# A New Class of Meromorphic Functions Associated with Spirallike Functions

Authors: Lei Shi; Zhi-Gang Wang; Jing-Ping Yi
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We introduce a new class of meromorphic functions associated with spirallike functions. Such results as subordination property, integral representation, convolution property, and coefficient inequalities are proved.


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## Body

## 1. Introduction

LetΣ denote the class of functions f of the form
(1.1)f(z)=1z+∑k=0∞‍akzk,
which are analytic in the punctured open unit disk
(1.2)𝕌*:={z:z∈ℂ,0<|z|<1}=:𝕌∖{0}.Let𝒫 denote the class of functions p given by
(1.3)p(z)=1+∑k=1∞‍pkzk(z∈𝕌),
which are analytic in 𝕌 and satisfy the condition
(1.4)Re(p(z))>0(z∈𝕌).Letf,g∈Σ, where f is given by (1.1) and g is defined by
(1.5)g(z)=1z+∑k=0∞‍bkzk,
then the Hadamard product (or convolution) f*g is defined by
(1.6)(f*g)(z):=1z+∑k=0∞‍akbkzk=:(g*f)(z).For two functionsf and g, analytic in 𝕌, we say that the function f is subordinate to g in 𝕌 and write
(1.7)f(z)≺g(z)(z∈𝕌),
if there exists a Schwarz function ω, which is analytic in 𝕌 with
(1.8)ω(0)=0,|ω(z)|<1(z∈𝕌),
such that
(1.9)f(z)=g(ω(z))(z∈𝕌).
Indeed, it is known that
(1.10)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:
(1.11)f(z)≺g(z)(z∈𝕌)⟺f(0)=g(0),f(𝕌)⊂g(𝕌).A functionf∈Σ is said to be in the class ℳ𝒮*(β) of meromorphic starlike functions of orderβ if it satisfies the inequality
(1.12)Re(zf′(z)f(z))<-β(z∈𝕌;0≦β<1).For the real numberβ(0<β<1), we know that
(1.13)|f(z)zf′(z)+12β|<12β⟺Re(zf′(z)f(z))<-β.
If the complex number α satisfies the condition
(1.14)|α-12|<12,
it can be easily verified that
(1.15)|f(z)zf′(z)+12α|<12|α|⟺Re(-1αzf′(z)f(z))>1.We now introduce and investigate the following class of meromorphic functions.Definition 1.1.
A functionf∈Σ is said to be in the class ℳ𝒮α if it satisfies the inequality
(1.16)Re(-1αzf′(z)f(z))>1(z∈𝕌;|α-12|<12).Remark 1.2.
For0<α<1, the class ℳ𝒮α is the familiar class of meromorphic starlike functions of order α.Remark 1.3.
Ifα=|α|eiψ(-π/2<ψ<π/2), then the condition (1.16) is equivalent to
(1.17)Re(e-iψzf′(z)f(z))<-|α|(z∈𝕌),
which implies that f belongs to the class of meromorphic spirallike functions. Thus, the class of meromorphic spirallike functions is a special case of the class ℳ𝒮α.
For some recent investigations on spirallike functions and related functions, see, for example, the earlier works [1–9] and the references cited in each of these earlier investigations.Remark 1.4.
The function(1.18)f(z)=z-1(1-z)2α[Re(1/α)-1](z∈𝕌*;|α-12|<12)
belongs to the class ℳ𝒮α.
It is clear that(1.19)Re(1α)>1(|α-12|<12).
Then, for the function f given by (1.18), we know that
(1.20)Re(-1αzf′(z)f(z))=Re(1α+2[Re(1α)-1]z1-z)>Re(1α)-Re(1α)+1=1,
which implies that f∈ℳ𝒮α.
In this paper, we aim at deriving the subordination property, integral representation, convolution property, and coefficient inequalities of the function classℳ𝒮α.


## 2. Preliminary Results

In order to derive our main results, we need the following lemmas.Lemma 2.1.
Letλ be a complex number. Suppose also that the sequence {Ak}k=0∞ is defined by
(2.1)A0=2λ,Ak+1=2λk+2(1+∑l=0k‍Al)(k∈ℕ0:=ℕ∪{0}).
Then
(2.2)Ak=1(k+1)!∏j=0k‍(2λ+j)(k∈ℕ0).Proof.
From (2.1), we know that
(2.3)(k+2)Ak+1=2λ(1+∑l=0k‍Al),(k+1)Ak=2λ(1+∑l=0k-1‍Al).
By virtue of (2.3), we find that
(2.4)Ak+1Ak=k+1+2λk+2(k∈ℕ0).
Thus, for k≧1, we deduce from (2.4) that
(2.5)Ak=AkAk-1⋯A3A2·A2A1·A1A0·A0=1(k+1)!∏j=0k‍(2λ+j).
By virtue of (2.1) and (2.5), we get the desired assertion (2.2) of Lemma 2.1.Lemma 2.2 (Jack's Lemma [10]).
Letϕ be a nonconstant regular function in 𝕌. If |ϕ| attains its maximum value on the circle |z|=r<1 at z0, then
(2.6)z0ϕ′(z0)=tϕ(z0),
for some real number t(t≧1).


## 3. Main Results

We begin by deriving the following subordination property of functions belonging to the classℳ𝒮α.Theorem 3.1.
A functionf∈ℳ𝒮α if and only if
(3.1)-zf′(z)f(z)≺1+2α[Re(1α)-1]z1-z(z∈𝕌*;|α-12|<12).Proof.
Suppose that(3.2)h(z):=-(1/α)(zf′(z)/f(z))-1-iIm(1/α)Re(1/α)-1(z∈𝕌;f∈ℳ𝒮α).
We easily know that h∈𝒫, which implies that
(3.3)-(1/α)(zf′(z)/f(z))-1-iIm(1/α)Re(1/α)-1=1+ω(z)1-ω(z)(z∈𝕌;f∈ℳ𝒮α),
where ω is analytic in 𝕌 with ω(0)=0 and |ω(z)|<1(z∈𝕌).
It follows from (3.3) that
(3.4)-zf′(z)f(z)=1+2α[Re(1α)-1]ω(z)1-ω(z)(z∈𝕌),
which is equivalent to the subordination relationship (3.1).
On the other hand, the above deductive process can be converse. The proof of Theorem3.1 is thus completed.Theorem 3.2.
Letf∈ℳ𝒮α. Then
(3.5)f(z)=1z·exp(-2α[Re(1α)-1]∫0z‍ω(t)t(1-ω(t))dt)(z∈𝕌*),
where ω is analytic in 𝕌 with ω(0)=0 and |ω(z)|<1(z∈𝕌).Proof.
Forf∈ℳ𝒮α, by Theorem 3.1, we know that (3.1) holds true. It follows that
(3.6)-zf′(z)f(z)=1+2α[Re(1α)-1]ω(z)1-ω(z)(z∈𝕌),
where ω is analytic in 𝕌 with ω(0)=0 and |ω(z)|<1(z∈𝕌).
We now find from (3.6) that
(3.7)f′(z)f(z)+1z=-2α[Re(1α)-1]ω(z)z(1-ω(z))(z∈𝕌*),
which, upon integration, yields
(3.8)log(zf(z))=-2α[Re(1α)-1]∫0z‍ω(t)t(1-ω(t))dt(z∈𝕌).
The assertion (3.5) of Theorem 3.2 can be easily derived from (3.8).Theorem 3.3.
Letf∈ℳ𝒮α. Then
(3.9)f(z)*(1-eiθ)z+2α[Re(1/α)-1]eiθ(1-z)z(1-z)2≠0(z∈𝕌*;0<θ<2π).Proof.
Assume thatf∈ℳ𝒮α. By Theorem 3.1, we know that (3.1) holds, which implies that
(3.10)-zf′(z)f(z)≠1+2α[Re(1α)-1]eiθ1-eiθ(z∈𝕌*;0<θ<2π).
It is easy to see that the condition (3.10) can be written as follows:
(3.11)(1-eiθ)zf′(z)+(1-eiθ+2α[Re(1α)-1]eiθ)f(z)≠0.
We note that
(3.12)f(z)=f(z)*(1z+1+z1-z)=f(z)*1z(1-z),-zf′(z)=f(z)*(1z-z(1-z)2)=f(z)*1-2zz(1-z)2.
Thus, by substituting (3.12) into (3.11), we get the desired assertion (3.9) of Theorem 3.3.Theorem 3.4.
Letλ=[Re(1/α)-1]|α|. If f∈ℳ𝒮α, then
(3.13)|ak|≦1(k+1)!∏j=0k‍(2λ+j)(k∈ℕ0).
The inequality (3.13) is sharp for the function given by
(3.14)f(z)=1z(1-z)2-2α(0<α<1).Proof.
Suppose that(3.15)h(z):=-(1/α)(zf′(z)/f(z))-1-iIm(1/α)Re(1/α)-1.
We easily know that h∈𝒫.
If we put(3.16)h(z)=1+h1z+h2z2+⋯,
it is known that
(3.17)|hk|≦2(k∈ℕ).
From (3.15), we have
(3.18)-1αzf′(z)f(z)-1-iIm(1α)=[Re(1α)-1]h(z).
We now set
(3.19)A:=1+iIm(1α),B:=Re(1α)-1.
It follows from (3.18) that
(3.20)-zf′(z)=[αA+αBh(z)]f(z).
Combining (1.1), (3.16), and (3.20), we obtain
(3.21)-z(-1z2+a1+2a2z+⋯+kakzk-1+⋯)=(1+αBh1z+⋯+αBhkzk+⋯)(1z+a0+a1z+⋯+akzk+⋯).
In view of (3.21), we get
(3.22)a0+αBh1=0,(3.23)-kak=ak+αB(ak-1h1+ak-2h2+⋯+a0hk+hk+1)(k∈ℕ).
From (3.17) and (3.22), we obtain
(3.24)|a0|≦2|α|B=2λ.
Moreover, we deduce from (3.17) and (3.23) that
(3.25)|ak|≦2|α|Bk+1(1+∑l=0k-1‍|al|)=2λk+1(1+∑l=0k-1‍|al|)(k∈ℕ).
Next, we define the sequence{Ak}k=0∞ as follows:
(3.26)A0=2λ,Ak+1=2λk+2(1+∑l=0k‍Al)(k∈ℕ0).
In order to prove that(3.27)|ak|≦Ak,
we make use of the principle of mathematical induction. By noting that
(3.28)|a0|≦2λ=A0.
Therefore, assuming that
(3.29)|al|≦Al(l=0,1,2,…,k;k∈ℕ0).
Combining (3.25) and (3.26), we get
(3.30)|ak+1|≦2λk+2(1+∑l=0k‍|al|)≦2λk+2(1+∑l=0k‍Al)=Ak+1.
Hence, by the principle of mathematical induction, we have
(3.31)|ak|≦Ak(k∈ℕ0)
as desired.
By means of Lemma2.1 and (3.26), we know that
(3.32)Ak=1(k+1)!∏j=0k‍(2λ+j)(k∈ℕ0).
Combining (3.31) and (3.32), we readily get the coefficient estimates asserted by Theorem 3.4.
For the sharpness, we consider the functionf given by (3.14). A simple calculation shows that
(3.33)Re(-zf′(z)f(z))=Re(1+(2α-3)z1-z)≧2-α>α.
Thus, the function f belongs to the class ℳ𝒮α. Since 0<α<1, we have
(3.34)λ=1-α.
Then f becomes
(3.35)f(z)=z-1(1-z)-2λ=z-1(∑n=0∞‍(-2λn)(-z)n)=1z+∑n=0∞2λ(2λ+1)⋯(2λ+n)(n+1)!zn.
This completes the proof of Theorem 3.4.Theorem 3.5.
Iff∈Σ satisfies the inequality
(3.36)∑k=0∞‍(k+|k+2α|)|ak|≦1-|2α-1|(|α-12|<12),
then f∈ℳ𝒮α.Proof.
To provef∈ℳ𝒮α, it suffices to show that
(3.37)|f(z)zf′(z)+12α|<12|α|(z∈𝕌),
which is equivalent to
(3.38)|zf′(z)+2αf(z)zf′(z)|<1(z∈𝕌*).
From (3.36), we know that
(3.39)1-∑k=0∞‍k|ak|≧|2α-1|+∑k=0∞‍|k+2α||ak|>0.
Now, by the maximum modulus principle, we deduce from (1.1) and (3.39) that
(3.40)|zf′(z)+2αf(z)zf′(z)|=|(2α-1)+∑k=0∞(k+2α)akzk+1-1+∑k=0∞kakzk+1|≦|2α-1|+∑k=0∞|k+2α||ak||z|k+11-∑k=0∞k|ak||z|k+1<|2α-1|+∑k=0∞|k+2α||ak|1-∑k=0∞k|ak|≦1,
which implies that the assertion of Theorem 3.5 holds.Theorem 3.6.
Iff∈Σ satisfies the condition
(3.41)|1+zf′′(z)f′(z)-zf′(z)f(z)|<1-α2α(12<α<1),
then f∈ℳ𝒮α.Proof.
Define the functionφ by
(3.42)φ(z):=(zf′(z)/f(z))+1(zf′(z)/f(z))+2α-1(z∈𝕌).
Then we see that φ is analytic in 𝕌 with φ(0)=0.
It follows from (3.42) that
(3.43)-zf′(z)f(z)=1+(1-2α)φ(z)1-φ(z).
By differentiating both sides of (3.43) logarithmically, we obtain
(3.44)1+zf′′(z)f′(z)-zf′(z)f(z)=2(1-α)zφ′(z)[1+(1-2α)φ(z)](1-φ(z)).
From (3.41) and (3.44), we find that
(3.45)|1+zf′′(z)f′(z)-zf′(z)f(z)|=|2(1-α)zφ′(z)[1+(1-2α)φ(z)](1-φ(z))|<1-α2α.
Next, we claim that |φ(z)|<1. Indeed, if not, there exists a point z0∈𝕌 such that
(3.46)max|z|≦|z0||φ(z)|=|φ(z0)|=1.
By Lemma 2.2, we have
(3.47)φ(z0)=eiθ,z0φ′(z0)=teiθ(t≧1).
Moreover, for z=z0, we find from (3.44) and (3.47) that
(3.48)|1+z0f′′(z0)f′(z0)-z0f′(z0)f(z0)|=|2(1-α)teiθ(1+(1-2α)eiθ)(1-eiθ)|=2(1-α)t1+2(1-2α)cosθ+(1-2α)2·2-2cosθ≧1-α2α(12<α<1).
But (3.48) contradicts to (3.45). Therefore, we conclude that |φ(z)|<1, that is
(3.49)|(zf′(z)/f(z))+1(zf′(z)/f(z))+2α-1|<1,
which shows that f∈ℳ𝒮α.



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