We introduced a new subclass of univalent harmonic functions defined by the shear construction in the present paper. First, we showed that the convolutions of two special subclass harmonic mappings are convex in the horizontal direction. Secondly, we proved a necessary and sufficient condition for the above subclass of harmonic mappings to be convex in the horizontal direction. We also presented some basic examples of univalent harmonic functions explaining the behavior of the image domains.

1. Introduction

Let f=u+iv be a continuous complex-valued harmonic mapping in the open unit disc 𝔻={z∈ℂ:|z|<1}, where u and v are real-valued harmonic functions in 𝔻. Such functions can be expressed as f=h+g¯; here h is known as the analytic part and g the coanalytic part of f, respectively. The Jacobian of the mapping f=h+g¯ is given by 𝒥f=|h′|2-|g′|2. A necessary and sufficient condition (see [1] or [2]) for f to be locally univalent and sense-preserving in 𝔻 is that |g′|<|h′|, or equivalently, there exists an analytic complex dilatation ω(z) of f such that
(1)|ω(z)|=|g′(z)h′(z)|<1(h′(z)≠0,z∈𝔻).

We denote by 𝒮H the class of harmonic, sense-preserving, and univalent mappings in 𝔻, normalized by the conditions f(0)=0 and fz(0)=1. Thus, a harmonic mapping in the class 𝒮H can be expressed as f=h+g¯, where
(2)h(z)=z+∑n=2∞anzn,g(z)=∑n=1∞bnzn.

Let 𝒮H0 be the subclass of 𝒮H whose members f satisfy the additional condition fz¯(0)=0. Let 𝒦H0 and 𝒞H0 be the subclasses of 𝒮H0 whose image domains are convex and close to convex, respectively.

A domain Ω∈ℂ is said to be convex in the direction γ, 0≤γ<π, if every line parallel to the line joining 0 and eiγ has a connected intersection with Ω. In particular, a domain is said to be convex in horizontal direction (CHD) if its intersection with each horizontal line is connected (or empty). In this paper, a function f=h+g¯∈𝒮H0 is called a CHD mapping if f maps 𝔻 onto a CHD domain. We denote by 𝒮CHD0 the class of all CHD mappings. Clearly, 𝒮CHD0⊂𝒞H0.

The shear construction is essential to the present work as it allows one to study harmonic functions through their related analytic functions (see [3]); the shear construction produces a univalent harmonic function that maps 𝔻 to a region, that is, CHD. This construction relies on the following theorem due to Clunie and Sheil-Small.

Theorem A (see [<xref ref-type="bibr" rid="B1">1</xref>]).

A harmonic function f=h+g¯ locally univalent in 𝔻 is a univalent mapping of 𝔻 onto a domain convex in the horizontal direction if and only if h-g is a conformal univalent mapping of 𝔻 onto a domain convex in the horizontal direction.

For two harmonic functions; f(z)=h(z)+g(z)¯=z+∑n=2∞anzn+∑n=1∞bn¯z¯n and F(z)=H(z)+G(z)¯=z+∑n=2∞Anzn+∑n=1∞Bn¯z¯n, their convolution is denoted by f*F and defined as follows:
(3)f*F=h*H+g*G¯=z+∑n=2∞anAnzn+∑n=2∞bnBn¯z¯n.

One can find the recent results involving harmonic convolutions in [4–8]. In [9, 10], explicit descriptions are given for half-plane and strip mappings. Specifically, the collection of the mappings f=h+g¯∈𝒮H0 that map 𝔻 onto the right half-plane, 𝕄={ω:Re(ω)>-1/2}. Such mappings satisfy the condition
(4)h(z)+g(z)=z1-z.

In [4], the following result was derived.

Theorem B.

Let f1=h1+g1¯, f2=h2+g2¯∈𝒦H0 be the right half-plane mappings. If f1*f2 is locally univalent and sense-preserving, then f1*f2∈𝒮CHD0.

Let f0=h0+g0¯ be the canonical right half-plane mapping with the dilatation ω0=g0′/h0′=-z; then
(5)h0=z-(1/2)z2(1-z)2,g0=-(1/2)z2(1-z)2.

Recently, Dorff et al. [5] obtained some results involving convolutions of f0 with right half-plane mappings and vertical strip mappings. They proved the following.

Theorem C.

Let f=h+g¯∈𝒦H0 with h+g=z/(1-z) and ω=g′/h′=eiθzn(n∈ℤ+;θ∈ℝ). If n=1,2, then f0*f∈𝒮CHD0.

Theorem D.

Let f=h+g¯∈𝒦H0 with h+g=(1/2isinα)log((1+zeiα)/(1+ze-iα)), where π/2≤α<π and ω=eiθzn. If n=1,2, then f0*f∈𝒮CHD0.

In this paper, we consider the harmonic mapping P=hp+gp¯ satisfies the condition
(6)hp-gp=z1-z,
and when ω=z, applying the shear construction, we have
(7)hp=z-(1/2)z2(1-z)2=12(z(1-z)2+z1-z),gp=(1/2)z2(1-z)2=12(z(1-z)2-z1-z).
Hence
(8)P(z)=12(z(1-z)2+z1-z)+12(z(1-z)2-z1-z)¯.
If F is analytic in 𝔻 and F(0)=0, then
(9)hp(z)*F(z)=12[zF′(z)+F(z)],gp(z)*F(z)=12[zF′(z)-F(z)].

We have known that P(𝔻)={u+iv:v2>-(u+(1/4))}. The image of 𝔻 under the harmonic mapping P(z) is shown in Figure 1.

Image of P(z).

Writing I(z)=z/(1-z), we can express P(z) as
(10)P(z)=zI′(z)+I(z)2+(zI′(z)-I(z)2)¯.

The generalization is done in the following way. For f:𝔻→ℂ a univalent analytic function with f(0)=f′(0)-1=0, define
(11)Pc[f](z)=czf′(z)+f(z)1+c+(czf′(z)-f(z)1+c)¯,(z∈𝔻,c>0).
Clearly, P1[I]=P.

Obviously,
(12)Pc[I](z)=H0(z)+G0(z)¯=czI′(z)+I(z)1+c+(czI′(z)-I(z)1+c)¯=Re{2cz(1+c)(1-z)2}+iIm{2z(1+c)(1-z)}≔U+iV.
Hence
(13)V2=-(2c1+cU+1(1+c)2).
So Pc[I](𝔻)={U+iV:V2>-((2c/(1+c))U+(1/(1+c)2)),c>0}.

If F is analytic in 𝔻 and F(0)=0, then we have
(14)H0(z)*F(z)=11+c[czF′(z)+F(z)],G0(z)*F(z)=11+c[czF′(z)-F(z)].

In [5], authors showed that Theorems C and D do not hold for n≥3. In the present paper, we construct a new subclass of harmonic mappings Pc[f] defined by (11). In Section 2, we show that convolutions of Pc[I] with fn=h+g¯ (where h-g=z/(1-z) and dilatation ω(z)=eiθzn(θ∈ℝ,n∈ℕ+)) are in the subclass 𝒮CHD0 for 0<c≤2/n and for all n∈ℕ+. In Section 3, we apply the transformation to a normalized analytic univalent function f and show that the sufficient and necessary condition for Pc[f]∈𝒮CHD0 is that f is convex. Furthermore, we present some basic examples of harmonic mappings satisfying the conditions of the theorems and illustrate them graphically with the help of the Mathematica software. With these examples we explain the behavior of the image domains.

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To prove our main results, we need several lemmas.

Lemma 1 (see [<xref ref-type="bibr" rid="B4">11</xref>]).

Let f be an analytic function in 𝔻 with f(0)=0 and f′(0)≠0, and let
(15)φ(z)=z(1+zeiθ1)(1+zeiθ2),
where θ1,θ2∈ℝ. If
(16)Re(zf′(z)φ(z))>0(z∈𝔻),
then f is convex in the horizontal direction.

Lemma 2.

Let Pc[I] be a mapping defined by (12) and f=h+g¯∈𝒮H with h-g=z/(1-z) and dilatation ω(z)=(g′(z)/h′(z))(h′(z)≠0,z∈𝔻). Then ω~1, the dilatation of Pc[I]*f, is given by
(17)ω~1=[(c-1)+(c+1)z]ω(1-ω)+czω′(1-z)[(c+1)+(c-1)z](1-ω)+czω′(1-z).

Proof.

Since h-g=z/(1-z) and g′=ωh′, then g′′=ω′h′+ωh′′. We immediately get
(18)h′=1(1-ω)(1-z)2,h′′=2(1-ω)+ω′(1-z)(1-ω)2(1-z)3.
From (14), we have
(19)ω~1=(G0*g)′(H0*h)′=(czg′-g)′(czh′+h)′=(c-1)g′+czg′′(c+1)h′+czh′′=(c-1)ωh′+cz(ω′h′+ωh′′)(c+1)h′+czh′′=[(c-1)+(c+1)z]ω(1-ω)+czω′(1-z)[(c+1)+(c-1)z](1-ω)+czω′(1-z).

Lemma 3.

Let f1=h1+g1¯∈𝒮H0 with h1-g1=z/(1-z). If Pc[I]*f1 is locally univalent, then Pc[I]*f1∈𝒮CHD0.

Proof.

Recall that Pc[I]=H0+G0¯ and
(20)H0-G0=2z(1+c)(1-z),h1-g1=z1-z.
Hence
(21)h1+g1=1+c2(H0-G0)*(h1+g1)=1+c2(H0*h1+H0*g1-G0*h1-G0*g1),H0+G0=(H0+G0)*(h1-g1)=(H0*h1-H0*g1+G0*h1-G0*g1).
Thus
(22)H0*h1-G0*g1=12[21+c(h1+g1)+(H0+G0)].
Next, we will show that (2/(1+c))(h1+g1)+(H0+G0) is convex in the horizontal direction.

Letting φ(z)=z/(1-z)2∈S*, we have(23)Re{z[(2/(1+c))(h1′+g1′)+(H0′+G0′)]φ}=Re{z[21+c(h1′-g1′)(h1′+g1′h1′-g1′)+(H0′-G0′)(H0′+G0′H0′-G0′)]++++×(φ)-121+c(h1′-g1′)(h1′+g1′h1′-g1′)}=Re{21+c(h1′-g1′)(h1′+g1′h1′-g1′)z[21+c(h1′-g1′)(1+ω11-ω1)+(H0′-G0′)(1+ω01-ω0)]++++×(φ)-121+c(h1′-g1′)(h1′+g1′h1′-g1′)}=21+cRe{(z/(1-z)2)[p1(z)+p0(z)]z/(1-z)2}=21+cRe{p1(z)+p0(z)}>0,where ω1=g1′/h1′, ω0=G0′/H0′ and p1(z)=(1+ω1)/(1-ω1), p0(z)=(1+ω0)/(1-ω0). So (2/(1+c))(h1+g1)+(H0+G0) is convex in the horizontal direction by Lemma 1, and then H0*h1-G0*g1 is convex in the horizontal direction. Finally, since we assumed that Pc[I]*f1 is locally univalent, we apply Theorem A to get the desired result.

Now we turn to the distribution theory of the roots of polynomial for the unit disc. Given a polynomial of degree n:
(24)p(z)=p0(z)=an,0zn+an-1,0zn-1+⋯+a1,0z+a0,0(an,0≠0)
with complex coefficients, the parallel algorithm for finding zeros of polynomial (24) inside the unit disc 𝔻 is worth studying. Let
(25)p*(z)=p0*(z)=znp(1z¯)¯=an,0¯+an-1,0¯z+⋯+a1,0¯zn-1+a0,0¯zn,
it is easy to verify that the zeros of (24) and (25) are inverse points with respect to the unit disc 𝔻. To derive the main judging theorem, we need the following lemma.

Lemma 4 (see [<xref ref-type="bibr" rid="B14">12</xref>]).

If all zeros of (24) are inside the unit disc 𝔻, then |a0,0|<|an,0|.

By Lemma 4, we can obtain the following result.

Lemma 5.

If |a0,0|≥|an,0|, then not all zeros of (24) are inside the unit disc 𝔻.

Next, we construct a function sequence
(26){pk(z)∣k=0,1,…,m,(m≤n)},
where
(27)pk(z)=an-k,kzn-k+an-k-1,kzn-k-1+⋯+a1,kz+a0,k,pk*(z)=an-k,k¯+an-k-1,k¯z+⋯+a1,k¯zn-k-1+a0,k¯zn-k,(28)pk+1(z)={pk(z)z,an-k,k=0,an-k,k¯pk(z)-a0,kpk*(z)z,an-k,k≠0,
where k=0,1,2,…,m-1. If |a0,k|≥|an-k,k|, then by Lemma 5 we know that not all roots of pk(z)=0 are inside the unit disc 𝔻; we would not need to construct the function pk+1(z). If |a0,k|<|an-k,k|, then use (28) to construct (26). The following lemma is a necessary and sufficient condition of all zeros of (24) lying inside the unit disc.

Lemma 6 (see [<xref ref-type="bibr" rid="B14">12</xref>]).

A necessary and sufficient condition that all zeros of (24) inside the unit disc 𝔻 is |a0,k|<|an-k,k|(k=0,1,…,n-1), where a0,k and an-k,k are given by (27).

The main result of this section is the following.

Theorem 7.

Let Pc[I] be a mapping defined by (12) and fn=h+g¯ with h-g=z/(1-z) and dilatation ω(z)=eiθzn(θ∈ℝ,n∈ℕ). Then Pc[I]*fn∈𝒮CHD0 for 0<c≤2/n.

Proof.

In view of Lemma 3, it suffices to show that the dilatation of Pc[I]*fn satisfies |ω~1(z)|<1, for all z∈𝔻. Setting ω(z)=eiθzn into (17), we get(29)ω~1=([(c-1)+(c+1)z]eiθzn(1-eiθzn)+cneiθzn(1-z))×([(c+1)+(c-1)z](1-eiθzn)+cneiθzn(1-z))-1=-e2iθzn×((1-(n+1)c1+ceiθzn+1)-1(zn+1-1-c1+czn-1-(n-1)c1+ce-iθz+1-(n+1)c1+ce-iθ)×(1-1-c1+cz-1-(n-1)c1+ceiθzn+1-(n+1)c1+ceiθzn+1)-1)=-e2iθznp(z)p*(z),where
(30)p(z)=zn+1-1-c1+czn-1-(n-1)c1+ce-iθz+1-(n+1)c1+ce-iθ
and p*(z)=zn+1p(1/z¯)¯.

Obviously, if z0 is a zero of p(z), the 1/z0¯ is a zero of p*(z). Hence, if A1,A2,…,An+1 are the zeros of p(z), we can write
(31)ω~1=-e2iθzn(z-A1)(1-A1¯z)·(z-A2)(1-A2¯z)⋯(z-An+1)(1-An+1¯z).

Now for |Ai|≤1, ((z-Ai)/(1-Ai¯z))(i=1,2,…,n+1) maps 𝔻¯ onto 𝔻¯. So in order to prove our result, we will show that all zeros of p(z) named A1,A2,…,An+1 lie inside the unit disc 𝔻¯ for 0<c≤2/n. In the following we divide our proof in two cases.

Case 1 (when c=2/n). In this case, substituting c=2/n into (29), we see that (32)|ω~1|=|-e2iθzn((n-2n+2)-1(zn+1-n-2n+2zn+n-2n+2e-iθz-e-iθ)-e2iθzn+i×(1-n-2n+2z+n-2n+2eiθzn-eiθzn+1)-1)|=|eiθzn((1-n-2n+2z+n-2n+2eiθzn-eiθzn+1)-1(-eiθzn+1+n-2n+2eiθzn-n-2n+2z+1)-e2iθzn+×(1-n-2n+2z+n-2n+2eiθzn-eiθzn+1)-1)|=|eiθzn|<1.Case 2 (when 0<c<2/n). From (31), it is enough to show that all zeros of (30) are inside the unit disc 𝔻¯ for 0<c<2/n. Since |a0,0|=|((1-(n+1)c)/(1+c))e-iθ|<1=|an+1,0|, we can apply (28) to p(z); thus we have
(33)p1(z)=an+1,0¯p(z)-a0,0p*(z)z=p(z)-((1-(n+1)c)/(1+c))e-iθp*(z)z=(n+2)c(2-nc)(1+c)2zn-nc(2-nc)(1+c)2zn-1-2c(2-nc)(1+c)2e-iθ=(n+2)c(2-nc)(1+c)2(zn-nn+2zn-1-2n+2e-iθ)=(n+2)c(2-nc)(1+c)2q1(z),
where q1(z)=zn-(n/(n+2))zn-1-(2/(n+2))e-iθ and q1*(z)=1-(n/(n+2))z-(2/(n+2))eiθzn.

Since |a0,1|=|-(2/(n+2))e-iθ|=2/(n+2)<1=|an,1| for n≥1, by using (28) on q1(z) again, we get
(34)p2(z)=an,1¯q1(z)-a0,1q1*(z)z=q1(z)+(2/(n+2))e-iθq1*(z)z=n(n+4)(n+2)2(zn-1-n+2n+4zn-2-2n+4e-iθ)=n(n+4)(n+2)2q2(z),
where q2(z)=zn-1-((n+2)/(n+4))zn-2-(2/(n+4))e-iθ and q2*(z)=1-((n+2)/(n+4))z-(2/(n+4))eiθzn-1.

Since |a0,2|=|-(2/(n+4))eiθ|=2/(n+4)<1=|an-1,2|; we have
(35)p3(z)=an-1,2¯q2(z)-a0,2q2*(z)z=q1(z)+(2/(n+4))e-iθq1*(z)z=(n+2)(n+6)(n+4)2(zn-2-n+4n+6zn-3-2n+6e-iθ)=(n+2)(n+6)(n+4)2q3(z).
Continuing in this manner we derive that
(36)pk(z)=an+1-k,k¯qk(z)-a0,kqk*(z)z=[n+2(k-2)](n+2k)[n+2(k-1)]2×(zn+1-k-n+2(k-1)n+2kzn-k-2n+2ke-iθ)=[n+2(k-2)](n+2k)[n+2(k-1)]2qk(z)(k=2,3,…,n),
where qk(z)=zn+1-k-((n+2(k-1))/(n+2k))zn-k-(2/(n+2k))e-iθ and then |a0,k|=|-(2/(n+2k))e-iθ|=2/(n+2k)<1=|an+1-k,k|. Hence, let k=n, we have
(37)pn(z)=3n(3n-4)(3n-2)2(z-3n-23n-23ne-iθ).
If θ≠2mπ, m∈ℕ, then |a0,n|=|(3n-2+2e-iθ)/3n|<1=|a1,n|; by Lemma 6, thus, A1,A2,…,An+1 lie inside unit disc.

If θ=2mπ,m∈ℕ, then p(z)=zn+1-((1-c)/(1+c))zn-((1-(n-1)c)/(1+c))z+((1-(n+1)c)/(1+c)). As z=1 is a zero of p(z), therefore we can write
(38)p(z)=(z-1)×[zn+(1-1-c1+c)zn-1+(1-1-c1+c)zn-2+++⋯+(1-1-c1+c)z-1-(n+1)c1+c]=(z-1)Q(z).
It suffices to show that zeros of Q(z) lie inside unit disc. Since |-(1-(n+1)c)/(1+c)|<1 whenever 0<c<2/n, by applying (28) on Q(z), we obtain analogously that all zeros of Q(z) lie inside unit disc.

To sum up, we showed that all zeros of p(z) lie inside or on the unit circle |z|=1 for 0<c≤2/n. The proof of our theorem is now completed.

Next we give an example showing given values of θ, n, and c in Theorem 7.

Example 8.

In Theorem 7, if we take θ=0, n=1, then h1-g1=z/(1-z) and ω=z, and we get f1=h1+g1¯=P(z)=((z-(1/2)z2)/(1-z)2)+((1/2)z2/(1-z)2¯). Let c=1; then P1[I]=P(z) and P1[I]*f1=P(z)*P(z)=H1+G1¯. From (7) and (9), we have
(39)H1=hp*hp=12[zhp′+hp]=4z-3z2+z34(1-z)3,G1=gp*gp=12[zgp′-gp]=z2+z34(1-z)3.

The images of concentric circles inside 𝔻 under the harmonic mapping P(z) and concentric circles under the convolution map P(z)*P(z) are shown in Figures 1 and 2, respectively.

Image of P(z)*P(z).

3. Harmonic Univalent Mappings Convex in One Direction

For the local univalence of Pc[f] defined by (11), we have the following.

Lemma 9.

The function Pc[f] defined by (11) is locally univalent if and only if f is convex.

Proof.

Write Pc[f]=F=H+G¯. By (1), F is locally univalent if and only if |ω(z)|=|G′(z)/H′(z)|<1 for all z∈𝔻, and F will be locally univalent if and only if
(40)|ω(z)|=|(c-1)f′(z)+czf′′(z)(c+1)f′(z)+czf′′(z)|<1,
or equivalently, since c>0 and f∈𝒮,
(41)|(1+zf′′(z)f′(z))-1c|<|(1+zf′′(z)f′(z))+1c|.
It can be easily seen that the inequality above is equivalent to
(42)Re(1+zf′′(z)f′(z))>0,
which is the analytic condition for convexity. Thus, Pc[f] is locally univalent if and only if f is convex.

We immediately have the following result.

Theorem 10.

The function Pc[f]∈𝒮CHD0 is defined by (11) if and only if f is convex.

Proof.

Again write Pc[f]=F=H+G¯. Since H-G=2f/(1+c) is convex, particularly, H-G is convex in the horizontal direction; by Lemma 9 and Theorem A, the proof is complete.

The following example satisfies the condition of Theorem 10.

Example 11.

Consider the univalent analytic function f(z)=z+(z2/4); obviously, f(0)=f′(0)-1=0, and
(43)Re(1+zf′′(z)f′(z))=Re(1+z/21+(z/2))>0,forz∈𝔻.
So, f(z) is convex analytic in 𝔻. Therefore, in view of Theorem 10, the function Pc[f](z)∈SCHD0. However
(44)Pc[f](z)=Re(2czf′(z)1+c)+iIm(2f(z)1+c).
Now, we let c=1; then
(45)P1[f](z)=Re(z+z22)+iIm(z+z24).
The images of 𝔻 under f(z) and P1[f](z) are shown in Figures 3 and 4, respectively. Obviously, we see that the image of 𝔻 under P1[f](z) is convex in the horizontal direction.

Image of f(z)=z+z2/4.

Image of P1[f](z).

Conflict of Interests

The authors declare that they have no competing interests.

Authors’ Contribution

The authors completed the paper together. They also read and approved the final paper.

Acknowledgments

The present investigation was supported by the National Natural Science Foundation under Grants 11301008 and 11226088, the Key Courses Construction of Honghe University underGrant ZDKC1003 and the Scientific Research Fund of Yunnan Province under Grant 2010ZC150.

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