# On Certain New Subclass of Meromorphic Close-to-Convex Functions

Authors: Jing-Ping Yi; Zhi-Gang Wang; Ming-Hua Zeng
Journal: Journal of Complex Analysis
 (2013)

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## Abstract

We introduce a certain new subclass of meromorphic close-to-convex functions. Such results as inclusion relationship, coefficient inequalities, distortion, and growth theorems for this class of functions are derived.


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## Body

## 1. Introduction

LetΣ denote the class of functions of the form
(1)f(z)=1z+∑n=1∞‍anzn,
which are analytic in the punctured open unit disk:
(2)𝕌*∶={z:z∈ℂand0<|z|<1}∶=𝕌∖{0}.Let𝒫 denote the class of functions p given by
(3)p(z)=1+∑n=1∞‍pnzn(z∈𝕌),
which are analytic and convex in 𝕌 and satisfy the following condition:
(4)ℜ(p(z))>0(z∈𝕌).A functionf∈Σ is said to be in the class ℳ𝒮*(α) of meromorphic starlike functions of orderα if it satisfies the following inequality:
(5)ℜ(-zf′(z)f(z))>α(0≦α<1;z∈𝕌).
Moreover, a function f∈Σ is said to be in the class ℳ𝒞 of meromorphic close-to-convex functions if it satisfies the following condition:
(6)ℜ(-zf′(z)g(z))>0(z∈𝕌;g∈ℳ𝒮*(0)=:ℳ𝒮*).For two functionsf and g analytic in 𝕌, we say that the function f(z) is subordinate to g(z) in 𝕌 and write f(z)≺g(z)(z∈𝕌), if there exists a Schwarz function ω(z), analytic in 𝕌 with ω(0)=0 and |ω(z)|<1 such that f(z)=g(ω(z)). Indeed, it is well 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).Recently, Wang et al. [1] introduced and investigated the class ℳ𝒦 of meromorphic close-to-convex functions which satisfy the inequality
(9)ℜ(f′(z)g(z)g(-z))>0(z∈𝕌),
where g∈ℳ𝒮*(1/2). Kowalczyk and Leś-Bomba [2] discussed the class 𝒦s(γ) of analytic functions related to the starlike functions; a function f(z)=z+∑n=2∞‍anzn which is analytic in 𝕌 is said to be in the class 𝒦s(γ)(0≦γ<1), if it is satisfies the following inequality:
(10)ℜ(z2f′(z)g(z)g(-z))>γ(z∈𝕌),
where g∈𝒮*(1/2). Şeker [3] discussed the class 𝒦s(k)(γ) of analytic functions which satisfy the following condition:
(11)ℜ(zkf′(z)gk(z))>γ(z∈𝕌),
where 0≦γ<1,gk(z)=∏v=0k-1‍ε-vg(εvz)(εk=1;k≧1), and g∈𝒮*((k-1)/k).Motivated essentially by the classesℳ𝒦,𝒦s(γ), and 𝒦s(k)(γ), we introduce and study the following more generalized class ℳ𝒦(k)(γ) of meromorphic functions.Definition 1.
A functionf∈Σ is said to be in the class ℳ𝒦(k)(γ) if it satisfies the following inequality:
(12)ℜ(-f′(z)zk-2gk(z))>γ(z∈𝕌),
where 0≦γ<1,k≧1 is a fixed positive integer, and gk(z) is given by
(13)gk(z)=∏v=0k-1‍εvg(εvz)(ε=e(2πi)/k;g∈ℳ𝒮*(k-1k)).We observe that the inequality (12) is equivalent to
(14)|f′(z)zk-2gk(z)+1|<|f′(z)zk-2gk(z)+2γ-1|(z∈𝕌).Sinceℳ𝒦(2)(0)=ℳ𝒦, the class ℳ𝒦(k)(γ) is a generalization of the class ℳ𝒦.For some recent investigations on the class of close-to-convex functions, one can find them in [4–7] and the references cited therein. In the present paper, we aim at proving that the class ℳ𝒦(k)(γ) is a subclass of meromorphic close-to-convex functions. Furthermore, some interesting results of the class ℳ𝒦(k)(γ) are derived.


## 2. Preliminary Results

To prove our main results, we need the following lemmas.Lemma 2.
Letφj∈ℳ𝒮*(αj), where 0≦αj<1(j=0,1,…,k-1). Then for k-1≦∑j=0k-1‍αj<k, one has
(15)zk-1∏j=0k-1‍φj(z)∈ℳ𝒮*(∑j=0k-1‍αj-(k-1)).Proof.
Sinceφj∈ℳ𝒮*(αj)(j=0,1,…,k-1), we have
(16)ℜ(-zφ0′(z)φ0(z))>α0,ℜ(-zφ1′(z)φ1(z))>α1,…,ℜ(-zφk-1′(z)φk-1(z))>αk-1.
We now let
(17)F(z)=zk-1φ0(z)φ1(z)⋯φk-1(z).
Differentiating (17) logarithmically, we obtain
(18)-zF′(z)F(z)=-zφ0′(z)φ0(z)-zφ1′(z)φ1(z)-⋯-zφk-1′(z)φk-1(z)-(k-1).
From (18) together with (16), we can get
(19)ℜ(-zF′(z)F(z))=ℜ(-zφ0′(z)φ0(z))+ℜ(-zφ1′(z)φ1(z))+⋯+ℜ(-zφk-1′(z)φk-1(z))-(k-1)>∑j=0k-1‍αj-(k-1).
Thus, if 0≦∑j=0k-1‍αj-(k-1)<1, we know that
(20)F(z)=zk-1∏j=0k-1‍φj(z)∈ℳ𝒮*(∑j=0k-1‍αj-(k-1)).Lemma 3 (see [8]).
Let-1≦B2≦B1<A1≦A2≦1. Then
(21)1+A1z1+B1z≺1+A2z1+B2z.Lemma 4 (see [9]).
Suppose thatg∈ℳ𝒮*. Then
(22)(1-r)2r≦|g(z)|≦(1+r)2r(|z|=r;0<r<1).Lemma 5 (see [10, page 105]).
If the function(23)p(z)=1+∑n=1∞‍pnzn(z∈𝕌)
analytic and convex in 𝕌 and satisfies the condition
(24)ℜ(p(z))>γ(z∈𝕌),
then
(25)1-(1-2γ)r1+r≦|p(z)|≦1+(1-2γ)r1-r(|z|=r<1).Lemma 6 (see [10]).
If the functionp∈𝒫 is given by (3), then
(26)|pn|≦2(n∈ℕ),1-r1+r≦|p(z)|≦1+r1-r(|z|=r;0<r<1).Lemma 7 (see [11]).
Suppose that(27)g(z)=1z+∑n=1∞‍cnzn∈ℳ𝒮*.
Then
(28)|cn|≦2n+1(n∈ℕ∶={1,2,…}).
Each of these inequalities is sharp, with the extremal function given by
(29)g(z)=z-1(1+zn+1)2/(n+1).Lemma 8 (see [12]).
Let(30)f(z)=1+∑n=1∞‍cnzn
be analytic in 𝕌 and let
(31)g(z)=1+∑n=1∞‍dnzn
be analytic and convex in 𝕌. If f≺g, then
(32)|cn|≦|d1|(n∈ℕ∶={1,2,…}).Lemma 9.
Ifp(z)≺(1+(1-2γ)z)/(1-z), where p(z) is given by (3), then
(33)|pn|≦2(1-γ).Proof.
By Lemma8, we easily get the assertion of Lemma 9.


## 3. Main Results

We first give the following result.Theorem 10.
Letg(z)=1/z+∑n=1∞‍bnzn∈ℳ𝒮*((k-1)/k), Then
(34)Gk(z)=zk-1gk(z)=1z+∑n=1∞‍Bnzn∈ℳ𝒮*,
where gk(z) is given by (13).Proof.
From (13), we know that
(35)zk-1gk(z)=zk-1∏v=0k-1‍εvg(εvz)=zk-1∏v=0k-1‍εv(1εvz+∑n=1∞‍bn(εvz)n)=zk-1∏v=0k-1‍(1z+∑n=1∞‍bnε(n+1)vzn).
Now, suppose that
(36)g(z)=1z+∑n=1∞‍bnzn∈ℳ𝒮*(k-1k).
Then, by Lemma 2 and (35), we get the assertion of Theorem 10 easily.Remark 11.
From Theorem10 and Definition 1, we know that if f∈ℳ𝒦k(γ), then f is a meromorphic close-to-convex function. So the class ℳ𝒦k(γ) is a subclass of meromorphic close-to-convex functions.Now, we prove a sufficient condition for functions to belong to the classℳ𝒦k(γ).Theorem 12.
Letf∈Σ,g∈ℳ𝒮*((k-1)/k). If for 0≦γ<1, one has
(37)2∑n=1∞‍n|an|+(|1-2γ|+1)∑n=1∞‍|Bn|≦2(1-γ),
where the coefficients Bn (n=1,2,…) are given by (34), then f∈ℳ𝒦k(γ).Proof.
We set forf given by (1) and gk defined by (13)
(38)Δ=|-zf′(z)-zk-1gk(z)|-|-zf′(z)+(1-2γ)zk-1gk(z)|=|-∑n=1∞‍nanzn-∑n=1∞‍Bnzn|-|2-2γz-∑n=1∞‍nanzn+(1-2γ)∑n=1∞‍Bnzn|.
For |z|=r(0<r<1), from inequality (37), we have
(39)Δ≦∑n=1∞‍n|an||z|n+∑n=1∞‍|Bn||z|n-((2-2γ)1|z|-∑n=1∞‍n|an||z|n-|1-2γ|∑n=1∞‍|Bn||z|n)=-(2-2γ)1|z|+2∑n=1∞‍n|an||z|n+(|1-2γ|+1)×∑n=1∞‍|Bn||z|n<(-(2-2γ)+2∑n=1∞‍n|an|+(|1-2γ|+1)∑n=1∞‍|Bn|)1|z|≦0.
Thus, we have
(40)|-f′(z)zk-2gk(z)-1|<|-f′(z)zk-2gk(z)+1-2γ|(z∈𝕌),
that is, f∈ℳ𝒦(k)(γ). This completes the proof of Theorem 12.Next, we give the inclusion relationship for classf∈ℳ𝒦(k)(γ).Theorem 13.
Let0≦γ1≦γ2<1. Then one has
(41)ℳ𝒦(k)(γ2)⊂ℳ𝒦(k)(γ1).Proof.
Suppose thatf∈ℳ𝒦(k)(γ2). By Definition 1, we have
(42)-f′(z)zk-2gk(z)≺1+(1-2γ2)z1-z.
Since 0≦γ1≦γ2<1, we get
(43)-1<1-2γ2≦1-2γ1≦1.
Thus, by Lemma 3, we obtain
(44)-f′(z)zk-2gk(z)≺1+(1-2γ1)z1-z,
that is, f∈ℳ𝒦(k)(γ1). This means that ℳ𝒦(k)(γ2)⊂ℳ𝒦(k)(γ1). Hence the proof is completed.In what follows, we derive the coefficient inequality for the classℳ𝒦(k)(γ).Theorem 14.
Suppose that(45)f(z)=1z+∑n=1∞‍anzn∈ℳ𝒦(k)(γ).
Then
(46)|a1|≦1,(47)|an|≦2n(n+1)+(2-2γ)n(1+∑j=1n-1‍2j+1)(n∈ℕ∖{1}).Proof.
Suppose thatf∈ℳ𝒦(k)(γ); we know that
(48)ℜ(-zf′(z)Gk(z))>γ,
where Gk is given by (34). If we set
(49)q(z)∶=-zf′(z)Gk(z),
it follows that
(50)q(z)=1+∑n=1∞‍dnzn.
In view of Definition 1 and Lemma 9, we know that
(51)|dn|≦(2-2γ)(n∈ℕ).
By substituting the series expressions of functions f,Gk, and q into (49), we obtain
(52)(1+∑n=1∞‍dnzn)(1z+∑n=1∞‍Bnzn)=(1z-∑n=1∞‍nanzn).
Since f is univalent in 𝕌*, it is well known that
(53)|a1|≦1.
On the other hand, we find from (52) that
(54)-nan=Bn+dn+1+∑j=1n-1‍Bjdn-j(n∈ℕ).
Combining (28), (51), and (54), we have
(55)n|an|≦2n+1+(2-2γ)+(2-2γ)∑j=1n-1‍2j+1.
Thus, the assertion (47) of Theorem 14 follows directly from (55).Finally, we give the distortion and growth theorems for the function classℳ𝒦(k)(γ).Theorem 15.
Iff∈ℳ𝒦(k)(γ), then
(56)(1-r)2[1-(1-2γ)r]r2(1+r)≦|f′(z)|≦(1+r)2[1+(1-2γ)r]r2(1-r)(|z|=r<1),(57)∫0r‍(1-t)2[1-(1-2γ)t]t2(1+t)dt≦|f(z)|≦∫0r‍(1+t)2[1+(1-2γ)t]t2(1-t)dt.Proof.
Iff∈ℳ𝒦(k)(γ), then there exists a function g∈ℳ𝒮*((k-1)/k) such that (12) holds. It follows from Theorem 10 that the function given by (34) is a meromorphic starlike function. Therefore, by Lemma 4, we have
(58)(1-r)2r≦|Gk(z)|≦(1+r)2r(|z|=r;0<r<1).
Let q be defined by (49); by Lemma 5, we know that
(59)1-(1-2γ)r1+r≦|q(z)|≦1+(1-2γ)r1-r(|z|=r<1).
Thus, from (49), (58), and (59), we readily get (56). Upon integrating (56) from 0 to r, we get (57). The proof of Theorem 15 is thus completed.



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