≤α<p). For the detail on the subject of meromorphic spiral-like functions and related topics, we refer the work of Liu and Srivastava [1], Goyal and Prajapat [2], Raina and Srivastava [3], Xu and Yang [4], and Spacek [5] and Robertson [6].Analogous to the subclass of∑‍(1,1) for meromorphic univalent functions studied by Wang et al. [7] and Nehari and Netanyahu [8], we define a subclass ∑Nλ‍(p,n,α) of ∑‍(p,n) consisting of functions f(z) satisfying
(1.3)-Reeiλ((zf′(z))′f′(z))<αcosλ(α>p,z∈𝕌).For more details of the above classes see also [9, 10].Motivated from the work of Frasin [11], we introduce the following integral operator of multivalent meromorphic functions ∑‍(p)(1.4)Hm,p(z)=1zp+1∫0z‍∏j=1m‍(upfj(u))αjdu.Forp=1, (1.4) reduces to the integral operator introduced and studied by Mohammed and Darus [12, 13]. Similar integral operators for different classes of analytic, univalent, and multivalent functons in the open unit disk are studied by various authors, see [14–19].In this paper, first, we find sufficient conditions for the classes∑λ*‍(p,n,α) and ∑cλ‍(p,n,α) and then study some mapping properties of the integral operator given by (1.4).We will assume throughout our discussion, unless otherwise stated, thatλ is real with |λ|<π/2, 0≤α<p, p,n∈ℕ, αj>0 for j∈{1,…m}.To obtain our main results, we need the following Lemmas.Lemma 1.1 (see [20]).
Ifq(z)∈∑‍(1,n) with n≥1 and satisfies the condition
(1.5)|z2q′(z)+1|<nn2+1(z∈𝕌),
then
(1.6)q(z)∈∑0*‍(1,n,0).Lemma 1.2 (see [21]).
LetΩ be a set in the complex plane ℂ and suppose that Ψ is a mapping from ℂ2×𝕌 to ℂ which satisfies Ψ(ix,y,z)∉Ω for z∈𝕌, and for all real x,y such that y≤(-n/2)(1+x2). If h(z)=1+cnzn+⋯ is analytic in 𝕌 and Ψ(h(z),zh′(z),z)∈Ω for all z∈𝕌, then Reh(z)>0.


## 2. Some Properties of the Classes∑λ*‍(p,n,α) and ∑cλ‍(p,n,α)

Theorem 2.1.
Iff(z)∈∑‍(p,n) satisfies
(2.1)|(zpf(z))eiλ/(p-α)cosλ{eiλzf′(z)f(z)+αcosλ+ipsinλ}+(p-α)cosλ|<nn2+1(p-α)cosλ(z∈𝕌),
then f(z)∈∑λ*‍(p,n,α).Proof.
Let us set a functionh(z) by
(2.2)h(z)=1z(zpf(z))eiλ/(p-α)cosλ=1z+eiλan(p-α)cosλzn+⋯
for f(z)∈∑‍(p,n). Then clearly (2.2) shows that h(z)∈∑‍(1,n).
Differentiating (2.2) logarithmically, we have
(2.3)h′(z)h(z)=eiλ(p-α)cosλ[f′(z)f(z)+pz]-1z
which gives
(2.4)|z2h′(z)+1|=|(zpf(z))eiλ/(p-α)cosλ1(p-α)cosλ{eiλzf′(z)f(z)+αcosλ+ipsinλ}+1|.
Thus using (2.1), we have
(2.5)|z2h′(z)+1|≤nn2+1(z∈𝕌).
Hence, using Lemma1.1, we have h(z)∑0*(1,n,0).
From (2.3), we can write
(2.6)zh′(z)h(z)=1(p-α)cosλ[eiλzf′(z)f(z)+(αcosλ+ipsinλ)].
Sinceh(z)∈∑0*‍(1,n,0), it implies that Re(-zh′(z)/h(z))>0. Therefore, we get
(2.7)1(p-α)cosλ[Re(-eiλzf′(z)f(z))-αcosλ]=Re(-zh′(z)h(z))>0
or
(2.8)Re(-eiλzf′(z)f(z))>αcosλ,
and this implies that f(z)∈∑λ*‍(p,n,α).
If we takeλ=0, we obtain the following result.Corollary 2.2.
Iff(z)∈∑‍(p,n) satisfies
(2.9)|(zpf(z))1/(p-α){eiλzf′(z)f(z)+α}+(p-α)|<12(p-α)(z∈𝕌),
then f(z)∈∑*(p,n,α).Theorem 2.3.
Iff(z)∈∑‍(p,n) satisfies
(2.10)|(zp+1f′(z)-p)eiλ/(p-α)cosλ{eiλ(zf′′(z)f′(z)+1)+αcosλ+ipsinλ}+(p-α)cosλ|<(n+1)(p-α)cosλ(n+1)2+1(z∈𝕌),
then f(z)∈∑cλ‍(p,n,α).Proof.
Let us set(2.11)h(z)=-∫0z‍1t2(-tp+1f′(t)p)eiλ/(p-α)cosλdt=1z+n-p+1npeiλan(p-α)cosλzn+⋯.
Also let(2.12)g(z)=-zh′(z)=1z(-zp+1f′(z)p)eiλ/(p-α)cosλ=1z+p-n-1peiλan(p-α)cosλzn+⋯.
Then clearlyh(z) and g(z)∈∑‍(1,n). Now
(2.13)g(z)=1z(-zp+1f′(z)p)eiλ/(p-α)cosλ.
Differentiating logarithmically and then simple computation gives us(2.14)|z2g′(z)+1|=|(zp+1f′(z)-p)e^iλ/(p-α)cosλ1(p-α)cosλ{eiλ(zf′′(z)f′(z)+1)+αcosλ+ipsinλ}+1|<nn2+1.
Therefore, by using Lemma1.1, we have
(2.15)g(z)=-zh′(z)∈∑0*‍(1,n,0)
which implies that h(z)∈∑c0‍(1,n,0). Since
(2.16)1+zh′′(z)h′(z)=eiλ(p-α)cosλ{zf′′(z)f′(z)+(p+1)}-1,
therefore
(2.17)Re(1+zh′′(z)h′(z))=1(p-α)cosλRe{eiλ(1+zf′′(z)f′(z))+peiλ-(p-α)cosλ}=1(p-α)cosλ{Reeiλ(1+zf′′(z)f′(z))+αcosλ}.
Since h(z)∈∑c0‍(1,n,0), so
(2.18)-1(p-α)cosλ{Reeiλ(1+zf′′(z)f′(z))+αcosλ}>0,
or
(2.19)-Reeiλ(1+zf′′(z)f′(z))>αcosλ.
It follows thatf(z)∈∑cλ‍(p,n,α).Theorem 2.4.
Iff(z)∈∑‍(p,n) satisfies
(2.20)Re(eiλzf′(z)f(z))(αzf′′(z)f′(z)-1)>M24L+N(z∈𝕌),
thenf(z)∈∑λ*‍(p,n,β), where 0≤α≤1, 0≤β<p and
(2.21)L=α(p-β)[n2+(β-p)cos2λ]cosλ,M=α(β-p)(1-βcosλ)sin2λcosλ,N=α(β2cos2λ+sin2λ-n2(β-p))cosλ+(1+α)βcosλ.Proof.
Let us set(2.22)eiλzf′(z)f(z)=[(β-p)h(z)-β]cosλ-isinλ.
Thenh(z) is analytic in 𝕌 with p(0)=1.
Taking logarithmic differentiation of (2.22) and then by simple computation, we obtain
(2.23)eiλzf′(z)f(z)(αzf′′(z)f′(z)-1)=Azh′(z)+Bh2(z)+Ch(z)+D=Ψ(h(z),zh′(z),z)
with
(2.24)A=(β-p)αcosλ,B=αe-iλ(β-p)2cos2λ,C=(p-β)[αe-iλ(2βcos2λ+isin2λ)+(1+α)e-iλcosλ],D=αe-iλ(β2cos2λ-sin2λ+iβsin2λ)+(1+α)(βcosλ-isinλ).
Now for all realx and y satisfying y≤-(n/2)(1+x2), we have
(2.25)Ψ(ix,y,z)=Ay-Bx2+C(ix)+D.
Reputing the values ofA, B, C, and D and then taking real part, we obtain
(2.26)ReΨ(ix,y,z)≤-Lx2+Mx+N=-(Lx-M2L)2+M24L+N<M24L+N,
where L, M, and N are given in (2.21).
LetΩ={w:Rew>(M2/4L)+N}. Then Ψ(h(z),zh′(z),z)∈Ω and Ψ(ix,y,z)∉Ω, for all real x and y satisfying y≤-(n/2)(1+x2), z∈𝕌. By using Lemma 1.2, we have Reh(z)>0, that is f(z)∈∑λ*‍(p,n,β).
If we putλ=0, we obtain the following result.Corollary 2.5.
Iff(z)∈∑‍(p,n) satisfies
(2.27)Rezf′(z)f(z)(αzf′′(z)f′(z)-1)>α(β2-n2(β-p))+(1+α)β(z∈𝕌),
then f(z)∑*(p,n,β), where 0≤α≤1, 0≤β<p.Theorem 2.6.
Forj∈{1,…m}, let fj(z)∈∑‍(p,n) and satisfy (2.9). If
(2.28)∑j=1m‍αj<p+1p-β,
then Hm,p(z)∈∑N‍(p,n,ζ), where ζ>p.Proof.
From (1.4), we obtain
(2.29)zp+1Hm,p′(z)+(p+1)zpHm,p(z)=∏j=1m‍(zpfj(z))αj.
Differentiating again logarithmically and then by simple computation, we get(2.30)zHm,p′′(z)Hm,p′(z)+1+(p2-1)Hm,p(z)zHm,p′(z)+2p=[1+(p+1)Hm,p(z)zHm,p′(z)][∑j=1m‍αj(zfj′(z)fj(z)+p)-1],
or, equivalently we can write
(2.31)-(zHm,p′′(z)Hm,p′(z)+1)=Hm,p(z)zHm,p′(z)[(p+1)(∑j=1m‍αj(-zfj′(z)fj(z)-p)+1)+(p2-1)]+∑j=1m‍αj(-zfj′(z)fj(z)-p)+(1+2p).
Now taking real part on both sides, we obtain
(2.32)-Re(zHm,p′′(z)Hm,p′(z)+1)=ReHm,p(z)zHm,p′(z)[(p+1)(∑j=1m‍αj(-zfj′(z)fj(z)-p)+1)+(p2-1)]+∑j=1m‍αj(-Rezfj′(z)fj(z)-p)+(1+2p).
This further implies that
(2.33)-Re(zHm,p′′(z)Hm,p′(z)+1)≤|Hm,p(z)zHm,p′(z)[(p+1)(∑j=1m‍αj(-zfj′(z)fj(z)-p)+1)+(p2-1)]|+∑j=1m‍αj(-Rezfj′(z)fj(z)-p)+(1+2p).
Let(2.34)ζ=|Hm,p(z)zHm,p′(z)[(p+1)(∑j=1m‍αj(-zfj′(z)fj(z)-p)+1)+(p2-1)]|+[∑j=1m‍αj(-Rezfj′(z)fj(z)-p)+(1+2p)].
Clearly we have
(2.35)ζ>[∑j=1m‍αj(-Rezfj′(z)fj(z)-p)+(1+2p)].
Then by using (2.28) and Corollary 2.2, we obtain
(2.36)ζ>∑j=1m‍αj(β-p)+(1+2p)>p.
Therefore Hm,p(z)∈∑N‍(p,n,ζ) with ζ>p.
Making use of (2.27) and Corollary 2.5, one can prove the following result.Theorem 2.7.
Forj∈{1,…m}, let fj(z)∈∑‍(p,n) and satisfy (2.27). If
(2.37)∑j=1m‍αj<p+1p-β,
then Hm,p(z)∈∑N‍(p,n,ζ), where ζ>p.



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*Source: 837913-2012-09-12.xml*