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\title{On quasiconformal extension of harmonic mappings with nonzero pole}
\alttitle{Sur l'extension quasiconforme des applications harmoniques Ã&nbsp; pÃ´le non nul}

\author{\firstname{Bappaditya} \lastname{Bhowmik}\IsCorresp}
\address{Department of Mathematics, Indian Institute of Technology Kharagpur, Kharagpur -- 721302, India}
\email{bappaditya@maths.iitkgp.ac.in}

\author{\firstname{Goutam} \lastname{Satpati}}
\address[1]{Department of Mathematics, Indian Institute of Technology Kharagpur, Kharagpur -- 721302, India}
\email{goutam.satpati@gmail.com}

\subjclass{31A05, 30C62, 30C55}
\keywords{\kwd{Quasiconformal mappings} \kwd{harmonic mappings} \kwd{linearly connected domain} \kwd{quasidisk}}
\altkeywords{\kwd{Applications quasiconformes} \kwd{applications harmoniques} \kwd{domaine linÃ©airement connectÃ©} \kwd{quasidisque}}

\begin{abstract}
Let $\Sigma_H^k(p)$ be the class of sense-preserving univalent harmonic mappings defined on the open unit disk $\mathbb{D}$ of the complex plane with a simple pole at $z=p \in (0,1)$ that have $k$-quasiconformal extensions ($0\leq k&lt;1$) to the extended complex plane. We first derive a sufficient condition for harmonic mappings defined on $\mathbb{D}$ with pole at $z=p\in (0,1)$ to belong in the class $\Sigma_H^k(p)$. As a consequence of this, we derive a convolution result involving functions in $\smash{\Sigma_H^{k_i}(p)}$, $0\leq k_i&lt;1$ for $i=1,2$. We also consider harmonic mappings with a nonzero pole defined on a linearly connected domain $\Omega\subset\mathbb{D}$ and prove criteria for univalence and quasiconformal extensions for such mappings.
\end{abstract}

\begin{altabstract}
Soit $\Sigma_H^k(p)$ la classe des applications harmoniques injectives prÃ©servant l'orientation, dÃ©finies sur le disque unitÃ© ouvert $\mathbb{D}$ du plan complexe ayant un pÃ´le simple en $z=p \in (0,1)$ et qui ont une extension $k$-quasiconforme ($0\leq k&lt;1$) au plan complexe Ã©tendu. Nous dÃ©crivons d'abord une condition suffisante pour qu'une application harmonique appartienne Ã&nbsp; la classe $\Sigma_H^k(p)$. En consÃ©quence, nous obtenons un rÃ©sultat de convolution impliquant des applications dans $\smash{\Sigma_H^{k_i}(p)}$, $0\leq k_i&lt;1$ pour $i=1,2$. Nous considÃ©rons Ã©galement des applications harmoniques avec un pÃ´le non nul dÃ©finies sur un domaine linÃ©airement connectÃ© $\Omega\subset\mathbb{D}$ et dÃ©crivons des critÃ¨res d'univalence et d'extensions quasiconformes pour de telles applications.
\end{altabstract}

\thanks{The first author of this article would like to thank SERB, India for its financial support through Core Research Grant (Ref.~No.~CRG/2022/001835). The second author of this article would like to thank NBHM, DAE, Govt.~of India (Ref.~No.~0204/3/2021/R\&amp;D-II/7248) for its financial support}
\CDRGrant[SERB, India]{CRG/2022/001835}
\CDRGrant[NBHM, DAE, Govt.~of India]{0204/3/2021/R\&amp;D-II/7248}

\begin{document}

\maketitle

\section{Introduction}

Let $\IC$ be the finite complex plane, $\ID \coloneqq\braces[\big]{z\in \IC:\abs{z}&lt;1}$ be the open unit disk, and $\sphere$ be the extended complex plane $\C\cup\{\infty\}$. In order to describe the aim of this article, first we need to be familiar with the quasiconformal mappings and quasiconformal extensions of conformal mappings. We briefly present these below. Let $\Omega \subset \sphere$ be a domain (an open and connected set). A sense-preserving homeomorphism $f \colon \Omega \to \sphere$ is called \emph{$k$-quasiconformal}, $k\in[0,1)$, if it has locally $L^2$-derivatives on $\Omega \setminus \braces[\big]{f\inv (\infty), \infty}$ (in the sense of distribution) satisfying
\[
\abs{f_{\bar{z}}} \leq k \abs{f_z} \ \text{almost everywhere on $\Omega \setminus \braces[\big]{f\inv (\infty), \infty}$},
\]
where $f_z\coloneqq\partial f/\partial z$ and $f_{\overline{z}}\coloneqq\partial f/\partial \bar{z}$. The quantity $\mu_f=f_{\bar{z}}/f_z$ is called \emph{complex dilatation} of~$f$. If $f$ is $k$-quasiconformal, then it is also called $K$-quasiconformal, where $K=(1+k)/(1-k) \in [1,\infty)$. If $k=0$ or $K=1$, then by virtue of Weyl's lemma, $f$ will be conformal on $\Omega$. The study of conformal maps that have quasiconformal extensions to the extended plane is an interesting problem in function theory that attracted many researchers around the globe over the last few decades (see f.i.~\cite{BSS, BS, Hotta, Lehto71, Sugawa}). One important aspect to study such kind of mappings is that they can be identified with the elements of the universal Teichm{\"u}ller space (see e.g.~\cite[Chapter~III]{Lehto2}).

\begin{defi}
Let $k\in [0,1)$. We say that a univalent (holomorphic or meromorphic) function $f$ defined on a domain $\Omega \subset \sphere$ admits a $k$-quasiconformal extension to $\sphere$ if there exists a $k$-quasiconformal mapping $F \colon \sphere \to \sphere$ such that $\restr{F}{\Omega}=f$. Moreover, a holomorphic function $f \colon \Omega \to \IC$ ($\Omega \subset \IC$) is said to admit a $k$-quasiconformal extension to $\IC$ if there exists a $k$-quasiconformal mapping $F \colon \IC \to \IC$ such that $\restr{F}{\Omega}=f$.
\end{defi}

If a conformal map $f$ has a quasiconformal extension $F$ to $\IC$ such that $F(z) \to \infty$ when $z \to \infty$, then we also say that $f$ has a quasiconformal extension to $\sphere$ and vice-versa. In order to describe our main goal of this article, we first start with meromorphic univalent mappings with a nonzero pole at the point $z=p\in (0,1)$. Let $\Sigma(p)$ be the class of all meromorphic univalent functions on $\ID$ having a simple pole at $p\in(0,1)$ with residue $1$, such that any function $f \in \Sigma(p)$ has the following Laurent expansion
\[
f(z)= \frac{1}{z-p} + \sum_{n=1}^{\infty} a_n z^n, \quad z \in \ID\setminus \{p\}.
\]
In~\cite{BSS}, the authors of the present article along with T.~Sugawa studied the class $\Sigma_k(p)$ which consists of all functions in $\Sigma(p)$ that admit $k$-quasiconformal extension to $\sphere$. In particular, we proved a sufficient condition for a meromorphic function with pole at $z=p \in (0,1)$ to belong to the class $\Sigma_k(p)$ (see f.i.~\cite[Theorem~2]{BSS}). This result actually generalized a result of Krzy{\.z} (compare~\cite[Theorem~1]{Krzyz}) for functions with pole at the origin. We state this result below.

\begin{theo}\label{thA}
Let $0\leq k&lt;1$ and $0&lt;p&lt;1$. Suppose that $\omega$ is an analytic function on\/ $\ID$ such that $\abs[\big]{\omega'(z)}\leq k(1+p)^{-2}$, for $z\in\ID$, then the function $f$ given by
\[
f(z)=\frac{1}{z-p}+\omega(z), \quad z\in \ID,
\]
is a member of $\Sigma_k(p)$. A $k$-quasiconformal extension of $f$ to\/ $\sphere$ is given by setting
\[
f(z)=\frac{1}{z-p}+\omega(1/\bar z), \quad \text{for $z\in \ID^{*}$}.
\]
\end{theo}

In this article, we aim to prove a harmonic analogue of the above result. To this end, we start with a brief overview on harmonic mappings. A real-valued function defined on a domain $\Omega \subset \IC$ is said to be harmonic if it has continuous second-order partial derivatives and satisfies Laplace's equation. A planar harmonic mapping is a complex-valued function defined on $\Omega$ such that both its real and imaginary parts are harmonic on $\Omega$. If $\Omega$ is a simply connected domain in $\IC$, then any planar harmonic mapping $f$ has the canonical representation as $f=h+ \bar{g}$, where $h$ and $g$ are analytic functions on $\Omega$. By Lewy's theorem, we know that $f$ is locally univalent on $\Omega$ if and only if $J_f\coloneqq\abs{h'}^2-\abs{g'}^2$ is non-vanishing on $\Omega$. In this case, either $J_f(z)&gt; 0$ throughout on $\Omega$ and therefore $f$ is sense-preserving, or $J_f(z)&lt; 0$ on $\Omega$ and therefore $f$ is sense-reversing. For a locally univalent sense-preserving harmonic mapping $f$ defined on $\ID$, the quotient $w(z)\coloneqq g'(z)/h'(z)$, $z\in \ID$, is called the \emph{second complex dilatation} of $f$. Here, $w$ is analytic on $\ID$ and maps $\ID$ onto itself; and therefore $\norm{w}_{\infty}= \sup_{z\in \ID} \abs[\big]{w(z)} \leq 1$. We note that, if $f$ is a sense-preserving homeomorphism and harmonic on $\ID$ with its dilatation $w$ satisfying $\norm{w}_{\infty}\leq k&lt;1$, then $f$ is $k$-quasiconformal on $\ID$. Let $f$ be a sense-preserving harmonic mapping on $\Omega \subsetneq \sphere$, then by a quasiconformal extension of $f$ we mean that there exists a sense-preserving homeomorphism $F \colon \sphere \to \sphere$ such that $\restr{F}{\Omega}=f$, and $F$ is quasiconformal on $\sphere$.

In~\cite{Schober}, Hengartner and Schober studied sense-preserving univalent harmonic mappings $f$ defined on $\ID$ having a simple pole at the origin, with the representation of the form
\begin{equation}\label{eq3}
f(z)= \frac{1}{z} + h(z) + \bar{g(z)} = \frac{1}{z} + \sum_{n=1}^{\infty} a_n z^n + \sum_{n=1}^{\infty} \bar{b_nz^n}, \quad z\in \ID\setminus\{0\}.
\end{equation}
We denote the class of all such functions as $\Sigma_H$. In fact, in~\cite{Schober}, the authors considered harmonic mappings having a logarithmic singularity at the origin. But, in this article, we avoid this possibility. In an analogous way, we consider the class of all sense-preserving univalent harmonic mappings on $\ID$ that have simple pole at $p\in (0,1)$. We denote this class of functions as $\Sigma_H(p)$. Thus, each function $f\in \Sigma_H(p)$ can be represented as
\begin{equation}\label{eq1}
f(z)= \frac{1}{z-p} + h(z)+ \bar{g(z)} = \frac{1}{z-p} + \sum_{n=1}^{\infty} a_n z^n + \sum_{n=1}^{\infty} \bar{b_nz^n}, \quad z\in\ID\setminus\{p\}.
\end{equation}

In~\cite{Mateljevic}, Mateljevi{\'c} studied the class $\Sigma_H^k$, which consists of all functions in $\Sigma_H$ that have $k$-quasiconformal extension to $\sphere$, for some $k\in [0,1)$. Similarly, in this article, we consider the subclass $\Sigma_H^k(p)$ of all functions in $\Sigma_H(p)$ that have $k$-quasiconformal extensions to $\sphere$.

It should be noted that, if $b_n=0$ for all $n$ (or, equivalently $g \equiv 0$) in~\eqref{eq1}, then the classes $\Sigma_H^k(p)$ and $\Sigma_H(p)$ reduce to $\Sigma_k(p)$ and $\Sigma(p)$, respectively. It is also true that $\Sigma_H^k(0)=\Sigma_H^k$.

We refer to a recent article~\cite{BS1} of ours for more information and results on functions in $\Sigma_H^k(p)$. In the present article, we first prove a sufficient condition for harmonic mappings of the form~\eqref{eq1} to belong in the class $\Sigma_H^k(p)$, which generalizes Theorem~\ref{thA}. This is the content of Theorem~\ref{th2}. We also derive several interesting consequences of the obtained result. One of the consequences of the aforementioned theorem yields a result on convolution for functions in $\Sigma_H^{k_i}(p),i=1,2$, which is the content of Theorem~\ref{th3}. In order to describe the remaining goal of this article, we need to have the notion of a \emph{linearly connected domain} (cf.~\cite[p.~103]{Pommerenke}).
\begin{defi}
A simply connected domain $\Omega \subset \IC$ is said to be linearly connected if there exists a constant $1 \leq M &lt;\infty$, such that any two distinct points $z_1,z_2 \in \Omega$ are joined by a curve $\gamma \subset \Omega$ of length $l(\gamma) \leq M\abs{z_1-z_2}$, or equivalently, $\diam(\gamma)\leq M\abs{z_1-z_2}$.
\end{defi}

If $\Omega \subset \IC$ is a linearly connected domain, then we call the constant $M$ as the \emph{domain constant} for $\Omega$. It should be noted that if $\Omega$ is a convex domain then we can choose $M=1$. In this article, our aim is to further generalize Theorem~\ref{th2}, when a harmonic mapping of the form~\eqref{eq1} is defined on a linearly connected domain in $\ID$. In fact, we will see that the univalence of the harmonic mapping is achieved when it is defined on a linearly connected domain $\Omega \subset \ID$, but for quasiconformal extensibility, we have to assume $\Omega$ to be a quasidisk. This will be proved in Theorem~\ref{th4} of the next section.

We clarify here that a simply connected domain $\Omega \subset \IC$ is called \emph{$k$-quasidisk} if it is the image of a disk or a half-plane under a $k$-quasiconformal self-mapping of $\sphere$. The domain $\Omega$ is a quasidisk if it is a $k$-quasidisk for some $k \in [0,1)$. The Jordan curve which bounds a quasidisk is called a \emph{quasicircle}. It is also well known that every quasidisk is a linearly connected domain (compare~\cite{Pommerenke}). There is a close relationship between quasiconformal extensibility and the quasidisks as described in~\cite[Theorem~8.3, p.~98]{LV}. It states that a quasiconformal mapping between two quasidisks admits a quasiconformal extension to the whole plane $\sphere$. In~\cite{Ahlfors} (see also~\cite{Hernandez}), Ahlfors proved a criterion for quasicircle. Indeed, he proved that a Jordan curve $\gamma \subset \sphere$ is a quasicircle, if and only if, for all points $w_1,w_2,w_3,w_4 \in \gamma$, such that $w_1$ and $w_4$ separate $w_2$ and $w_3$ with
\begin{equation}\label{eq2}
\abs[\big]{(w_1,w_2,w_3,w_4)} \leq C,
\end{equation}
for some real positive constant $C$, where $(w_1,w_2,w_3,w_4)$ is the \emph{cross ratio} defined by
\[
(w_1,w_2,w_3,w_4)\coloneqq \parens[\Bigg]{\frac{w_1-w_3}{w_1-w_4}}\parens[\Bigg]{\frac{w_2-w_4}{w_2-w_3}}.
\]
We will apply the above results to prove Theorem~\ref{th4}.

\section{Main Results}

Our first result provides a sufficient condition for harmonic mappings to belong to the class $\Sigma_H^k(p)$.

\begin{theo}\label{th2}
Let $0\leq k &lt;1$ and $0 &lt; p &lt;1$. Let $f$ be a harmonic mapping on $\ID$ of the form~\eqref{eq1} that satisfies
\begin{equation}\label{eq13b}
\abs[\big]{h'(z)} + \abs[\big]{g'(z)} \leq \frac{k}{(1+p)^2}, \quad z\in \ID,
\end{equation}
then $f \in \Sigma_H^k(p)$. The quasiconformal extension of $f$ is given by
\begin{equation}\label{eq14}
F(z)=
\begin{cases}
f(z),																							&amp; z\in \ID, \\
\frac{1}{z-p} + h\parens[\big]{\frac{1}{\bar{z}}} + \bar{g\parens[\big]{\frac{1}{\bar{z}}}},		&amp; z\in \IDsb.
\end{cases}
\end{equation}
\end{theo}

\begin{proof}
If $k=0$, then~\eqref{eq13b} implies that $f$ has the form $f(z)=1/(z-p)+c$, where $c$ is a constant. Since $f$ is conformal on $\sphere$, $f$ has a trivial 0-quasiconformal extension on $\sphere$. Thus, we assume $k \in (0,1)$. We follow the technique used in~\cite[Theorem~1]{Krzyz} to establish our result. We first show that $f$ has a continuous extension to the closed disk $\IDb$. If $z_1,z_2 \in \ID$ with $z_1 \neq z_2$, then~\eqref{eq13b} gives
\begin{equation}\label{eq14a}
\begin{split}
\abs[\big]{h(z_1)-h(z_2)+ \bar{g(z_1)-g(z_2)}}		&amp; = \abs*{\int_{[z_2,z_1]} h'(z)\,\dd z + \bar{g'(z)\,\dd z}} \\
&amp; \leq \int_{[z_2,z_1]} \parens[\big]{\abs[\big]{h'(z)}+ \abs[\big]{g'(z)}}\,\abs{\dd z} \\
&amp; \leq \frac{k\abs{z_1-z_2}}{(1+p)^2}.
\end{split}
\end{equation}
Hence, the function $f(z)- 1/(z-p)$ has a continuous (more precisely, harmonic) extension to $\IDb$, for which the inequality~\eqref{eq14a} still holds good. This in turn implies, $f$ has a continuous extension to $\IDb$. Since
\[
\abs[\Bigg]{\frac1{z_1-p}-\frac1{z_2-p}} = \frac{\abs{z_1-z_2}}{\abs{z_1-p}\abs{z_2-p}}\geq \frac{\abs{z_1-z_2}}{(1+p)^2},
\]
where $z_1,z_2 \in \IDb \setminus \{p\}$ with $z_1 \neq z_2$; using~\eqref{eq14a}, we get
\[
\begin{split}
\abs[\big]{f(z_1)-f(z_2)} 		&amp; \geq \frac{\abs{z_1-z_2}}{(1+p)^2} - \abs[\big]{h(z_1)-h(z_2)+ \bar{g(z_1)-g(z_2)}} \\
&amp; \geq \frac{(1-k)\abs{z_1 - z_2}}{(1+p)^2}&gt;0.
\end{split}
\]
Hence, $f$ is univalent on $\IDb$ (in particular, $f \in \Sigma_H(p)$) and the image curve $f(\partial\ID)$ is a Jordan curve. Also, for $z_1,z_2 \in \partial \ID$ with $z_1 \neq z_2$, using~\eqref{eq14a} once again, we obtain
\[
\begin{split}
\abs[\big]{f(z_1)-f(z_2)} 	&amp; \leq \frac{\abs{z_1-z_2}}{\abs{z_1-p}\abs{z_2-p}} + \abs[\big]{h(z_1)-h(z_2)+ \bar{g(z_1)-g(z_2)}}\\
&amp; \leq \frac{\abs{z_1 - z_2}}{(1-p)^2} + \frac{k\abs{z_1 - z_2}}{(1+p)^2}\\
&amp; &lt; \frac{(1+k)\abs{z_1 - z_2}}{(1-p)^2},
\end{split}
\]
which implies that $f(\partial\ID)$ is a rectifiable Jordan curve of length at most $2\pi(1+k)/(1-p)^2$. Next, we show that $f$ is also sense-preserving on $\ID$ and $\abs{\mu_f}&lt;k$ (hence $\norm{w}_{\infty}\leq k&lt;1$) on $\ID \setminus \{p\}$. Since $f$ has the form~\eqref{eq1}, we have
\[
f_{\overline{z}}(z)= \bar{g'(z)} \quad\text{and}\quad f_z(z)= -\frac{1}{(z-p)^{2}} + h'(z), \quad z \in \ID \setminus \{p\}.
\]
Also, the inequality~\eqref{eq13b} yields
\[
\abs[\big]{h'(z)} +\abs[\big]{g'(z)} &lt; \frac{1}{(1+p)^{2}} &lt; \frac{1}{\abs{z-p}^{2}},
\]
which implies
\[
\abs{f_z}= \abs[\Bigg]{-\frac{1}{(z-p)^{2}} + h'(z)} \geq \frac{1}{\abs{z-p}^{2}} - \abs[\big]{h'(z)} &gt; \abs[\big]{g'(z)}=\abs{f_{\overline{z}}},
\]
for $z \in \ID \setminus \{p\}$. Thus, $J_f(z)=\abs{f_z}^2-\abs{f_{\overline{z}}}^2&gt;0$ on $\ID \setminus \{p\}$ and $J_f(p)= \infty$. As a result, $f$ is sense-preserving on $\ID$. Since sense-preserving harmonic mappings are necessarily open maps (see f.i.~\cite[p.~10]{Duren1}), it also follows that $f$ is a homeomorphism on $\IDb$. Moreover, in view of the inequality~\eqref{eq13b}, we also see that, for $z \in \ID \setminus \{p\}$, the complex dilatation of $f$ satisfies
\[
\abs[\big]{\mu_f(z)}= \frac{\abs[\big]{f_{\overline{z}}(z)}}{\abs[\big]{f_z(z)}}= \frac{\abs[\big]{g'(z)}}{\abs[\big]{-\frac{1}{(z-p)^{2}} +h'(z)}} \leq
\frac{\abs[\big]{g'(z)}}{\frac{1}{\abs{z-p}^{2}}-\abs[\big]{h'(z)}} &lt; \frac{\abs[\big]{g'(z)}}{\frac{1}{(1+p)^{2}}-\abs[\big]{h'(z)}}&lt;k.
\]
Therefore, $f$ is $k$-quasiconformal on $\ID$.

Next, we note that the extension $F$ of $f$ proved in~\eqref{eq14} such a way that it agrees with $f$ on $\partial \ID$ and hence, continuous on $\sphere$. Using a similar kind of argument as before, we now prove that $F$ is sense-preserving on $\IDs$. To deal with the point at infinity as an ordinary point, we make a change of variable $\zeta=1/z$. For $\zeta \in \ID$, we denote $F(1/\zeta)$ as $\tilde{F}(\zeta)$, then from~\eqref{eq14} we get
\begin{equation}\label{eq15}
\tilde{F}(\zeta) = \frac{\zeta}{1-p\zeta} + h(\bar{\zeta}) + \bar{g(\bar{\zeta})}, \quad \zeta \in \ID.
\end{equation}
Since $\zeta \in \ID$, from~\eqref{eq13b} we have
\[
\abs{\tilde{F}_{\zeta}}=\abs[\Bigg]{\frac{1}{(1-p\zeta)^{2}} + \bar{g'(\bar{\zeta})}} &gt; \frac{1}{(1+p)^2} - \abs[\big]{g'(\bar{\zeta})} &gt; \abs[\big]{h'(\bar{\zeta})} = \abs{\tilde{F}_{\overline{\zeta}}},
\]
which gives
\[
J_F(1/\zeta)\abs{\zeta}^{-4} = J_{\tilde{F}}(\zeta) = \abs{\tilde{F}_{\zeta}}^2 - \abs{\tilde{F}_{\overline{\zeta}}}^2&gt;0.
\]
This implies that $F$ is sense-preserving on $\IDs$. We have already proved that $f$ is sense-preserving on $\ID$ and univalent on the closed disk $\IDb$. Since $F \equiv f$ on $\IDb$ and $J_F$ is a continuous function, $F$ is sense-preserving on the whole domain $\sphere$. Next, we observe that $F$ is also harmonic on $\IDs$. As $f(\partial\ID)$ is a Jordan curve, if we take an arbitrary point $a$, not lying on $f(\partial\ID)$, then the winding number $n\parens[\big]{f(\partial\ID),a}$ is either $0$ or $-1$. Now, using the argument principle for sense-preserving harmonic mappings (compare~\cite[p.~9]{Duren1}), it follows that $F$ is univalent on $\IDs$ and covers the bounded component $\sphere \setminus f(\IDb)$. Hence, $F \colon \sphere \to \sphere$ is a sense-preserving bijective continuous mapping. Since $\sphere$ is both compact and $T_2$, finally it follows that $F \colon \sphere \to \sphere$ is a homeomorphism. It now remains to show that $\abs{\mu_f} \leq k$ on $\IDs$. This will prove that $F$ is a $k$-quasiconformal extension of $f$. For $\zeta=1/z \in \ID$, using once again~\eqref{eq13b}, we get from~\eqref{eq15} that
\[
\abs[\big]{\mu_f(z)} = \abs[\big]{\mu_{\tilde{F}}(\zeta)} = \frac{\abs[\big]{{\tilde F}_{\overline{\zeta}}(\zeta)}}{\abs[\big]{{\tilde F}_{\zeta}(\zeta)}} =
\frac{\abs[\big]{h'(\bar{\zeta})}}{\abs[\bigg]{\frac{1}{(1-p\zeta)^{2}} + \bar{g'(\bar{\zeta})}}}
\leq \frac{\abs[\big]{h'(\bar{\zeta})}}{\frac{1}{\abs{1-p\zeta}^{2}} - \abs[\big]{g'(\bar{\zeta})}} &lt; \frac{\abs[\big]{h'(\bar{\zeta})}}{\frac{1}{(1+p)^{2}} - \abs[\big]{g'(\bar{\zeta})}} &lt; k.
\]
Therefore, $F$ is also a $k$-quasiconformal mapping on $\IDs$. Since the unit circle $\partial \ID$ is removable for quasiconformality of $F$ (see~\cite[p.~205]{LV}), it follows that $F$ is $k$-quasiconformal on $\sphere$, and hence $\restr{F}{\ID} \equiv f \in \Sigma_H^k(p)$. The proof of the theorem is now complete.
\end{proof}

\begin{rema}
For $p=0$, Theorem~\ref{th2} gives us a sufficient condition for a harmonic mapping $f$ with pole at the origin to belong in the class $\Sigma_H^k$. We state this precisely as follows. Let $f$ be a harmonic mapping having expansion of the form~\eqref{eq3} and satisfies $\abs[\big]{h'(z)}+\abs[\big]{g'(z)} \leq k$ on $\ID$, then $f$ is univalent on $\ID$ and admits $k$-quasiconformal extension as~\eqref{eq14} with $p=0$.
\end{rema}

By applying Theorem~\ref{th2}, we can create many examples of functions in $\Sigma_H^k(p)$. We present these examples below. This also shows that the class of functions that satisfy condition~\eqref{eq13b} is not too small.
\begin{exem}
Let
\[
f_{\theta}(z)= \frac{1}{z-p} + k\parens[\Bigg]{\frac{1-p}{1+p}}^2 \Real\parens[\Bigg]{\frac{e^{i\theta}z}{1-pe^{i\theta}z}}, \quad z\in \ID, \ \theta \in [0,2\pi).
\]
It is easy to verify that $f$ is harmonic on $\ID$ such that it has the form~\eqref{eq1}, where
\[
h(z)=g(z)= \frac{k}{2} \parens[\Bigg]{\frac{1-p}{1+p}}^2 \frac{e^{i \theta}z}{1-pe^{i \theta}z}.
\]
We see that, for $z \in \ID$,
\[
\abs[\big]{h'(z)}+\abs[\big]{g'(z)}= k\parens[\Bigg]{\frac{1-p}{1+p}}^2 \frac{1}{\abs[\big]{1-pe^{i \theta}z}^2} &lt; \frac{k}{(1+p)^2}.
\]
By Theorem~\ref{th2}, we get $f\in \Sigma_H^k(p)$ and its quasiconformal extension is given by
\[
F_{\theta}(z)=
\begin{cases}
f(z),																											&amp; z\in \ID, \\
\frac{1}{z-p} + k\parens[\bigg]{\frac{1-p}{1+p}}^2 \Real\parens[\bigg]{\frac{e^{-i \theta}}{z - pe^{-i \theta}}},	&amp; z\in \IDsb.
\end{cases}
\]
For $z\in \IDs$, one can also easily check that
\[
\abs[\big]{\mu_{F_0}(z)}= \frac{\frac{k}{2} \parens[\bigg]{\frac{1-p}{1+p}}^2}{1 + \frac{k}{2} \parens[\bigg]{\frac{1-p}{1+p}}^2} &lt; k
\qquad\text{and}\qquad \abs[\big]{\mu_{F_{\theta}}(z)}&lt; \frac{k}{2-k}&lt;k, \quad \text{for $\theta \in (0,2\pi)$}.
\]
\end{exem}

As a corollary of~\cite[Theorem~1]{Krzyz}, Krzy{\.z} proved that if $f$ is a meromorphic function on $\ID$ with pole at the origin and satisfies $\abs[\big]{z^2f'(z)+1}\leq k\abs{z}^2$, for $z \in \ID$, then $f$ is univalent on $\ID$ and admits a $k$-quasiconformal extension to $\sphere$ (see f.i.~\cite[Corollary~2, p.~102]{Krzyz}). This criterion is known as Krzy{\.z}'s criterion for univalence and quasiconformal extension and has been used by many mathematicians. An application of Theorem~\ref{th2} will give us the analogous result for harmonic mappings with nonzero pole.
\begin{coro}
Let $f$ be a harmonic mapping having expansion of the form~\eqref{eq1} and satisfies
\[
\abs[\Bigg]{f_z(z)+ \frac{1}{(z-p)^{2}}} + \abs[\big]{f_{\overline{z}}(z)} \leq \frac{k}{(1+p)^2}, \quad z\in \ID,
\]
then $f \in \Sigma_H^k(p)$.
\end{coro}
\begin{proof}
The proof immediately follows from Theorem~\ref{th2}, since~\eqref{eq1} yields
\[
f_z(z)= -(z-p)^{-2} + h'(z) \qquad\text{and}\qquad f_{\overline{z}}(z)=\bar{g'(z)}.\qedhere
\]
\end{proof}

Another immediate consequence of Theorem~\ref{th2} yields the following result.

\begin{coro}\label{cor2}
Let $f$ be a harmonic mapping having expansion of the form~\eqref{eq1}. If
\[
\sum_{n=1}^{\infty} n\parens[\big]{\abs{a_n}+ \abs{b_n}} \leq \frac{k}{(1+p)^2},
\]
then $f \in \Sigma_H^k(p)$.
\end{coro}
\begin{proof} Since $f$ is of the form~\eqref{eq1}, we have $h(z)=\sum_{n=1}^{\infty} a_n z^n$ and $g(z)=\sum_{n=1}^{\infty} b_n z^n$, $z \in \ID$. The proof now easily follows from Theorem~\ref{th2}, because
\[
\abs[\big]{h'(z)}+\abs[\big]{g'(z)} \leq \sum_{n=1}^{\infty} n\parens[\big]{\abs{a_n}+\abs{b_n}}\abs{z}^{n-1} &lt; \sum_{n=1}^{\infty} n\parens[\big]{\abs{a_n}+\abs{b_n}}
\leq \frac{k}{(1+p)^2}. \qedhere
\]
\end{proof}

In order to state and prove our next theorem, we need some preparations. Let $f_1,f_2 \in \Sigma_H(p)$ with the expansions
\begin{equation}\label{eq17}
f_1(z)=\frac{1}{z-p} + \sum_{n=1}^{\infty} a_n z^n + \sum_{n=1}^{\infty} \bar{b_nz^n} \qquad\text{and} \qquad f_2(z)= \frac{1}{z-p} + \sum_{n=1}^{\infty} A_n z^n + \sum_{n=1}^{\infty} \bar{B_nz^n},
\end{equation}
$z \in \ID$, respectively; then the convolution $f_1 \star f_2$ of $f_1$ and $f_2$ is defined as
\[
f_1 \star f_2(z)= \frac{1}{z-p} + \sum_{n=1}^{\infty} a_n A_n z^n + \sum_{n=1}^{\infty} \bar{b_nB_nz^n}, \quad z\in \ID.
\]
In~\cite[Theorem~3]{BSS}, we proved a convolution property for the class $\Sigma_k(p)$ using the area theorem. Here we prove the following regarding convolution property for the class $\Sigma_H^k(p)$.

\begin{theo}\label{th3}
Let $0\leq k_1,k_2 &lt;1$, $p\in (0,1)$ and $f_1 \in \Sigma_H^{k_1}(p)$, $f_2 \in \Sigma_H^{k_2}(p)$ with the expansions stated in~\eqref{eq17} on\/ $\ID$. If
\[
\sum_{n=1}^{\infty} n\parens[\big]{\abs{a_n}+ \abs{b_n}} \leq \frac{k_1}{(1+p)^2} \qquad\text{and}\qquad \sum_{n=1}^{\infty} n\parens[\big]{\abs{A_n}+ \abs{B_n}} \leq \frac{k_2}{(1+p)^2},
\]
then $f_1 \star f_2 \in \Sigma_H^{\sqrt{k_1k_2}}(p)$.
\end{theo}
\begin{proof}
Since $k_1,k_2 \in [0,1)$ and $p \in (0,1)$, the inequalities assumed in the hypothesis of the theorem imply that $\abs{a_n}+\abs{b_n} \leq 1$ and $\abs{A_n}+\abs{B_n} \leq 1$, for all $n\geq 1$. Now, using these inequalities along with the Cauchy--Schwarz inequality, we get
\[
\begin{split}
\sum_{n=1}^{\infty} n\parens[\big]{\abs{a_n A_n}+\abs{b_n B_n}}	&amp; \leq \sum_{n=1}^{\infty} \parens[\bigg]{\sqrt{n}\parens[\big]{\abs{a_n}+\abs{b_n}}} \parens[\bigg]{\sqrt{n}\parens[\big]{\abs{A_n}+\abs{B_n}}}\\
&amp; \leq \parens*{\sum_{n=1}^{\infty} n\parens[\big]{\abs{a_n}+\abs{b_n}}^2}^{1/2} \parens*{\sum_{n=1}^{\infty} n\parens[\big]{\abs{A_n}+\abs{B_n}}^2}^{1/2}\\
&amp; \leq \parens*{\sum_{n=1}^{\infty} n\parens[\big]{\abs{a_n}+\abs{b_n}}}^{1/2} \parens*{\sum_{n=1}^{\infty} n\parens[\big]{\abs{A_n}+\abs{B_n}}}^{1/2}\\
&amp; \leq \frac{\sqrt{k_1 k_2}}{(1+p)^2}.
\end{split}
\]
Hence, we have $f_1 \star f_2 \in \Sigma_H^{\sqrt{k_1k_2}}(p)$ by an application of Corollary~\ref{cor2}.
\end{proof}

From the proof of the first part of Theorem~\ref{th2}, we observe that convexity of the unit disk plays a vital role for univalence of the harmonic mappings that satisfy the inequality~\eqref{eq13b}. In the next theorem, we will see that the domain constant plays the same role whenever the harmonic mapping is defined on any linearly connected domain in $\ID$. In particular, if the harmonic mapping is defined on any quasidisk in $\ID$, then it admits a quasiconformal extension to $\sphere$ under a similar condition.

\begin{theo}\label{th4}
Let $\Omega$ be a linearly connected domain in $\ID$ with the domain constant $M \in [1,\infty)$, and contains the point $p\in (0,1)$. Let $f$ be a harmonic mapping on $\Omega$ with the expansion of the form
\begin{equation}\label{eq17a}
f(z)=\frac{1}{z-p} + h(z) + \bar{g(z)}, \quad z\in \Omega \setminus \{p\},
\end{equation}
where $h$ and $g$ are analytic on $\Omega$. If
\begin{equation}\label{eq20}
\abs[\big]{h'(z)} + \abs[\big]{g'(z)} \leq \frac{1}{M(1+p)^2}, \quad z \in \Omega,
\end{equation}
then $f$ is univalent on\/ $\Omega$. Moreover, if\/ $\Omega$ is a quasidisk in\/ $\ID$ and $f$ satisfies
\begin{equation}\label{eq20a}
\abs[\big]{h'(z)} + \abs[\big]{g'(z)} \leq \frac{k}{M(1+p)^2}, \quad z \in \Omega,
\end{equation}
for some $k \in [0,1)$, then $f$ admits a quasiconformal extension to\/ $\sphere$.
\end{theo}
\begin{proof} Since $f$ has the expansion of the form~\eqref{eq17a}, proceeding in a similar fashion as in the first part of the proof of Theorem~\ref{th2}, we get
\begin{equation}\label{eq21}
\abs[\big]{f(z_1)-f(z_2)} \geq \frac{\abs{z_1-z_2}}{\abs{z_1-p}\abs{z_2-p}} - \int_{z_2}^{z_1} \parens[\big]{\abs[\big]{h'(z)}+ \abs[\big]{g'(z)}}\,\abs{\dd z},
\end{equation}
where $z_1 \neq z_2 \in \Omega$. Since $\Omega$ is a linearly connected domain (with domain constant $M$), we choose here the curve $\gamma$ joining $z_1,z_2$, as the path of integration, for which
\begin{equation}\label{eq22}
l(\gamma) \leq M\abs{z_1 - z_2}.
\end{equation}
Hence, using~\eqref{eq20} and~\eqref{eq22}, we get from~\eqref{eq21} that
\[
\abs[\big]{f(z_1)-f(z_2)} \geq \frac{\abs{z_1-z_2}}{\abs{z_1-p}\abs{z_2-p}} - \frac{l(\gamma)}{M(1+p)^2}
\geq \frac{\abs{z_1-z_2}}{\abs{z_1-p}\abs{z_2-p}} - \frac{\abs{z_1-z_2}}{(1+p)^2} &gt;0,
\]
which implies $f$ is univalent on $\Omega$.

To prove the second part of the theorem, we first observe that the case $k=0$ is trivial, hence we assume $k\in (0,1)$. Next we see that, if $f$ satisfies~\eqref{eq20a}, then for $z_1 \neq z_2 \in \Omega$, we have
\begin{equation}\label{eq23}
\begin{split}
\abs[\big]{h(z_1)-h(z_2)+ \bar{g(z_1)-g(z_2)}} 	&amp; = \abs*{\int_{z_2}^{z_1} h'(z)\,\dd z + \bar{g'(z)\,\dd z}} \\
&amp; \leq \int_{z_2}^{z_1} \parens[\big]{\abs[\big]{h'(z)}+ \abs[\big]{g'(z)}}\,\abs{\dd z} \\
&amp; \leq \frac{k~l(\gamma)}{M(1+p)^2} \\
&amp; \leq \frac{k\abs{z_1-z_2}}{(1+p)^2},
\end{split}
\end{equation}
where $\gamma$ is the curve joining $z_1,z_2$, which satisfies the inequality~\eqref{eq22}. Thus, as in the proof of Theorem~\ref{th2}, we can extend $f$ continuously to the boundary $\partial \Omega$, where the inequality~\eqref{eq23} remains valid. Also following the proof of Theorem~\ref{th2}, we conclude that $f$ is a sense-preserving homeomorphism on $\bar{\Omega}$ ($\coloneqq\Omega \cup \partial \Omega$) with $\abs{\mu_f}&lt;k$ on $\Omega \setminus \{p\}$. As a result, $f$ is $k$-quasiconformal on~$\Omega$.

We now use Ahlfors's criterion to prove that $f(\partial \Omega)$ is a quasidisk. In order to do so, let us consider four distinct points $w_i$, $i=1,2,3,4$ on $f(\partial \Omega)$ such that $w_1$ and $w_4$ separate $w_2$ and $w_3$, and show that the cross-ratio defined in~\eqref{eq2} is uniformly bounded by a constant. Since $f$ is a homeomorphism on $\bar{\Omega}$, there exist exactly four distinct points $z_i$ on $\partial \Omega$ such that $f(z_i)=w_i$ for $i=1,2,3,4$.

Now, for any two distinct points $z_i$ and $z_j$ on $\partial \Omega$, since $\Omega \subset \ID$, using~\eqref{eq23} we have
\[
\abs[\big]{f(z_i)-f(z_j)} \geq \frac{\abs{z_i-z_j}}{\abs{z_i-p}\abs{z_j-p}} - \frac{k\abs{z_i - z_j}}{(1+p)^2} &gt; \frac{(1 - k)\abs{z_i-z_j}}{\abs{z_i-p}\abs{z_j-p}}.
\]
Similarly, we obtain
\[
\abs[\big]{f(z_i)-f(z_j)} \leq \frac{\abs{z_i-z_j}}{\abs{z_i-p}\abs{z_j-p}} + \frac{k\abs{z_i - z_j}}{(1+p)^2} &lt; \frac{(1+k)\abs{z_i-z_j}}{\abs{z_i-p}\abs{z_j-p}}.
\]
The last two inequalities hold for all $i,j=1,2,3,4$ with $i \neq j$. Using these inequalities, we obtain
\[
\begin{split}
\abs[\big]{(w_1,w_2,w_3,w_4)}	&amp; = \abs[\Bigg]{\parens[\Bigg]{\frac{w_1-w_3}{w_1-w_4}}\parens[\Bigg]{\frac{w_2-w_4}{w_2-w_3}}} \\
&amp; = \frac{\abs[\big]{f(z_1)-f(z_3)}}{\abs[\big]{f(z_1)-f(z_4)}} \frac{\abs[\big]{f(z_2)-f(z_4)}}{\abs[\big]{f(z_2)-f(z_3)}} \\
&amp; &lt; \parens[\Bigg]{\frac{1+k}{1-k}}^2 \abs[\big]{(z_1,z_2,z_3,z_4)}.
\end{split}
\]
Each $z_i$ lies on $\partial \Omega$, which is a quasicircle. Thus, by~\eqref{eq2} we conclude that $\abs[\big]{(z_1,z_2,z_3,z_4)}$ is finite. Hence $\abs[\big]{(w_1,w_2,w_3,w_4)}&lt; C$, for some constant $C&gt;0$. Therefore, by applying Ahlfors's criteria, it follows that $f(\partial \Omega)$ is also a quasicircle. Hence, $f$ is a $k$-quasiconformal mapping from the quasidisk $\Omega$ onto the quasidisk $f(\Omega)$. Now by~\cite[Theorem~8.3, p.~98]{LV}, we conclude that $f$ admits a quasiconformal extension to $\sphere$. This ends the proof of the theorem.
\end{proof}

\begin{remas}
\mbox{}
\begin{enumerate}
\item Since the unit circle $\partial \ID$ is a quasicircle, we may assume the domain $\Omega$ to be $\ID$ in Theorem~\ref{th4} and deduce Theorem~\ref{th2} as a special case, because the domain constant $M$ for $\ID$ is $1$. But, Theorem~\ref{th4} does not provide us explicit quasiconformal extension of the harmonic mapping, likewise we got in Theorem~\ref{th2}.
\item Theorem~\ref{th4} can be proved for harmonic mappings with pole at the origin (i.e., for $p =0$), and also for meromorphic mappings (considering $g \equiv 0$) defined on a linearly connected domain $\Omega \subset \ID$. But, we are not presenting here the details, as they easily follow from Theorem~\ref{th4}.
\end{enumerate}
\end{remas}

\section*{Declaration of interests}

The authors do not work for, advise, own shares in, or receive funds from any organization that could benefit from this article, and have declared no affiliations other than their research organizations.

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