# A New Subclass of Meromorphic Close-to-Convex Functions

Authors: Amit Soni; Shashi Kant
Journal: Journal of Complex Analysis
 (2013)

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## Abstract

A new subclassMK(t, A, B) of meromorphic close-to-convex functions, defined by means of subordination, is investigated. Some results such as inclusion relationship, coefficient inequality, convolution property, and distortion property for this class are derived. The results obtained here are extension of earlier known work.


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## Body

## 1. Introduction

Let∑‍ denote the class of functions f of the form
(1)f(z)=1z+∑n=1∞‍anzn,
which are analytic in the punctured open unit disk 𝔻={z:z∈ℂ:0<|z|<1}. For any two analytic function f and g in 𝕌=𝔻⋃‍{0}, we say that f is subordinate to g in 𝕌, written as f(z)≺g(z), if there exists a Schwarz function w(z) such that f(z)=g(w(z)), for z∈𝕌. In particular, if g is univalent in 𝕌, then f is subordinate to g, if and only if f(0)=g(0)andf(𝕌)⊂g(𝕌). If f∈∑‍ is given by (1) and g∈∑‍ is given by
(2)g(z)=1z+∑n=1∞‍bnzn,
then the Hadamard product (or convolution) of the function f and g is defined by
(3)(f*g)(z)=1z+∑n=1∞‍anbnzn=(g*f)(z).
A function f∈∑‍ is said to be in the class MS*(α) of meromorphic starlike functions of order α if it satisfies the inequality
(4)R(zf′(z)f(z))<-α,(z∈𝕌,0≤α<1).
Moreover, a function f∈∑‍ is said to be in the class MC  of meromorphic close to convex function if it satisfies the condition
(5)R(zf′(z)g(z))<0(z∈𝕌,g∈MS*(0)=MS*).
Recently, Wang et al. [1] introduced and discussed the class MK of meromorphic functions f(z)∈∑‍ which satisfies the inequality
(6)R(f′(z)g(z)g(-z))>0(z∈𝕌),
where g(z)∈MS*(1/2).The classMK is very close to the interesting subclass Ks of close-to-convex function for analytic function introduced and studied by Gao and Zhou [2]. Many classes related to the class Ks have been further studied by some authors. Especially, Wang et al. [3, 4], Kowalczyk and Leś-Bomba [5], Xu et al. [6], Şeker [7], and Cho et al. [8] introduced generalization of Ks and obtained some properties for analytic functions in each class.More recently, by means of subordination, Sim and Kwon [9] discussed a subclass ∑‍(A,B) of the class MK. A function f(z)∈∑‍ is said be in the class ∑‍(A,B) if it satisfies the following subordination relation:
(7)f′(z)g(z)g(-z)≺1+Az1+Bz,
where -1≤B<A≤1 and g(z)∈MS*(1/2). Here, the assumption that g(z) is meromorphic starlike function of order 1/2 makes the function -zg(z)g(-z) meromorphic starlike. So, instead of -zg(z)g(-z) in (6) and (7), we can considertg(z)g(tz),0<|t|≤1, because if g(z)∈MS*(1/2), then tg(z)g(tz) is also a meromorphic starlike function, which motivates us to define a new subclass MK(t,A,B) of meromorphic close-to-convex functions as follows.Definition 1.
A functionf∈∑‍ is said to be in the class MK(t,A,B) if there exists g∈MS*(1/2), such that
(8)-f'(z)tg(z)g(tz)≺1+Az1+Bz,(z∈𝕌,0<|t|≤1,-1≤B<A≤1).
The class MK(t,A,B) provides a generalization of the classes given by Sim and Kwon [9] and Wang et al. [1]. The transformation (1+Az)/(1+Bz) involved in the class MK(t,A,B) is analytic and convex univalent in 𝕌. Moreover,
(9)0≤1-A1-B<R(1+Az1+Bz)<1+A1+B,(-1<B<A≤1;z∈𝕌).In this paper, we aim at proving results such as inclusion relationship, coefficient bounds, and distortion theorem for the classMK(t,A,B).


## 2. Properties of Meromorphic Starlike Functions

In beginning, we prove the following result of meromorphic starlike functions.Theorem 2.
Suppose thatϕ∈MS*(α1) and ψ∈MS*(α2) with 0≤α1+α2-1<1. Then,
(10)zt1t2ϕ(t1z)ψ(t2z)∈MS*(α1+α2-1),where0<|t1|,|t2|≤1.Proof.
Letϕi∈MS*(α1)andψ∈MS*(α2). By definition, we know that
(11)R(t1zϕ′(t1z)ϕ(t1z))<-α1,R(t2zψ′(t2z)ψ(t2z))<-α2.
Next, we suppose that
(12)h(z)=zt1t2ϕ(t1z)ψ(t2z).
Then, we easily find that
(13)zh′(z)h(z)=1+t1zϕ′(t1z)ϕ(t1z)+t2zψ′(t2z)ψ(t2z).
It follows that
(14)R(zh′(z)h(z))<-(α1+α2-1).
By noting that0≤(α1+α2-1)<1, which implies that h(z)∈MS*(α1+α2-1), it completes the proof of Theorem 2.Forα1=α2=1/2 and t1=1, Theorem 2 gives the following corollary.Corollary 3.
Ifg(z)∈MS*(1/2) and 0<|t|≤1, then tzg(z)g(tz)∈MS*.Lemma 4 (see [10]).
Let-1≤B2≤B1<A1≤A2≤1. Then,
(15)1+A1z1+B1z≺1+A2z1+B2z.Theorem 5.
Let-1≤B2≤B1<A1≤A2≤1. Then,
(16)MK(t,A1,B1)⊂MK(t,A2,B2).Proof.
Suppose thatf∈MK(t,A1,B1). Then,
(17)-f′(z)tg(z)g(tz)≺1+A1z1+B1z.
Since -1≤B2≤B1<A1≤A2≤1, by Lemma 4, we have
(18)-f′(z)tg(z)g(tz)≺1+A1z1+B1z≺1+A2z1+B2z.
It implies that f(z)∈MK(t,A2,B2), which completes the proof.


## 3. Coefficient Estimates

In this section,we obtain the coefficient estimates of functions belonging to the classMK(t,A,B). For this, we require the following Lemmas.Lemma 6 (see [11]).
Let(19)h(z)=1+∑n=1∞‍hnzn
be analytic in 𝕌 and
(20)k(z)=1+∑n=1∞‍knzn
be analytic and convex in 𝕌. If h(z)≺k(z), then
(21)|hn|≤|k1|,(n∈N).Lemma 7.
Letg(z)∈MS*(1/2) and be given by (2). Then,
(22)|bntn+1+b1bn-2tn-1+b2bn-3tn-2+⋯+bn-3b2t3+b1bn-2t2+bn|≤2n+1,(0<|t|≤1).Proof.
According to Corollary3, we have tzg(z)g(tz)∈MS*,and if
(23)G(z)=tzg(z)g(tz)=1z+∑n=1∞‍cnzn,
then it is well known that
(24)|cn|≤2n+1.
Substituting the series expressions of G(z) and g(z) in (23) and comparing the coefficients of both sides of this equation, we get
(25)cn=bntn+1+b1bn-2tn-1+b2bn-3tn-2+⋯+bn-3b2t3+b1bn-2t2+bn.
On substituting the value from (25) in (24), we get the result.Theorem 8.
Let-1≤B<A≤1. If
(26)f(z)=1z+∑n=1∞‍anzn∈MK(t,A,B),
then
(27)|a1|≤1,|an|≤A-Bn(1+∑k=2n‍2k)+2n(n+1)(n∈N∖{1}).Proof.
Suppose thatf∈MK(t,A,B). Then, we know that
(28)-zf′(z)G(z)≺1+Az1+Bz,
where G(z)=tzg(z)g(tz).
If we set(29)q(z)=-zf′(z)G(z),
it follows that
(30)q(z)=1+∑n=1∞‍dnzn.
In view of Lemma 6, we know that
(31)|dn|≤A-B(n∈N).
By substituting the series expression of functions f,G, and q in (29), we obtain
(32)(1+d1z+d2z2+⋯+dnzn+⋯)×(1z+c1z+c2z2+⋯+cnzn+⋯)=1z-a1z-2a2z2-⋯-nanzn-⋯.
Since f is univalent in 𝔻, it is well known that
(33)|a1|≤1.
On comparing the coefficient of zn in both sides of (32), we have
(34)-nan=cn+d1cn-1+d2cn-2+⋯+dn-1c1+dn+1(n∈N).
Substituting the values from (24) and (31) to (34), we have
(35)n|an|≤(A-B)(1+22+23+⋯+2n-1+2n)+2n+1,(36)|an|≤(A-B)n(1+∑k=2n‍2k)+2n(n+1).
This evidently completes the proof of Theorem 8.Theorem 9.
Let-1≤B<A≤1. If
(37)g(z)=1z+∑n=1∞‍bnzn∈MS*(12)
and f∈∑‍ given by (1), satisfying the condition
(38)(1+|B|)∑n=1∞‍n|an|+(1+|A|)∑n=1∞‍|cn|≤(A-B),
where cn is given by (25), then f∈MK(t,A,B).Proof.
To provef∈MK(t,A,B), it suffices to show that
(39)|zf′(z)/G(z)+1A+B(zf′(z)/G(z))|<1,
where G(z) is given by (23). From (38), we know that
(40)(A-B)-|A|∑n=1∞‍|cn|-|B|∑n=1∞‍n|an|≥∑n=1∞‍n|an|+∑n=1∞‍|cn|.
Now, by maximum modulus principle and (40), we deduce that
(41)|zf′(z)/G(z)+1A+B(zf′(z)/G(z))|=|∑n=1∞nanzn+1+∑n=1∞cnzn+1(A-B)+A∑n=1∞cnzn+1+B∑n=1∞nanzn+1|≤∑n=1∞n|an|+∑n=1∞|cn||A-B|-|A|∑n=1∞|cn|-|B|∑n=1∞n|an|≤1.
This evidently completes the proof of Theorem 9.Theorem 10.
Let-1≤B<A≤1. If f(z)∈MK(t,A,B), then
(42)f(z)*((1+Beiθ)(2z-1)z(1-z)2)+G(z)*((1+Aeiθ)(z2-z+1)z(1-z))≠0,(z∈𝔻,0<θ<2π).Proof.
Suppose thatf∈MK(t,A,B). Then, we know that
(43)-zf′(z)G(z)≠1+Aeiθ1+Beiθ(z∈𝔻,0<θ<2π),
where G is given by (23). It is easy to see that the condition (43) can be written as
(44)zf′(z)(1+Beiθ)+G(z)(1+Aeiθ)≠0,(z∈𝔻,0<θ<2π).
Note that
(45)zf′(z)=f(z)*2z-1z(1-z)2,G(z)=G(z)*z2-z+1z(1-z).
By substituting (45) into (44), we get the desired assertion (42) of Theorem 10.


## 4. Distortion Theorem

Lemma 11 (see [12]).
Letϕ∈MS*. Then,
(46)(1-r)2r≤|ϕ(z)|≤(1+r)2r(|z|=r,0<r<1).Theorem 12.
Let-1≤B<A≤1. If f(z)∈MK(t,A,B), then
(47)(1-Ar)(1-r)2(1-Br)r2≤|f′(z)|≤(1+Ar)(1+r)2(1+Br)r2,(|z|=r;0<r<1).Proof.
Suppose thatf(z)∈MK(t,A,B). By definition, we have the transformation
(48)-zf′(z)G(z)=1+Aw(z)1+Bw(z),w(z)∈𝕌,
which maps |w(z)|≤r in the closed disk
(49)|-zf′(z)G(z)-1-ABr21-B2r2|≤(A-B)r1-B2r2,|z|=r.
This implies that
(50)1-Ar1-Br≤|-zf′(z)G(z)|≤1+Ar1+Br.
Thus, by virtue of Lemma 11, we readily get the desired result.Remark. Taking A=1,B=-1, and t=-1, we obtain the results such as inclusion relation, coefficient inequality, convolution property, and distortion theorem derived by Wang et al. [1].



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