# On a New Criterion for Meromorphic Starlike Functions

Authors: Lei Shi; Zhi-Gang Wang
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

The main purpose of this paper is to derive a new criterion for meromorphic starlike functions of orderα.


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## Body

## 1. Introduction and Preliminaries

LetΣn denote the class of functions of the form
(1)f(z)=1z+∑k=n∞‍ak-1zk-1(n∈ℕ:={1,2,…}),
which areanalytic in thepunctured open unit disk
(2)𝕌*:={z:z∈ℂand0<|z|<1}=:𝕌∖{0}.A functionf∈Σn is said to be in the class ℳ𝒮n*(α) ofmeromorphic starlike functions of orderα if it satisfies the condition
(3)ℜ(zf′(z)f(z))<-α(z∈𝕌;0≦α<1).
For simplicity, we write ℳ𝒮n*(0)=:ℳ𝒮n*.For two functionsf and g, analytic in 𝕌, we say that the function f is subordinate to g in 𝕌 and write
(4)f(z)≺g(z)(z∈𝕌),
if there exists a Schwarz function ω, which is analytic in 𝕌 with
(5)ω(0)=0,|ω(z)|<1(z∈𝕌),
such that
(6)f(z)=g(ω(z))(z∈𝕌).
Indeed, it is known that
(7)f(z)≺g(z)(z∈𝕌)⟹f(0)=g(0),f(𝕌)⊂g(𝕌).
Furthermore, if the function g is univalent in 𝕌, then we have the following equivalence:
(8)f(z)≺g(z)(z∈𝕌)⟺f(0)=g(0),f(𝕌)⊂g(𝕌).In a recent paper, Miller et al. [1] proved the following result.Theorem A.
Letn∈ℕ, 0≦λ≦1, and
(9)M0(λ,n)=n+1-λ(n+1-λ)2+λ2+1-λ.
If f∈Σn satisfies the condition
(10)|z2f′(z)+(1-λ)zf(z)+λ|<M0(λ,n)(z∈𝕌),
then f∈ℳ𝒮n*.More recently, Catas [2] improved Theorem A as follows.Theorem B.
Letn∈ℕ, 0≦λ<1, and
(11)M(λ,n)=max⁡{M0(λ,n),M1(λ,n)},
where M0(λ,n) is given by (9) and (12)M1(λ,n)=2(n+1-λ)(1-λ)(1-λ)(n-1)+(n+1-λ)2(1-λ)+[(n-1)(1-λ)]2. If f∈Σn satisfies the condition
(13)|z2f′(z)+(1-λ)zf(z)+λ|<M(λ,n)(z∈𝕌),
then f∈ℳ𝒮n*.In this paper, we aim at finding the conditions for starlikeness of the expression|z2f′(z)+λzf(z)+1-λ| for λ>1.For some recent investigations of meromorphic functions, see, for example, the works of [3–12] and the references cited therein.In order to prove our main results, we require the following subordination result due to Hallenbeck and Ruscheweyh [13].Lemma 1.
Letϕ be a convex function with ϕ(0)=1, and let γ≠0 be a complex number with ℜ(γ)≧0. If a function
(14)𝔭(z)=1+𝔭nzn+𝔭n+1zn+1+⋯
satisfies the condition
(15)𝔭(z)+1γz𝔭′(z)≺ϕ(z),
then
(16)𝔭(z)≺χ(z):=γnzγ/n∫0z‍ϕ(t)t(γ/n)-1dt≺ϕ(z).


## 2. Main Results

We begin by stating the following result.Theorem 2.
Letn∈ℕ, λ>1, and 0≦α<1. If f∈Σn satisfies the inequality
(17)|z2f′(z)+λzf(z)+1-λ|<M,
where
(18)M:=M(λ,α,n)=(1-α)(λ+n-1)λ-α+(1-λ)2+(λ+n-1)2,
then f∈ℳ𝒮n*(α).Proof.
Suppose that(19)q(z):=zf(z)(z∈𝕌).
It follows from (19) that
(20)zq′(z)=zf(z)+z2f′(z).
By combining (17), (19), and (20), we easily get
(21)|q(z)+1λ-1zq′(z)-1|<Mλ-1,
or equivalently
(22)q(z)+1λ-1zq′(z)≺1+Mλ-1z.
An application of Lemma 1 yields
(23)q(z)≺λ-1nz(λ-1)/n∫0z‍(1+Mλ-1t)t[(λ-1)/n]-1dt=1+Mλ+n-1z.
The subordination (23) is equivalent to
(24)|q(z)-1|<Mλ+n-1=:N.
From (18) and (24), we know that
(25)N<1-αλ-α<1.
We suppose that(26)-zf′(z)f(z):=(1-α)p(z)+α.
By virtue of (19) and (26), we get
(27)z2f′(z)=-q(z)[(1-α)p(z)+α],
which implies that (17) can be written as
(28)|q(z)[(1-α)p(z)+α-λ]+λ-1|<M=(λ+n-1)N.
We now only need to show that (28) implies ℜ(p(z))>0 in 𝕌. Indeed, if this is false, since p(0)=1, then there exists a point z0∈𝕌 such that p(z0)=βi, where β is a real number. Thus, in order to show that (28) implies ℜ(p(z))>0 in 𝕌, it suffices to obtain the contradiction from the inequality
(29)|q(z0)[(1-α)βi+α-λ]+λ-1|≧(λ+n-1)N(β∈ℝ).
By setting
(30)q(z0)=u+iv(u,v∈ℝ),
we have
(31)E=|q(z0)[(1-α)βi+α-λ]+λ-1|2=(u2+v2)[(1-α)2β2+(α-λ)2]-2(1-λ)ℜ((u+iv)[(1-α)βi+α-λ])+(1-λ)2=(u2+v2)(1-α)2β2+2(1-λ)(1-α)βv+|(u+iv)(α-λ)-(1-λ)|2.
By means of (24), we obtain
(32)|(u+iv)(α-λ)-(1-λ)|=|(u+iv)(α-λ)-(α-λ)+α-λ-1+λ|=|(α-λ)(u+iv-1)-(1-α)|≧1-α-(λ-α)|u+iv-1|≧1-α-(λ-α)N.
It follows from (31) and (32) that
(33)E≧(u2+v2)(1-α)2β2+2(1-λ)(1-α)βv+[1-α-(λ-α)N]2.
We now set(34)F(β)≔E-M2≧(u2+v2)(1-α)2β2+2(1-λ)(1-α)vβ+[1-α-(λ-α)N]2-(λ+n-1)2N2.
If F(β)≧0, then (29) holds true. Since (u2+v2)(1-α)2>0, the inequality F(β)≧0 holds if the discriminant Δ≦0; that is,
(35)Δ=(1-α)2×{(1-λ)2v2-(u2+v2)×[(1-α-(λ-α)N)2-(λ+n-1)2N2]}≦0,
and the last inequality is equivalent to
(36)v2[(1-λ)2-(1-α-(λ-α)N)2+(λ+n-1)2N2]≦u2[(1-α-(λ-α)N)2-(λ+n-1)2N2].
Furthermore, in view of (24) and (36), after a geometric argument, we deduce that
(37)v2u2≦N21-N2≦(1-α-(λ-α)N)2-(λ+n-1)2N2(1-λ)2-(1-α-(λ-α)N)2+(λ+n-1)2N2.
It follows from (37) that Δ≦0, which implies that F(β)≧0. But this contradicts (28). Therefore, we know that ℜ(p(z))>0 in 𝕌. By virtue of (26), we conclude that
(38)ℜ(zf′(z)f(z))<-ℜ((1-α)p(z)+α)<-α.
This evidently completes the proof of Theorem 2.Takingα=0 in Theorem 2, we obtain the following result.Corollary 3.
Letn∈ℕ and λ>1. If f∈Σn satisfies the inequality
(39)|z2f′(z)+λzf(z)+1-λ|<(λ+n-1)λ+(1-λ)2+(λ+n-1)2,
then f∈ℳ𝒮n*.



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*Source: 346162-2014-03-19.xml*