Show Summary Details
In This Section

# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

1 Issue per year

IMPACT FACTOR 2016 (Open Mathematics): 0.682
IMPACT FACTOR 2016 (Central European Journal of Mathematics): 0.489

CiteScore 2016: 0.62

SCImago Journal Rank (SJR) 2015: 0.521
Source Normalized Impact per Paper (SNIP) 2015: 1.233

Mathematical Citation Quotient (MCQ) 2015: 0.39

Open Access
Online
ISSN
2391-5455
See all formats and pricing
In This Section

# Convolutions of harmonic right half-plane mappings

YingChun Li
• College of Mathematics, Honghe University, Mengzi 661199, Yunnan, China
• Email:
/ ZhiHong Liu
• Corresponding author
• College of Mathematics, Honghe University, Mengzi 661199, Yunnan, China
• School of Mathematics and Econometrics, Hunan University, Changsha 410082, Hunan, China
• Email:
Published Online: 2016-10-20 | DOI: https://doi.org/10.1515/math-2016-0069

## Abstract

We first prove that the convolution of a normalized right half-plane mapping with another subclass of normalized right half-plane mappings with the dilatation $-z\left(a+z\right)/\left(1+az\right)$ is CHD (convex in the horizontal direction) provided $a=1$ or $-1\le a\le 0$. Secondly, we give a simply method to prove the convolution of two special subclasses of harmonic univalent mappings in the right half-plane is CHD which was proved by Kumar et al. [1, Theorem 2.2]. In addition, we derive the convolution of harmonic univalent mappings involving the generalized harmonic right half-plane mappings is CHD. Finally, we present two examples of harmonic mappings to illuminate our main results.

MSC 2010: 30C45; 58E20

## 1 Introduction and main results

Assume that $f=u+iv$ is a complex-valued harmonic function defined on the open unit disk $\mathbb{U}=\left\{z\in \mathbb{C}\phantom{\rule{thickmathspace}{0ex}}:\phantom{\rule{thickmathspace}{0ex}}|z|<1\right\}$, where u and v are real harmonic functions in $\mathbb{U}$. Such function can be expressed as $f=h+\overline{\mathrm{g}}$, where $h(z)=z+∑n=1∞anznandg(z)=∑n=1∞bnzn$

are analytic in $\mathbb{U},h$ and g are known as the analytic part and co-analytic part of f, respectively. A harmonic mapping $f=h+\overline{g}$ is locally univalent and sense-preserving if and only if the dilatation of f defined by $\omega \left(z\right)={g}^{\prime }\left(z\right)/{h}^{\prime }\left(z\right)$, satisfies $\left|\omega \left(z\right)\right|<1$ for all $z\in \mathbb{U}$.

We denote by ${\mathcal{S}}_{H}$ the class of all harmonic, sense-preserving and univalent mappings $f=h+\overline{g}$ in $\mathbb{U}$, which are normalized by the conditions $h\left(0\right)=g\left(0\right)=0$ and ${h}^{\prime }\left(0\right)=1$. Let ${\mathcal{S}}_{H}^{0}$ be the subset of all $f\in {\mathcal{S}}_{H}$ in which ${g}^{\prime }\left(0\right)=0$. Further, let ${\mathcal{K}}_{H},{\mathcal{C}}_{H}$ (resp. ${\mathcal{K}}_{H}^{0},{\mathcal{C}}_{H}^{0}$) be the subclass of ${\mathcal{S}}_{H}$ (resp. ${\mathcal{S}}_{H}^{0}$) whose images are convex and close-to-convex domains. A domain Ω is said to be convex in the horizontal direction (CHD) if every line parallel to the real axis has a connected intersection with Ω.

For harmonic univalent functions $f(z)=h(z)+g(z)¯=z+∑n=2∞anzn+∑n=1∞bn¯z¯n$

and $F(z)=H(z)+G(z)¯=z+∑n=2∞Anzn+∑n=1∞Bn¯z¯n,$

the convolution (or Hadamard product) of them is given by $(f∗F)(z)=(h∗H)(z)+(g∗G)(z)¯=z+∑n=2∞anAnzn+∑n=1∞bnBn¯z¯n.$

Many research papers in recent years have studied the convolution or Hadamard product of planar harmonic mappings, see [2-12]. However, corresponding questions for the class of univalent harmonic mappings seem to be difficult to handle as can be seen from the recent investigations of the authors [13-17]. In [1], Kumar et al. constructed the harmonic functions ${f}_{a}={h}_{a}+\overline{{g}_{a}}\in {\mathcal{K}}_{H}$ in the right half-plane, which satisfy the conditions ${h}_{a}+{g}_{a}=z/\left(1-z\right)$ with ${\omega }_{a}\left(z\right)=\left(a-z\right)/\left(1-az\right)\left(-1 . By using the technique of shear construction (see [18]), we have $ha(z)=11+az−12z2(1−z)2andga(z)=a1+az−12z2(1−z)2.$(1)

Obviously, for $a=0,{f}_{0}\left(z\right)={h}_{0}\left(z\right)+\overline{{g}_{0}\left(z\right)}\in {\mathcal{K}}_{H}^{0}$ is the standard right half-plane mapping, where $h0(z)=z−12z2(1−z)2andg0(z)=−12z2(1−z)2.$(2)

Recently Dorff et al. studied the convolution of harmonic univalent mappings in the right half-plane (cf. [14, 15 They proved that:

Theorem A: Let ${f}_{1}={h}_{1}+\overline{{g}_{1}},\phantom{\rule{thickmathspace}{0ex}}{f}_{2}={h}_{2}+\overline{{g}_{2}}\in {\mathcal{S}}_{H}^{0}$ with ${h}_{i}+{\mathrm{g}}_{i}=z/\left(1-z\right)$ for i = 1,2. If f1 * f2 it is locally univalent and sense-preserving, then ${f}_{1}\ast {f}_{2}\in {\mathcal{S}}_{H}^{0}$ and convex in the horizontal direction.

Theorem B: Let $\phantom{\rule{thickmathspace}{0ex}}{f}_{n}=h+\overline{g}\in {\mathcal{S}}_{H}^{0}$ with $h+g=z+\left(1-\mathrm{z}\right)$ and $\omega \left(z\right)={g}^{\prime }\left(z\right)/{h}^{\prime }\left(z\right)={e}^{i\theta }{z}^{n}\left(\theta \in \mathbb{R},\phantom{\rule{thickmathspace}{0ex}}n\in {\mathbb{N}}^{+}\right)$. If n = 1,2, then ${f}_{0}\ast {f}_{n}\in {S}_{H}^{0}$ and is convex in the horizontal direction.

Theorem C: Let $f=h+\overline{g}\in {\mathcal{K}}_{H}^{0}$ with $h\left(z\right)+g\left(z\right)=z/\left(1-z\right)$ and $\omega \left(z\right)=\left(z+a\right)/\left(1+az\right)$ with $a\in \left(-1,1\right)$. Then ${f}_{0}\ast f\in {S}_{H}^{0}$ and is convex in the horizontal direction.

We now begin to state the elementary result concerning the convolutions of f0 with the other special subclass harmonic mappings.

Theorem 1.1: Let $f=h+\overline{g}\in {S}_{H}^{0}$ with $h\left(z\right)+g\left(z\right)=z/\left(1-z\right)$ and $\omega \left(z\right)=-z\left(z+a\right)/\left(1+az\right)$, then ${f}_{0}\ast f\in {\mathcal{S}}_{H}^{0}$ and is convex in the horizontal direction for $a=1\phantom{\rule{thickmathspace}{0ex}}or-1\le a\le 0$.

The following generalized right half-plane harmonic univalent mappings were introduced by Muir [7]: $Lc(z)=Hc(z)+Gc(z)¯=11+cZ1−z+CZ(1−z)2+11+cZ1−z−CZ(1−z)2¯(z∈U;c>0).$(3)

Clearly, ${L}_{1}\left(z\right)={f}_{0}\left(z\right)$, it was proved in [7] that ${L}_{c}\left(\mathbb{U}\right)=\left\{\mathrm{R}\mathrm{e}\left(\omega \right)>-1/\left(1+c\right)\right\}$ for each c > 0. Moreover, if $f=h+\overline{g}\in {\mathcal{S}}_{H}$, then the above representation gives that $Lc∗f=h+czh′1+c+g−czg′1+c.¯$(4)

The following Cohn’s Rule is helpful in proving our main results.

Cohn’s Rule: Given a polynomial $p(z)=p0(z)=an;0zn+an−1;0zn−1+⋯+a1;0z+a0;0(an;0≠0)$(5)of degree n, let $p∗(z)=p0∗(z)=znp(1/z¯)¯=an;0¯+an−1;0¯z+⋯+a1;0¯zn−1+a0;0¯zn$(6)Denote by r and s the number of zeros of p(z) inside the unit circle and on it, respectively. If $\left|{a}_{0,0}\right|<\left|{a}_{n,0}\right|$, then $p1(z)=an;0¯p(z)−a0;0p∗(z)Z$ is of degree n – 1 with r1 = r – 1 and s1 = s the number of zeros of p1 (z) inside the unit circle and on it, respectively.

It should be remarked that Cohn’s Rule and Schur-Cohn’s algorithm [19, p. 378] are important tools to prove harmonic mappings are locally univalent and sense-preserving. Some related works have been done on these topics, one can refer to [1, 3, 6, 13, 15, 16]. In [1], the authors proved the following result.

Theorem D: Let ${f}_{a}={h}_{a}+\overline{{g}_{a}}$ be given by (1). If ${f}_{n}=h+\overline{g}$ is the right half-plane mapping given by $h+g=z/\left(1-z\right)$ with $\omega \left(z\right)={e}^{i\theta }{z}^{n}\left(\theta \in \mathbb{R},\phantom{\rule{thickmathspace}{0ex}}n\in {\mathbb{N}}^{+}\right)$, then ${f}_{a}\ast {f}_{n}\in {\mathcal{S}}_{H}$ is CHD for $a\in \left[\frac{n-2}{n+2},1)\right$.

In this paper, we will use a new method to prove the above theorem. The main difference of our work from [1] is that we construct a sequence of functions for finding all zeros of polynomials which are in $\overline{\mathbb{U}}$, and we will show that the dilatation of fa * fn satisfies $|{\stackrel{~}{\omega }}_{1}\left(z\right)|\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}|{\left({g}_{a}\ast g\right)}^{\mathrm{\prime }}/\left({h}_{a}\ast h{\right)}^{\prime }|<1$ by using mathematical induction, which greatly simplifies the calculation compared with the proof of Theorem 2.2 in [1]. We also show that L_{c * fa is univalent and convex in the horizontal direction for $0, and derive the following theorem.

Theorem 1.2: Let ${L}_{c}={H}_{c}+\overline{{G}_{c}}$ be harmonic mappings given by (3). If ${f}_{a}={h}_{a}+{\overline{g}}_{a}$ is the right half-plane mappings given by (1), then Lc * fa is univalent and convex in the horizontal direction for $0.

Recently, Liu and Li [6] defined a subclass of harmonic mappings defined by $Pc(z)=Hc(z)−Gc(z)¯=11+cCZ(1−z)2+Z1−z+11+cCZ(1−z)2−Z1−z¯(z∈U;c>0).$(7)

They proved the following result.

Theorem E: Let Pc(z) be harmonic mappings defined by (7) and ${f}_{n}=h+\overline{g}$ with hg = z/(1–z) and dilatation $\omega \left(z\right)={e}^{i\theta }{z}^{n}\left(\theta \in \mathbb{R},\phantom{\rule{thickmathspace}{0ex}}n\in {\mathbb{N}}^{+}\right)$. Then Pc * fn is univalent and convex in the horizontal direction for $0.

Similar to the approach used in the proof of Theorem E, we get the following result.

Theorem 1.3: Let ${L}_{c}={H}_{c}+\overline{{G}_{c}}$ be harmonic mappings given by (3) and ${f}_{n}=h+\overline{g}$ with h + g = z/(1–z) and dilatation $\omega \left(z\right)={e}^{i\theta }{z}^{n}\left(\theta \in \mathbb{R},\phantom{\rule{thickmathspace}{0ex}}n\in {\mathbb{N}}^{+}\right)$. Then Lc * fn is univalent and convex in the horizontal direction for $0.

Remark 1.4: In Theorem 1.2 and Theorem 1.3, let c = 1, clearly, L1 = f0, so Theorem 1.2 and Theorem 1.3 are generalization of Theorem C and Theorem B, respectively. It also explains why Theorem B does not hold for n ≥ 3, since c = 1, according to Theorem 1.3, it follows that $0 hold for n = 1,2.

## 2 Preliminary results

In this section, we will give the following lemmas which play an important role in proving the main results.

Lemma 2.1: If $f=h+\overline{g}\in {\mathcal{S}}_{H}^{0}$ with h + g = z/(1–z) and dilatation $\omega \left(z\right)={g}^{\prime }\left(z\right)/{h}^{\prime }\left(z\right)$, then the dilatation of f0 * f is given by $ω¯(z)=−zω2+[ω−12ω′z]+12ω′1+[ω−12ωz]+12ωz2.$(8)

Lemma 2.2: Let ${f}_{a}={h}_{a}+\overline{{g}_{a}}$ be defined by (1) and $f=h+\overline{g}\in {\mathcal{S}}_{H}^{0}$ be right half-plane mapping, where h + g = z/(1–z) with dilatation $\omega \left(z\right)={g}^{\prime }\left(z\right)/{h}^{\prime }\left(z\right)\left({h}^{\prime }\left(z\right)\ne 0,\phantom{\rule{thickmathspace}{0ex}}z\in \mathbb{U}\right)$. Then ${\stackrel{~}{\omega }}_{1}$, the dilatation of fa * f, is given by $ω¯1(z)=2(a−z)ω(1+ω)+(a−1)ω′z(1−z)2(1−az)(1+ω)+(a−1)ωz(1−z).$(9)

Lemma 2.3: Let ${L}_{c}={H}_{c}+\overline{{G}_{c}}$ be defined by (3) and $f=h+\overline{g}\in {\mathcal{S}}_{H}^{0}$ be right half-plane mapping, where h + g = z/(1–z) with dilatation $\omega \left(z\right)={g}^{\prime }\left(z\right)/{h}^{\prime }\left(z\right)\left({h}^{\prime }\left(z\right)\ne 0,\phantom{\rule{thickmathspace}{0ex}}z\in \mathbb{U}\right)$ . Then ${\stackrel{~}{\omega }}_{2}$, the dilatation of Lc * f, is given by $ω~2(z)=[(1−c)−(1+c)z]ω(1+ω)−cω′z(1−z)[(1+c)−(1−c)z](1+ω)−cω′z(1−z)$(10)

Proof: By (4), we know that $ω¯2(z)=(g−czg′)′(h+czh′)′.$ Similar calculation as in the proof of [6, Lemma 7] gives (10). □

Lemma 2.4: A harmonic function $f=h+\overline{g}$ locally univalent in $\mathbb{U}$ is a univalent mapping of $\mathbb{U}$ onto a domain convex in the horizontal direction if and only if hg is a conformal univalent mapping of $\mathbb{U}$ onto a domain convex in the horizontal direction.

Lemma 2.5: Let f be an analytic function in $\mathbb{U}$ with f(0) = 0 and ${f}^{\prime }\left(0\right)\ne 0$, and let $φ(z)=Z(1+zeiθ1)(1+zeiθ2)$(11)where ${\theta }_{1},{\theta }_{2}\in \mathbb{R}$. If $Re(zf′(z)(z))>0(z∈U).$ Then f is convex in the horizontal direction.

Lemma 2.6: Let ${L}_{c}={H}_{c}+\overline{{G}_{c}}$ be a mapping given by (3) and $f=h+\overline{g}$ be the right half-plane mapping with h + g = z/(1–z) . If Lc * f is locally univalent, then ${L}_{c}\ast f\in {\mathcal{S}}_{H}^{0}$ and is convex in the horizontal direction.

Proof: Recalling that ${L}_{c}={H}_{c}+\overline{{G}_{c}}$ and$Hc+Gc=2z(1+c)(1−z),h+g=Z1−z.$Hence $h−g=1+c2(Hc+Gc)∗(h−g)=1+c2(Hc∗h−Hc∗g+Gc∗h−Gc∗g),Hc−Gc=(Hc−Gc)∗(h+g)=(Hc∗h+Hc∗g−Gc∗h−Gc∗g).$ Thus $Hc∗h−Gc∗g=1221+c(h−g)+(Hc−Gc).$(12)Next, we will show that $\frac{2}{1+c}\left(h-g\right)+\left({H}_{c}-{G}_{c}\right)$ is convex in the horizontal direction. Letting $\phi \left(z\right)=z/\left(1-z{\right)}^{2}\in {\mathcal{S}}^{\ast }$, we have $ReZφ21+c(h′−g′)+(Hc′−Gc′)=Rez21+c(h′+g′)(h′−g′h+g′)+(Hc′+Gc′)(Hc′−Gc′Hc′+Gc′)φ=Rez21+c(h′+g′)(1−ω1+ω)+(Hc′+Gc′)(1−ωc1+ωc)φ=21+cReZ(1−z)2[r(z)+rc(z)]z(1−z)2=21+cRe{r(z)+rc(z)}>0,$ where $r\left(z\right)=\frac{1-\omega \left(z\right)}{1+\omega \left(z\right)},{r}_{c}\left(z\right)=\frac{1-{\omega }_{\mathcal{C}}\left(z\right)}{1+{\omega }_{\mathcal{C}}\left(z\right)}$. Therefore, by Lemma 2.5 and the equation (12), we know that ${H}_{c}\ast h-{G}_{c}\ast g$ is convex in the horizontal direction.Finally, since we assumed that Lc * f is locally univalent and ${H}_{c}\ast h-{G}_{c}\ast g$ is convex in the horizontal direction, we apply Lemma 2.4 to obtain that ${L}_{c}\ast f={H}_{c}\ast h+\overline{{G}_{c}\ast g}$ is convex in the horizontal direction. □

## 3 Proofs of theorems

Proof of Theorem 1.1: By Theorem A, we know that f0f is convex in the horizontal direction. Now we need to establish that f0f is locally univalent.Substituting ω(z) = −z(z + a)/(1 + az) into (8) and simplifying, yields $ω~(z)=Zz3+2+3a2z2+(1+a)z+a21+2+3a2z+(1+a)z2+a2z3=Zq(z)q∗(z)=Z(z−A)(z−B)(z−C)(1−A¯z)(1−B¯z)(1−C¯z),$(13)If a = 1, then $q\left(z\right)={z}^{3}+\frac{5}{2}{z}^{2}+2z+\frac{1}{2}=\frac{1}{2}\left(1+z{\right)}^{2}\left(1+2z\right)$ has all its three zeros in $\overline{\mathbb{U}}$. By Cohn's Rule, so $|\stackrel{~}{\omega }\left(z\right)|<1$ for all $z\in \mathbb{U}$.If a = 0, it is clearly that $|\stackrel{~}{\omega }\left(z\right)|=|{z}^{2}|<1$ for all $z\in \mathbb{U}$.If −l ≤ a < 0, we apply Cohn’s Rule to $q\left(z\right)={z}^{3}+\frac{2+3a}{2}{z}^{2}+\left(1+a\right)z+\frac{a}{2}$. Note that $|\frac{a}{2}|<1$, thus we get $Q(z)=a3¯q(z)−a0q∗(z)Z=4−a24z2+2(2+2a−a2)4−a2z+4+2a−3a24−a2:=4−a24q1(z).$Since $4+2a−3a24−a2=1+2a(1−a)4−a2<1(as−l≤a<0),$we use Cohn’s Rule on q1(z) again, we get $q2(z)=q1(z)−4+2a−3a24−a2q1∗(z)Z=4a(a−1)(4+a−2a2)(4−a2)2Z+2+2a−a24+a−2a2.$Clearly, q2(z) has one zero at $z0=−2+2a−a24+a−2a2.$We show that |z0 ≤ 1, or equivalently, $|2+2a−a2|2−|4+a−2a2|2=(2+2a−a2)2−(4+a−2a2)2=3(4−a2)(a2−1)≤0.$Therefore, by Cohn’s Rule, q(z) has all its three zeros in $\overline{\mathbb{U}}$, that is $A,B,C\in \overline{\mathbb{U}}$ and so $|\stackrel{~}{\omega }\left(z\right)|<1$ for all $z\in \mathbb{U}$.□

A new proof of Theorem D: By Theorem A, it suffices to show that the dilatation of fafn satisfies $|\stackrel{~}{\omega }\left(z\right)|<1$ for all $z\in \mathbb{U}$. Setting $\omega \left(z\right)={e}^{i\theta }{z}^{n}$ in (9), we have $ω~1(z)=−zne2iθzn+1−azn+12(2+an−n)e−iθz+12(n−2a−an)e−iθ1−az+12(2+an−n)eiθzn+12(n−2a−an)eiθZn+1=−zne2iθp(z)p∗(z),$(14)where $p(z)=zn+1−azn+12(2+an−n)e−iθz+12(n−2a−an)e−iθ,$(15)and $p∗(z)=zn+1p(1/z¯)¯.$Firstly, we will show that $|{\stackrel{~}{\omega }}_{1}\left(z\right)|<1$ for $a=\frac{n-2}{n+2}$. In this case, substituting $a=\frac{n-2}{n+2}$ into (14), we have $|ω~1(z)|=−zne2iθzn+1−n−2n+2zn−n−2n+2e−iθz+e−iθ1−n−2n+2z−n−2n+2eiθzn+eiθZn+1=−zneiθeiθzn+1−n−2n+2eiθzn−n−2n+2z+11−n−2n+2z−n−2n+2eiθzn+eiθZn+1=|−zneiθ|<1.$Next, we will show that $|{\stackrel{~}{\omega }}_{1}\left(z\right)|<1$ for $\frac{n-2}{n+2}. Obviously, if z0 is a zero of p(z), then $1/\overline{{z}_{0}}$ is a zero of p(z). Hence, ifA1, A2, ⋯, An + 1 are the zeros of p(z) (not necessarily distinct), then we can write $ω~1(z)=−zne2iθ(z−A1)(1−A1¯z)(z−A2)(1−A2¯z)⋯(z−An+1)(1−An+1¯z).$Now for $\left|{A}_{j}\right|\le 1,\frac{z-{A}_{j}}{1-\overline{{A}_{j}}z}\left(j=1,2,\cdots ,n+1\right)$ maps $\mathbb{U}$ onto $\mathbb{U}$. It suffices to show that all zeros of (15) lie on $\overline{\mathbb{U}}$ for $\frac{n-2}{n+2}. Since $|{a}_{0,0}|=|\frac{1}{2}\left(n-2a-an\right){e}^{-i\theta }|<|{a}_{n+1,0}|=1$, then by using Cohn’s Rule on p(z), we get $p1(z)=an+1;0¯p(z)−a0;0p∗(z)Z=p(z)−12(n−2a−an)e−iθp∗(z)Z=(1−a)(2+n)[2(1+a)−(1−a)n]4(zn−nn+2zn−1+2n+2e−iθ).$Since $\frac{n-2}{n+2}, we have $\frac{\left(1-a\right)\left(2+n\right)\left[2\left(1+a\right)-\left(1-a\right)n\right]}{4}>0$. Let ${q}_{1}\left(z\right)={z}^{n}-\frac{n}{n+2}{z}^{n-1}+\frac{2}{n+2}{e}^{-i\theta }$, since $|{a}_{0,1}|=|\frac{2}{n+2}{e}^{-i\theta }|<1=|{a}_{n,1}|$, by using Cohn’s Rule on q1(z) again, we obtain $p2(z)=an;1¯q1(z)−a0;1q1∗(z)Z=q1(z)−2n+2e−iθq1∗(z)Z=n(n+4)(n+2)2zn−1−n+2n+4zn−2+2n+4e−iθ.$Let ${q}_{2}\left(z\right)={z}^{n-1}-\frac{n+2}{n+4}{z}^{n-2}+\frac{2}{n+4}{e}^{-i\theta }$, then $|{a}_{0,2}|=\frac{2}{n+4}<1=|{a}_{n-1,2}|$, we get $p3(z)=an−1;2¯q2(z)−a0;2q2∗(z)Z=q2(z)−2n+4e−iθq2∗(z)Z=(n+2)(n+6)(n+4)2zn−2−n+4n+6zn−3+2n+6e−iθ.$Let ${q}_{3}\left(z\right)={z}^{n-2}-\frac{n+4}{n+6}{z}^{n-3}+\frac{2}{n+6}{e}^{-i\theta }$, then $|{a}_{0,3}|=\frac{2}{n+6}<1=|{a}_{n-2,3}|$, we obtain $p4(z)=an−1;3¯q3(z)−a0;3q3∗(z)Z=q3(z)−2n+6e−iθq3∗(z)Z=(n+4)(n+8)(n+6)2zn−3−n+6n+8zn−4+2n+8e−iθ.$By using this manner, we claim that $Pk(z)=[n+2(k−2)](n+2k)[n+2(k−1)]2zn−k+1−n+2(k−1)n+2kzn−k+2n+2ke−iθ.$(16)where k = 2, 3, ⋯, n.To prove the equation (16) is correct for all $k\in {\mathbb{N}}^{+}\left(k\ge 2\right)$, it suffices to show $Pk+1(z)=[n+2(k−1)][n+2(k+1)](n+2k)2zn−k−n+2kn+2(k+1)zn−⋅k+1)+2n+2(k+1)e−iθ.$(17)Let ${q}_{k}\left(z\right)={z}^{n-k+1}-\frac{n+2\left(k-1\right)}{n+2k}{z}^{n-k}+\frac{2}{n+2k}{e}^{-i\theta }$, then ${q}_{k}^{\ast }\left(z\right)={z}^{n-k+1}\overline{{q}_{k}\left(1/\overline{z}\right)}=1-\frac{n+2\left(k-1\right)}{n+2k}z+\frac{2}{n+2k}{e}^{-i\theta }{z}^{n-k+1}$. Since $|{a}_{0,k}|=|\frac{2}{n+2k}{e}^{-i\theta }|=\frac{2}{n+2k}<1=|{a}_{n-k+1,k}|$, by using Cohn’s Rule on qk(z), we deduce that $Pk+1(z)=an−k+1,k¯qk(z)−a0,kqk∗(z)Z=qk(z)−2n+2ke−iθqk∗(z)Z=[n+2(k−1)][n+2(k+1)](n+2k)2zn−k−n+2kn+2(k+1)zn−(k+1)+2n+2(k+1)e−iθ.$Setting n = k(k ≥ 2) in (16), we have $Pn(z)=3n(3n−4)(3n−2)2(z−3n−2−2e−iθ3n).$Then ${z}_{0}=\frac{3n-2-2{e}^{-i\theta }}{3n}$ is a zero of pn(z), and $|z0|=3n−2−2e−iθ3n≤|3n−2|+|2e−iθ|3n=3n−2+23n=1.$So z0 lies inside or on the unit circle |z| = 1, by Lemma 1, we know that all zeros of (15) lie on $\overline{\mathbb{U}}$. This completes the proof.□

Proof of Theorem 1.2: In view of Lemma 2.6, it suffices to show that Lcfa is locally univalent and sense-preserving. Substituting ω(z) = ωa(z) = (az)/(1 − az) into (10), we have $ω~2(z)=[(1−c)−(1+c)z](a−z1−az)(1+a−z1−az)−c⋅a2−1)(1−az)2z(1−z)[(1+c)−(1−c)z](1+a−z1−az)−c⋅a2−1)(1−az)2z(1−z)=[(1−c)−(1+c)z][(a−z)(1−az)+(a−z)2]−(a2−1)cz(1−z)[(1+c)−(1−c)z][(1−az)2+(a−z)(1−az)]−(a2−1)cz(1−z)=−z3−2+a−c+2ac1+cz2+1+2a−2c+ac1+cz−a⋅1−c)1+c1−2+a−c+2ac1+cz+1+2a−2c+ac1+cz2−a⋅1−c)1+cz3$(18)Next we just need to show that $|{\stackrel{~}{\omega }}_{2}\left(z\right)|<1$ for $0, where −1 < a < 1. We shall consider the following two cases.Case 1. Suppose that a = 0. Then substituting a = 0 into (18) yields $ω~2(z)=−Zz2−2−c1+cz+1−2c1+c1−2−c1+cz+1−2c1+cz2=−Z(z−1)z−1−2c1+c(1−z)1−1−2c1+cz.$Then two zeros z1 = 1 and z2 = (1 − 2c)/(1 + c) of the above numerator lie in or on the unit circle for all 0 < c ≤ 2, so we have $|{\stackrel{~}{\omega }}_{2}\left(z\right)|<1.$Case 2. Suppose that a ≠ 0. From (18), we can write $ω~2(z)=−z3−2+a−c+2ac1+cz2+1+2a−2c+ac1+cz−a⋅1−c)1+c1−2+a−c+2ac1+cz+1+2a−2c+ac1+cz2−a⋅1−c)1+cz3=−p(z)p∗(z)=−(z−A)(z−B)(z−C)(1−A¯z)(1−B¯z)(1−C¯z).$We will show that $A,B,C\in \overline{\mathbb{U}}$ for 0 < c ≤ 2(1 + a)/(1 − a). Applying Lemma 1 to $p(z)=z3−2+a−c+2ac1+cz2+1+2a−2c+ac1+cz−a(1−c)1+c.$Note that $|-\frac{a\left(1-c\right)}{1+c}|<1$ for c > 0 and −1 < a < 1, we get $p1(z)=a3¯p(z)−a0p∗(z)Z=p(z)+a⋅1−c)1+cp∗(z)Z=(1+c+a−ac)(1+c−a+ac)(1+c)2z2+−2−c−6ac+c2+2a2−a2c−a2c2(1+c)2z+1−c+6ac−2c2−a2−a2c+2a2c2(1+c)2=(1+c+a−ac)(1+c−a+ac)(1+c)2z2+−2+c−2a−ac1+c+a−acZ+1−2c+a+2ac1+c+a−ac=(1+c+a−ac)(1+c−a+ac)(1+c)2(z−1)z−1+a−2c(1−a)1+a+c(1−a).$So p1(z) has two zeros ${z}_{1}^{\ast }=1$ and ${z}_{2}^{\ast }=\frac{1+a-2c\left(1-a\right)}{1+a+c\left(1-a\right)}$ which are in or on the unit circle for 0 < c ≤ 2(1 + a)/(1 - a). Thus, by Lemma 1, all zeros of p(z) lie on $\overline{\mathbb{U}}$, that is $A,B,C\in \overline{\mathbb{U}}$ and so $|{\stackrel{~}{\omega }}_{2}\left(z\right)|<1$ for all $z\in \mathbb{U}$. □

## 4 Examples

In this section, we give interesting examples resulting from Theorem 1.1 and Theorem 1.2.

Example 4.1: In Theorem 1.1, note that $f=h+\overline{g}\in {\mathcal{S}}_{H}^{0}$ with h + g = z/(1 - z) and the dilatation ω(z) = −z(a + z)/(1 + az) . By shearing $h′(z)+g′(z)=1(1−z)2andg′(z)=ω(z)h′(z).$Solving these equations, we get $h′(z)=1+az(1−z)3(1+z)andg′(z)=−z(a+z)(1−z)3(1+z).$Integration gives $h(z)=12Z1−z+1+a4Z(1−z)2+1−a8log1+z1−z,$(19)and then, $g(z)=12Z1−z−1+a4Z(1−z)2−1−a8log1+z1−z$(20)By the convolutions, we have $f0∗f=h0∗h+g0∗g¯=h(z)+zh′(z)2+g(z)−zg′(z)2¯.$So that $h0∗h=14Z1−z+1+a8Z(1−z)2+1−a16log1+z1−z+12z(1+az)(1−z)3(1+z)$and $g0∗g=14Z1−z−1+a8Z(1−z)2−1−a16log1+z1−z+12z2(a+z)(1−z)3(1+z).$Images of $\mathbb{U}$ under f0f for certain values of a are drawn in Figure 1 and Figure 2 (a)-(d) by using Mathematica. By Theorem 1.1, it follows that f0f is locally univalent and convex in the horizontal direction for a = −1, −0.5, 0,1, but it is not locally univalent for a = 0.2.

Fig. 1

lmages of $\mathbb{U}$ under f0f for a = 0.2.

Fig. 2

lmages of $\mathbb{U}$ under f0f for various values a.

Example 4.2: In Theorem 1.2, by (1) and (4), we have $Lc∗fa=11+cha(z)+czha′(z)+11+cga(z)−czga′(z)¯=11+c11+az−12z2(1−z)2+cz(1−az)(1+a)(1−z)3+11+ca1+az−12z2(1−z)2−cz(a−z)(1+a)(1−z)3¯=Rez(1+c)(1−z)+c(1−a)z(1+z)(1+c)(1+a)(1−z)3+iIm(1−a1+a+c)z(1+c)(1−z)2.$If we take a = 0.5, c = 6, in view of Theorem 1.2, we know that L6f0.5 is univalent and convex in the horizontal direction. The image of $\mathbb{U}$ under f0.5, L6 and L6f0.5 are shown in Figure 3 (a)-(c), respectively

Fig. 3

lmages of $\mathbb{U}$ under f0.5, L6 and L6f0.5.

If we take a = 0.5, c = 6.2, then $L6.2∗f0.5=Re17.2z1−z+3.110.8z(1+z)(1−z)3+iIm13+6.27.2z(1−z)2.$

The image of $\mathbb{U}$ under L6.2f0.5 is shown in Figure 4 (a). Figure 4(b) is a partial enlarged view of Figure 4(a) showing that the images of two outer most concentric circles in $\mathbb{U}$ are intersecting and so L6.2f0.5 is not univalent.

Fig. 4

lmages of the unit disk $\mathbb{U}$ under L6.2f0.5.

Remark 4.3: This example shows that the condition 0 < c ≤ 2(1 + a)/(1 - a) in Theorem 1.2 is sharp.

## Acknowledgement

The authors would like to thank the referees for their helpful comments. According to the hints of the referees the authors were able to improve the paper considerably.

The research was supported by the Project Education Fund of Yunnan Province under Grant No. 2015Y456, the First Bath of Young and Middle-aged Academic Training Object Backbone of Honghe University under Grant No. 2014GG0102.

## References

• [1]

Kumar R., Dorff M., Gupta S., Singh S., Convolution properties of some harmonic mappings in the right-half plane, Bull. Malays. Math. Sci. Soc., 2016, 39(1), 439-455

• [2]

Kumar R., Gupta S., Singh S., Dorff M., On harmonic convolutions involving a vertical strip mapping, Bull. Korean Math. Soc., 2015, 52(1), 105-123

• [3]

Li L., Ponnusamy S., Convolutions of slanted half-plane harmonic mappings, Analysis (Munich), 2013, 33(2), 159-176

• [4]

Li L., Ponnusamy S., Convolutions of harmonic mappings convex in one direction, Complex Anal. Oper. Theory, 2015, 9(1), 183–199

• [5]

Boyd Z., Dorff M., Nowak M., Romney M., Wo\}oszkiewicz M., Univalency of convolutions of harmonic mappings, Appl. Math. Comput., 2014, 234, 326-332

• [6]

Liu Z., Li y., The properties of a new subclass of harmonic univalent mappings, Abstr. Appl. Anal., 2013, Article ID 794108, 7 pages

• [7]

Muir S., Weak subordination for convex univalent harmonic functions, J. Math. Anal. Appl., 2008, 348, 689-692

• [8]

Muir S., Convex combinations of planar harmonic mappings realized through convolutions with half-strip mappings, Bull. Malays. Math. Sci. Soc., 2016, DO1:10.1007/s48-840-016-0336-0

• [9]

Nagpal S., Ravichandran y., Univalence and convexity in one direction of the convolution of harmonic mappings, Complex Var. Elliptic Equ., 2014, 59(9), 1328-1341

• [10]

Rosihan M. AIi., Stephen B. Adolf., Subramanian, K. G., Subclasses of harmonic mappings defined by convolution, Appl. Math. Lett., 2010, 23, 1243-1247

• [11]

Wang Z., Liu Z., Li y., On convolutions of harmonic univalent mappings convex in the direction of the real axis, J. Appl. Anal. Comput., 2016, 6, 145-155

• [12]

Sokół Janusz, Ibrahim Rabha W., Ahmad M. Z., Al-Janaby, Hiba F., lnequalities of harmonic univalent functions with connections of hypergeometric functions, Open Math., 2015, 13, 691-705

• [13]

Kumar R., Gupta S., Singh S., Dorff M., An application of Cohn's rule to convolutions of univalent harmonic mappings,

• [14]

Dorff M., Convolutions of planar harmonic convex mappins, Complex Var. Theorey Appl., 2001, 45, 263-271

• [15]

Dorff M., Nowak M., Wo\}oszkiewicz M., Convolutions of harmonic convex mappings, Complex var. EIIiptic Equ., 2012, 57(5), 489–503

• [16]

Li L., Ponnusamy S., Solution to an open problem on convolutions of harmonic mappings, Complex var. EIIiptic Equ., 2013, 58(12), 1647-1653

• [17]

Li L., Ponnusamy S., Sections of stable harmonic convex functions, Nonlinear Anal., 2015, 123-124, 178-190

• [18]

Duren P., Harmonic mappings in the PIane, Cambridge Tracts in Mathematics 156, Cambridge Univ. Press, Cambridge, 2004

• [19]

Rahman O. T., Schmeisser G., Analytic theory of polynomials, London Mathematical Society Monigraphs New Series 26, Oxford Univ. Press, Oxford, 2002

• [20]

Clunie J., Sheil-Small T., Harmonic univalent functions, Ann. Acad. Sci. Fenn. Ser. A I. Math., 1984, 9, 3-25

• [21]

Pommenrenke C., On starlike and close-to-convex functions, Proc. London Math. Soc., 1963, 13, 290-304

Accepted: 2016-07-27

Published Online: 2016-10-20

Published in Print: 2016-01-01

Citation Information: Open Mathematics, ISSN (Online) 2391-5455, Export Citation