New sufficient conditions, concerned with the coefficients of harmonic functions f(z)=h(z)+g(z)¯ in the open unit disk 𝕌 normalized by f(0)=h(0)=h′(0)−1=0, for f(z) to be harmonic close-to-convex functions are discussed. Furthermore, several illustrative examples and the image domains of harmonic close-to-convex functions satisfying the obtained conditions are enumerated.

1. Introduction

For a continuous complex-valued function f(z)=u(x,y)+iv(x,y)(z=x+iy), we say that f(z) is harmonic in the open unit disk 𝕌={z∈ℂ:|z|<1} if both u(x,y) and v(x,y) are real harmonic in 𝕌, that is, u(x,y) and v(x,y) satisfy the Laplace equationsΔu=uxx+uyy=0,Δv=vxx+vyy=0.
A complex-valued harmonic function f(z) in 𝕌 is given by f(z)=h(z)+g(z)¯ where h(z) and g(z) are analytic in 𝕌. We call h(z) and g(z) the analytic part and the coanalytic part of f(z), respectively. A necessary and sufficient condition for f(z) to be locally univalent and sense preserving in 𝕌 is |h′(z)|>|g′(z)| in 𝕌 (see [1] or [2]). Let ℋ denote the class of harmonic functions f(z) in 𝕌 with f(0)=h(0)=0 and h′(0)=1. Thus, every normalized harmonic function f(z) can be written byf(z)=h(z)+g(z)¯=z+∑n=2∞anzn+∑n=1∞bnzn¯∈H,
where a1=1 and b0=0, for convenience.

We next denote by 𝒮ℋ the class of functions f(z)∈ℋ that are univalent and sense preserving in 𝕌. Due to the sense-preserving property of f(z), we see that |b1|=|g′(0)|<|h′(0)|=1. If g(z)≡0, then 𝒮ℋ reduces to the class 𝒮 consisting of normalized analytic univalent functions. Furthermore, for every function f(z)∈𝒮ℋ, the functionF(z)=f(z)-b1f(z)¯1-|b1|2=z+∑n=2∞an-b1¯bn1-|b1|2zn+∑n=2∞bn-b1an1-|b1|2zn¯
is also a member of 𝒮ℋ. Therefore, we consider the subclass 𝒮ℋ0 of 𝒮ℋ defined asSH0={f(z)∈SH:b1=g′(0)=0}.
Conversely, if F(z)∈𝒮ℋ0, then f(z)=F(z)+b1F(z)¯∈𝒮ℋ for any b1(|b1|<1).

We say that a domain 𝔻 is a close-to-convex domain if the complement of 𝔻 can be written as a union of nonintersecting half-lines (except that the origin of one half-line may lie on one of the other half-lines). Let 𝒞, 𝒞ℋ, and 𝒞ℋ0 be the respective subclasses of 𝒮, 𝒮ℋ, and 𝒮ℋ0 consisting of all functions f(z), which map 𝕌 onto a certain close-to-convex domain.

Bshouty and Lyzzaik [3] have stated the following result.

Theorem 1.1.

If f(z)=h(z)+g(z)¯∈ℋ satisfies
g′(z)=zh′(z),Re(1+zh″(z)h′(z))>-12
for all z∈𝕌, then f(z)∈𝒞ℋ0⊂𝒮ℋ0.

A simple and interesting example is below.

Example 1.2.

The function
f(z)=1-(1-z)22(1-z)2+z22(1-z)2¯=z+∑n=2∞n+12zn+∑n=2∞n-12z¯n
satisfies the conditions of Theorem 1.1, and therefore f(z) belongs to the class 𝒞ℋ0. We now show that f(𝕌) is actually a close-to-convex domain. It follows that
f(z)=(z2(1-z)2+z2(1-z))+(z2(1-z)2-z2(1-z))¯=Re(z(1-z)2)+iIm(z1-z).
Setting
f(reiθ)=-2r2+r(1+r2)cosθ(1+r2-2rcosθ)2+rsinθ1+r2-2rcosθi=u+iv
for any z=reiθ∈𝕌(0≦r<1,0≦θ<2π), we see that
-4(u+v2)=4r(r-cosθ)(1-rcosθ)(1+r2-2rcosθ)2=4r(r-t)(1-rt)(1+r2-2rt)2≡ϕ(t)(-1≦t=cosθ≦1).
Since
ϕ′(t)=-4r(1-r2)2(1+r2-2rt)3≦0,
we obtain that
ϕ(t)≦ϕ(-1)=4r(1+r)2≡ψ(r).
Also, noting that
ψ′(r)=4(1-r)(1+r)3>0,
we know that
ψ(r)<ψ(1)=1,
which implies that
u>-v2-14.
Thus, f(z) maps 𝕌 onto the following close-to-convex domain as shown in Figure 1.

The image of f(z)=(1-(1-z)2)/2(1-z)2+z2/2(1-z)2¯.

Remark 1.3.

Let ℳ be the class of all functions satisfying the conditions of Theorem 1.1. Then, it was earlier conjectured by Mocanu [4, 5] that ℳ⊂𝒮ℋ0. Furthermore, we can immediately see that the function f(z) in Example 1.2 is a member of the class ℳ and it shows that f(z)∈ℳ is not necessarily starlike with respect to the origin in 𝕌 (f(z) is starlike with respect to the origin in 𝕌 if and only if tw∈f(𝕌) for all w∈f(𝕌) and t(0≦t≦1)).

Remark 1.4.

For the function f(z)=h(z)+g(z)¯∈ℋ given by
g′(z)=zn-1h′(z)(n=2,3,4,…),
letting w(t)=f(eit)=h(eit)+g(eit)¯(-π≦t<π), we know that
Im(w′′(t)w′(t))≦0(-π≦t<π),
which means that f(z) maps the unit circle ∂𝕌={z∈ℂ:|z|=1} onto a union of several concave curves (see [6, Theorem 2.1]).

Jahangiri and Silverman [7] have given the following coefficient inequality for f(z)∈ℋ to be in the class 𝒞ℋ.

Theorem 1.5.

If f(z)∈ℋ satisfies
∑n=2∞n∣an∣+∑n=1∞n∣bn∣≦1,
then f(z)∈𝒞ℋ.

Example 1.6.

The function
f(z)=z+15z¯5
belongs to the class 𝒞ℋ0⊂𝒞ℋ and satisfies the condition of Theorem 1.5. Indeed, f(z) maps 𝕌 onto the following hypocycloid of six cusps (cf. [8] or [6]) as shown in Figure 2.

The object of this paper is to find some sufficient conditions for functions f(z)∈ℋ to be in the class 𝒞ℋ. In order to establish our results, we have to recall here the following lemmas due to Clunie and Sheil-Small [1].

The image of f(z)=z+(1/5)z¯5.

Lemma 1.7.

If h(z) and g(z) are analytic in 𝕌 with |h′(0)|>|g′(0)| and h(z)+ɛg(z) is close-to-convex for each ɛ(|ɛ|=1), then f(z)=h(z)+g(z)¯ is harmonic close-to-convex.

Lemma 1.8.

If f(z)=h(z)+g(z)¯ is locally univalent in 𝕌 and h(z)+ɛg(z) is convex for some ɛ(|ɛ|≦1), then f(z) is univalent close-to-convex.

We also need the following result due to Hayami et al. [9].

Lemma 1.9.

If a function F(z)=z+∑n=2∞Anzn is analytic in 𝕌 and satisfies
∑n=2∞[|∑k=1n{∑j=1k(-1)k-jj(j+1)(αk-j)Aj}(βn-k)|+|∑k=1n{∑j=1k(-1)k-jj(j-1)(αk-j)Aj}(βn-k)|]≦2
for some real numbers α and β, then F(z) is convex in 𝕌.

2. Main Results

Our first result is contained in the following theorem.

Theorem 2.1.

If f(z)∈ℋ satisfies the following condition
∑n=2∞|nan-eiφ(n-1)an-1|+∑n=1∞|nbn-eiφ(n-1)bn-1|≦1
for some real number φ(0≦φ<2π), then f(z)∈𝒞ℋ.

Proof.

Let F(z)=z+∑n=2∞Anzn be analytic in 𝕌. If F(z) satisfies
∑n=2∞|nAn-eiφ(n-1)An-1|≦1,
then it follows that
|(1-eiφz)F′(z)-1|=|∑n=2∞(nAn-eiφ(n-1)An-1)zn-1|≦∑n=2∞|nAn-eiφ(n-1)An-1|⋅∣z∣n-1<∑n=2∞|nAn-eiφ(n-1)An-1|≦1(z∈U).
This gives us that
Re((1-eiφz)F′(z))>0(z∈U),
that is, F(z)∈𝒞. Then, it is sufficient to prove that
F(z)=h(z)+ɛg(z)1+ɛb1=z+∑n=2∞an+ɛbn1+ɛb1zn∈C
for each ɛ(|ɛ|=1) by Lemma 1.7. From the assumption of the theorem, we obtain that
∑n=2∞|nan+ɛbn1+ɛb1-eiφ(n-1)an-1+ɛbn-11+ɛb1|≦11-∣b1∣∑n=2∞[|nan-eiφ(n-1)an-1|+|nbn-eiφ(n-1)bn-1|]≦1-∣b1∣1-∣b1∣=1.
This completes the proof of the theorem.

Example 2.2.

The function
f(z)=-log(1-z)+(-mz-log(1-z))¯=z+∑n=2∞1nzn+(1-m)z¯+∑n=2∞1nz¯n(0<m≦1)
satisfies the condition of Theorem 2.1 with φ=0 and belongs to the class 𝒞ℋ. In particular, putting m=1, we obtain Figure 3.

By making use of Lemma 1.8 with ɛ=0 and applying Lemma 1.9, we readily obtain the next theorem.

The image of f(z)=-z¯-2log|1-z|.

Theorem 2.3.

If f(z)∈ℋ is locally univalent in 𝕌 and satisfies
∑n=2∞[|∑k=1n{∑j=1k(-1)k-jj(j+1)(αk-j)aj}(βn-k)|+|∑k=1n{∑j=1k(-1)k-jj(j-1)(αk-j)aj}(βn-k)|]≦2
for some real numbers α and β, then f(z)∈𝒞ℋ.

Putting α=β=0 in the above theorem, we arrive at the following result due to Jahangiri and Silverman [7].

Theorem 2.4.

If f(z)∈ℋ is locally univalent in 𝕌 with
∑n=2∞n2|an|≦1,
then f(z)∈𝒞ℋ.

Furthermore, taking α=1 and β=0 in the theorem, we have the following corollary.

Corollary 2.5.

If f(z)∈ℋ is locally univalent in 𝕌 and satisfies
∑n=2∞{n|(n+1)an-(n-1)an-1|+(n-1)|nan-(n-2)an-1|}≦2,
then f(z)∈𝒞ℋ.

Example 2.6.

The function
f(z)=-∫0zlog(1-t)tdt+(z+(1-z)log(1-z))¯=z+∑n=2∞1n2zn+∑n=2∞1n(n-1)z¯n
satisfies the conditions of Corollary 2.5 and belongs to the class 𝒞ℋ as shown in Figure 4.

The image of f(z)=-∫0z(log(1-t)/t)dt+(z+(1-z)log(1-z))¯.

3. Appendix

A sequence {cn}n=0∞ of nonnegative real numbers is called a convex null sequence if cn→0 as n→∞ andcn-cn+1≧cn+1-cn+2≧0
for all n(n=0,1,2,…).

The next lemma was obtained by Fejér [10].

Lemma 3.1.

Let {cn}k=0∞ be a convex null sequence. Then, the function
p(z)=c02+∑n=1∞cnzn
is analytic and satisfies Re(p(z))>0 in 𝕌.

Applying the above lemma, we deduce the following theorem.

Theorem 3.2.

For some b(|b|<1) and some convex null sequence {cn}n=0∞ with c0=2, the function
f(z)=h(z)+g(z)¯=z+∑n=2∞cn-1nzn+b(z+∑n=2∞cn-1nzn)¯
belongs to the class 𝒞ℋ.

Proof.

Let us define F(z) by
F(z)=h(z)+ɛg(z)1+ɛb=z+∑n=2∞cn-1nzn
for each ɛ(|ɛ|=1). Then, we know that
F′(z)=c02+∑n=1∞cnzn(c0=2).
By virtue of Lemmas 1.7 and 3.1, it follows that Re(F′(z))>0(z∈𝕌), that is, F(z)∈𝒞. Thus, we conclude that f(z)=h(z)+g(z)¯∈𝒞ℋ.

In the same manner, we also have the following theorem.

Theorem 3.3.

For some b(|b|<1) and some convex null sequence {cn}n=0∞ with c0=2, the function
f(z)=h(z)+g(z)¯=z+∑n=2∞1n(1+∑j=1n-1cj)zn+b(z+∑n=2∞1n(1+∑j=1n-1cj)zn)¯
belongs to the class 𝒞ℋ.

Proof.

Let us define F(z) by
F(z)=h(z)+ɛg(z)1+ɛb=z+∑n=2∞1n(1+∑j=1n-1cj)zn
for each ɛ(|ɛ|=1). Then, we know that
(1-z)F′(z)=c02+∑n=1∞cnzn(c0=2).
Therefore, by the help of Lemmas 1.7 and 3.1, we obtain that Re((1-z)F′(z))>0(z∈𝕌), that is, F(z)∈𝒞, which implies that f(z)=h(z)+g(z)¯∈𝒞ℋ.

Remark 3.4.

The sequence
{cn}n=0∞={2,1,23,…,2n+1,…}
is a convex null sequence because
limn→∞cn=limn→∞(2n+1)=0,cn-cn+1=2(n+1)(n+2)≧0,(cn-cn+1)-(cn+1-cn+2)=4(n+1)(n+2)(n+3)≧0(n=0,1,2,…).

Setting b=1/4 in Theorem 3.2 with the above sequence {cn}n=0∞, we derive the following example.

Example 3.5.

The function
f(z)=-z-2∫0zlog(1-t)tdt-14(z+2∫0zlog(1-t)tdt)¯=z+∑n=2∞2n2zn+14(z+∑n=2∞2n2zn)¯
is in the class 𝒞ℋ as shown in Figure 5.

The image of f(z) in Example 3.5.

Moreover, we know the following remark.

Remark 3.6.

The sequence
{cn}n=0∞={2,1,12,…,21-n,…}
is a convex null sequence because
limn→∞cn=limn→∞21-n=0,cn-cn+1=2-n≧0,(cn-cn+1)-(cn+1-cn+2)=2-(n+1)≧0(n=0,1,2,…).

Hence, letting b=1/4 in Theorem 3.3 with the sequence {cn}n=0∞={21-n}n=0∞, we have the following example.

Example 3.7.

The function
f(z)=-3log(1-z)+4log(1-z2)+(-34log(1-z)+log(1-z2))¯=z+∑n=2∞1n(1+∑j=1n-121-j)zn+14(z+∑n=2∞1n(1+∑j=1n-121-j)zn)¯
is in the class 𝒞ℋ as shown in Figure 6.

The image of f(z) in Example 3.7.

Dedication

This paper is dedicated to Professor Owa on the occasion of his retirement from Kinki University.

Acknowledgment

The author expresses his sincere thanks to the referees for their valuable suggestions and comments for improving this paper.

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