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

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 $\omega (z)={g}^{\prime}(z)/{h}^{\prime}(z)$, satisfies $\left|\omega (z)\right|<1$ 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 $h(0)=g(0)=0$ and ${h}^{\prime}(0)=1$. Let ${\mathcal{S}}_{H}^{0}$ be the subset of all $f\in {\mathcal{S}}_{H}$ in which ${g}^{\prime}(0)=0$. 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
$$f(z)=h(z)+\overline{g(z)}=z+\sum _{n=2}^{\mathrm{\infty}}{a}_{n}{z}^{n}+\sum _{n=1}^{\mathrm{\infty}}\overline{{b}_{n}}{\overline{z}}^{n}$$

and
$$F(z)=H(z)+\overline{G(z)}=z+\sum _{n=2}^{\mathrm{\infty}}{A}_{n}{z}^{n}+\sum _{n=1}^{\mathrm{\infty}}\overline{{B}_{n}}{\overline{z}}^{n},$$

the convolution (or Hadamard product) of them is given by
$$(f\ast F)(z)=(h\ast H)(z)+\overline{(g\ast G)(z)}=z+\sum _{n=2}^{\mathrm{\infty}}{a}_{n}{A}_{n}{z}^{n}+\sum _{n=1}^{\mathrm{\infty}}\overline{{b}_{n}{B}_{n}}{\overline{z}}^{n}.$$

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 ${h}_{a}+{g}_{a}=z/(1-z)$ with ${\omega}_{a}(z)=(a-z)/(1-az)(-1<a<1)$ . By using the technique of shear construction (see [18]), we have
$${h}_{a}(z)=\frac{\frac{1}{1+a}z-\frac{1}{2}{z}^{2}}{(1-z{)}^{2}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{g}_{a}(z)=\frac{\frac{a}{1+a}z-\frac{1}{2}{z}^{2}}{(1-z{)}^{2}}.$$(1)

Obviously, for $a=0,{f}_{0}(z)={h}_{0}(z)+\overline{{g}_{0}(z)}\in {\mathcal{K}}_{H}^{0}$ is the standard right half-plane mapping, where
$${h}_{0}(z)=\frac{z-\frac{1}{2}{z}^{2}}{(1-z{)}^{2}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{g}_{0}(z)=\frac{-\frac{1}{2}{z}^{2}}{(1-z{)}^{2}}.$$(2)

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

#### ([14, Theorem 5])

*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* ${h}_{i}+{\mathrm{g}}_{i}=z/(1-z)$ *for* *i* = 1,2. *If* *f*_{1} * *f*_{2} *it is locally univalent and sense-preserving, then* ${f}_{1}\ast {f}_{2}\in {\mathcal{S}}_{H}^{0}$ *and convex in the horizontal direction*.

#### ([15, Theorem 3])

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

#### ([15, Theorem 4])

Let $f=h+\overline{g}\in {\mathcal{K}}_{H}^{0}$ *with* $h(z)+g(z)=z/(1-z)$ *and* $\omega (z)=(z+a)/(1+az)$ *with* $a\in (-1,1)$. *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 *f*_{0} with the other special subclass harmonic mappings.

*Let* $f=h+\overline{g}\in {S}_{H}^{0}$ *with* $h(z)+g(z)=z/(1-z)$ *and* $\omega (z)=-z(z+a)/(1+az)$, *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]:
$${L}_{c}(z)={H}_{c}(z)+\overline{{G}_{c}(z)}=\frac{1}{1+c}\left[\frac{Z}{1-z}+\frac{\mathcal{C}Z}{(1-z{)}^{2}}\right]+\frac{1}{1+c}\overline{\left[\frac{Z}{1-z}-\frac{\mathcal{C}Z}{(1-z{)}^{2}}\right]}(z\in \mathbb{U};c>0).$$(3)

Clearly, ${L}_{1}(z)={f}_{0}(z)$, it was proved in [7] that ${L}_{c}(\mathbb{U})=\{\mathrm{R}\mathrm{e}(\omega )>-1/(1+c)\}$ for each *c* > 0. Moreover, if $f=h+\overline{g}\in {\mathcal{S}}_{H}$, then the above representation gives that
$${L}_{c}\ast f=\frac{h+cz{h}^{\prime}}{1+c}+\overline{\frac{g-cz{g}^{\prime}}{1+c}.}$$(4)

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

#### ([19, p.375])

*Given a polynomial*
$$p(z)={p}_{0}(z)={a}_{{n}_{;}0}{z}^{n}+{a}_{n-{1}_{;}0}{z}^{n-1}+\cdots +{a}_{{1}_{;}0}z+a{0}_{;}0({a}_{{n}_{;}0}\ne 0)$$(5)

*of degree* *n*, *let*
$${p}^{\ast}(z)={p}_{0}^{\ast}(z)={z}^{n}\overline{p(1/\overline{z})}=\overline{{a}_{{n}_{;}0}}+\overline{{a}_{n-{1}_{;}0}}z+\cdots +\overline{{a}_{{1}_{;}0}}{z}^{n-1}+\overline{{a}_{{0}_{;}0}}{z}^{n}$$(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*
$${p}_{1}(z)=\frac{\overline{{a}_{{n}_{;}0}}p(z)-{a}_{{0}_{;}0}{p}^{\ast}(z)}{Z}$$

*is of degree* *n* – 1 *with* *r*_{1} = r – 1 *and* *s*_{1} = *s* *the number of zeros of* *p*_{1} (*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.

*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* $h+g=z/(1-z)$ *with* $\omega (z)={e}^{i\theta}{z}^{n}(\theta \in \mathbb{R},\phantom{\rule{thickmathspace}{0ex}}n\in {\mathbb{N}}^{+})$, *then* ${f}_{a}\ast {f}_{n}\in {\mathcal{S}}_{H}$ *is* *CHD* *for* $a\in \left[\left.\frac{n-2}{n+2},1\right)\right.$.

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 *f*_{a} * *f*_{n} satisfies $|{\stackrel{~}{\omega}}_{1}(z)|\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}|\left.{({g}_{a}\ast g)}^{\mathrm{\prime}}/({h}_{a}\ast h{)}^{\prime}\right|<1$ 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} * *f*_{a} is univalent and convex in the horizontal direction for $0<c\le 2(1+a)/(1-a)$, and derive the following theorem.

*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* *L*_{c} * *f*_{a} *is univalent and convex in the horizontal direction for* $0<c\le 2(1+a)/(1-a)$.

Recently, Liu and Li [6] defined a subclass of harmonic mappings defined by
$${P}_{c}(z)={H}_{c}(z)-\overline{{G}_{c}(z)}=\frac{1}{1+c}\left[\frac{\mathcal{C}Z}{(1-z{)}^{2}}+\frac{Z}{1-z}\right]+\frac{1}{1+c}\overline{\left[\frac{\mathcal{C}Z}{(1-z{)}^{2}}-\frac{Z}{1-z}\right]}(z\in \mathbb{U};c>0).$$(7)

They proved the following result.

#### ([6, Theorem 7])

*Let* *P*_{c}(*z*) *be harmonic mappings defined by* (7) *and* ${f}_{n}=h+\overline{g}$ *with* *h* – *g* = *z*/(1–*z*) *and dilatation* $\omega (z)={e}^{i\theta}{z}^{n}(\theta \in \mathbb{R},\phantom{\rule{thickmathspace}{0ex}}n\in {\mathbb{N}}^{+})$. *Then* *P*_{c} * *f*_{n} *is univalent and convex in the horizontal direction for* $0<c\le 2/n$.

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

*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* $\omega (z)={e}^{i\theta}{z}^{n}(\theta \in \mathbb{R},\phantom{\rule{thickmathspace}{0ex}}n\in {\mathbb{N}}^{+})$. *Then* *L*_{c} * *f*_{n} *is univalent and convex in the horizontal direction for* $0<c\le 2/n$.

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