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\renewcommand{\spcDoi}{doi: 10.14419/gjma.v5i?.????}
\renewcommand{\spcJournalDesc}{Short communication}
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\begin{document}

\title{A criterion to justify a holomorphic function}

\author[1,2,3,*]{Feng Qi}
\author[4]{Bai-Ni Guo}
\affil[1]{Institute of Mathematics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China}
\affil[2]{College of Mathematics, Inner Mongolia University for Nationalities, Tongliao City, Inner Mongolia Autonomous Region, 028043, China}
\affil[3]{Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin City, 300387, China}
\affil[4]{School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China}
\affil[*]{Corresponding author E-mail: qifeng618@gmail.com}

\spcAbstract{In the paper, the authors present a criterion, free of considering differentiability and any partial derivative, to justify a holomorphic function and illustrate the criterion by several examples.}

\spcKeywords{criterion; justifying; holomorphic function; differentiability; partial derivative; example.}


\maketitle

\section{Main result}

In the theory of complex functions, the usually used criterion to justify a holomorphic function is considering differentiability of real and imaginary parts and solvability of system of the Cauchy-Riemann equations for a complex function.
\par
The main aim of this paper is to present a criterion which may be summarized as Theorems~\ref{anal-criteria-thm} and~\ref{anal-criteria-gen-thm} below, free of considering differentiability and any partial derivative, to justify a holomorphic function.

\begin{thm}\label{anal-criteria-thm}
Let $D\subseteq\mathbb{C}$ be a domain such that the origin $0\in D$. Then a complex function
\begin{equation}\label{f(z)=u=v-eq}
f(z)=f(x+iy)=u(x,y)+iv(x,y)
\end{equation}
is holomorphic on $D$ if and only if the functions
\begin{equation}\label{u-v-u-v-eq}
u(z,0), \quad v(z,0), \quad u(0,-iz), \quad\text{and} \quad v(0,-iz)
\end{equation}
are holomorphic and system of equations
\begin{equation}\label{f(z)=u-v-0-0-eq}
\begin{cases}
f(z)=u(z,0)+iv(z,0);\\
f(z)=u(0,-iz)+iv(0,-iz)
\end{cases}
\end{equation}
is valid on the domain $D$.
\end{thm}

\begin{thm}\label{anal-criteria-gen-thm}
Let $D\subseteq\mathbb{C}$ be a domain and $z_0=x_0+iy_0\in D$. Then a complex function $f(z)$ expressed in~\eqref{f(z)=u=v-eq} is holomorphic on $D$ if and only if the functions
\begin{equation}\label{u-v-u-v-z=0eq}
\begin{gathered}
u(z-iy_0,y_0), \quad v(z-iy_0,y_0), \\
u(x_0,i(x_0-z)), \quad v(x_0,i(x_0-z))
\end{gathered}
\end{equation}
are holomorphic and the system of equations
\begin{equation}\label{F(z)=u-v-0-0-eq}
\begin{cases}
f(z)=u(z-iy_0,y_0)+iv(z-iy_0,y_0);\\
f(z)=u(x_0,i(x_0-z))+iv(x_0,i(x_0-z))
\end{cases}
\end{equation}
is valid on the domain $D$.
\end{thm}

In next section, we will prove Theorems~\ref{anal-criteria-thm} and~\ref{anal-criteria-gen-thm}.
\par
In the final section, we will take several examples, which are much familiar for readers, to illustrate Theorems~\ref{anal-criteria-thm} and~\ref{anal-criteria-gen-thm}.
\par
Theorems~\ref{anal-criteria-thm} and~\ref{anal-criteria-gen-thm} generalize~\cite[pp.~182--183, Exercise~7]{Zhong-Yu-Quan}.

\section{Proofs of Theorems~\ref{anal-criteria-thm} and~\ref{anal-criteria-gen-thm}}

\begin{proof}[Proof of Theorem~\ref{anal-criteria-thm}]
Necessity. Taylor's theorem~\cite[p.~177]{Ahlfors-CA-2nd-1966} reads that, if $f(z)$ is holomorphic in the region $\Omega$, containing $z_0$, then the representation
\begin{equation}
f(z)=\sum_{k=0}^\infty\frac{f^{(k)}(z_0)}{k!}(z-z_0)^k
\end{equation}
is valid in the largest open disk of center $z_0$ contained in $\Omega$. Letting $z_0=0$ and assuming $\frac{f^{(k)}(0)}{k!}=A_k+i B_k$ lead to
\begin{equation*}
u(x,y)+iv(x,y)=\sum_{k=0}^\infty(A_k+i B_k)(x+iy)^k.
\end{equation*}
Taking $y=0$ and $x=0$ respectively in the above equation gives
\begin{equation*}
u(x,0)+iv(x,0)=\sum_{k=0}^\infty(A_k+i B_k)x^k
\end{equation*}
and
\begin{equation*}
u(0,y)+iv(0,y)=\sum_{k=0}^\infty(A_k+i B_k)(iy)^k.
\end{equation*}
Consequently, it follows that
\begin{gather*}
u(x,0)=\sum_{k=0}^\infty A_kx^k, \quad v(x,0)=\sum_{k=0}^\infty B_kx^k,\\
u(0,y)=\sum_{k=0}^\infty(-1)^kA_{2k}y^{2k}+\sum_{k=0}^\infty(-1)^{k+1}B_{2k+1}y^{2k+1},\\
v(0,y)=\sum_{k=0}^\infty(-1)^{k+1}A_{2k+1}y^{2k+1}+\sum_{k=0}^\infty(-1)^kB_{2k}y^{2k}.
\end{gather*}
This implies, by Taylor's theorem and the important uniqueness theorem (see~\cite[p.~279, Theorem~1]{Ahlfors-CA-2nd-1966}) that the four functions in~\eqref{u-v-u-v-eq} are holomorphic.
\par
Letting $y=0$ and $x=0$ respectively in~\eqref{f(z)=u=v-eq} reveals that the equation system
\begin{equation}\label{system-xy}
\begin{cases}
f(x)=u(x,0)+iv(x,0);\\
f(iy)=u(0,y)+iv(0,y)
\end{cases}
\end{equation}
is valid for $x,y\in(-\delta,\delta)$, where $\delta>0$. By the important uniqueness theorem and by holomorphic property of the four functions in~\eqref{u-v-u-v-eq}, we see that the system~\eqref{system-xy} is valid on the whole domain $D$. Consequently, replacing $x$ by $z$ and $y$ by $-iz$ in~\eqref{system-xy} figures out the equation system~\eqref{f(z)=u-v-0-0-eq}.
\par
The sufficiency is obvious. Theorem~\ref{anal-criteria-thm} is complete.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{anal-criteria-gen-thm}]
If the origin $0$ does not contain in the domain $D$ and the complex function $f(z)$ is holomorphic on $D$, then we may consider the function
\begin{gather*}
F(w)=f(w+z_0)=f(p+x_0+i(q+y_0))\\
=u(p+x_0,q+y_0)+iv(p+x_0,q+y_0)
=u_{z_0}(p,q)+iv_{z_0}(p,q)
\end{gather*}
for
\begin{equation*}
w=p+iq\in D_{z_0}=\{z-z_0:z_0,z\in D\}.
\end{equation*}
When we apply Theorem~\ref{anal-criteria-thm} to both the function $F(w)$ and the domain $D_{z_0}$, the equation system~\eqref{f(z)=u-v-0-0-eq} becomes
\begin{equation*}
\begin{cases}
F(w)=u_{z_0}(w,0)+iv_{z_0}(w,0);\\
F(w)=u_{z_0}(0,-iw)+iv_{z_0}(0,-iw)
\end{cases}
\end{equation*}
which is equivalent to
\begin{equation*}
\begin{cases}
f(w+z_0)=u(w+x_0,y_0)+iv(w+x_0,y_0);\\
f(w+z_0)=u(x_0,y_0-iw)+iv(x_0,y_0-iw)
\end{cases}
\end{equation*}
for $w=p+iq\in D_{z_0}$. Replacing $w$ by $z-z_0$ in the last equation system results in
\begin{equation*}
\begin{cases}
f(z)=u(z-z_0+x_0,y_0)+iv(z-z_0+x_0,y_0);\\
f(z)=u(x_0,y_0-i(z-z_0))+iv(x_0,y_0-i(z-z_0))
\end{cases}
\end{equation*}
for $z\in D$. Simplifying the above equations lead to~\eqref{F(z)=u-v-0-0-eq}. The proof of Theorem~\ref{anal-criteria-gen-thm} is complete.
\end{proof}

\section{Examples}

To illustrate the criterion, we list several examples below.

\subsection{Example}
It is common knowledge in the theory of complex functions that the complex functions
\begin{equation}\label{examp-4f(z)-eq}
\overline{z}=x-iy, \quad |z|=\sqrt{x^2+y^2}\,,\quad \Re(z)=x,\quad \text{and}\quad \Im(z)=y
\end{equation}
are not holomorphic everywhere on the whole complex plane $\mathbb{C}$. This may be verified by respectively considering solvability of the Cauchy-Riemann equations and differentiability of real and imaginary parts of the four functions in~\eqref{examp-4f(z)-eq}.
\par
With the help of Theorem~\ref{anal-criteria-thm}, due to the obvious fact that the equation systems
\begin{gather*}
\begin{cases}
\overline{z}=z;\\
\overline{z}=-z,
\end{cases}
\quad
\begin{cases}
|z|=\sqrt{z^2}\,;\\
|z|=\sqrt{-z^2}\,,
\end{cases}
\quad
\begin{cases}
\Re(z)=z;\\
\Re(z)=0,
\end{cases}
\end{gather*}
and
\begin{equation*}
\begin{cases}
\Im(z)=0;\\
\Im(z)=-iz
\end{cases}
\end{equation*}
are respectively valid only at the origin $0$, we immediately see that all of the four functions in~\eqref{examp-4f(z)-eq} are not holomorphic everywhere on the complex plane $\mathbb{C}$.
\par
The above arguments show that Theorem~\ref{anal-criteria-thm} provides a simpler criterion, without computing derivatives, to justify holomorphic functions.

\subsection{Example}
Let
\begin{equation}\label{examp-f(z)-eq}
f(z)=u(x,y)+iv(x,y)=x^2+axy+by^2+i\bigl(\alpha x^2+\beta xy+y^2\bigr)
\end{equation}
on $\mathbb{C}$. By virtue of the Cauchy-Riemann equations, we may obtain that, when
\begin{equation}\label{examp-solution-eq}
a=2,\quad b=-1, \quad \alpha=-1, \quad\text{and}\quad \beta=2,
\end{equation}
the function $f(z)$ is holomorphic on $\mathbb{C}$. See~\cite[p.~43, Example~2]{xi-an-CFB-v4}.
\par
Since
\begin{equation*}
u(z,0)=z^2, \quad v(z,0)=\alpha z^2, \quad u(0,-iz)=-bz^2,
\end{equation*}
and
\[
v(0,-iz)=-z^2,
\]
the equation system~\eqref{f(z)=u-v-0-0-eq} becomes
\begin{equation}\label{system-eqs}
\begin{cases}
x^2+axy+by^2+i\bigl(\alpha x^2+\beta xy+y^2\bigr)=(1+\alpha i)z^2;\\
x^2+axy+by^2+i\bigl(\alpha x^2+\beta xy+y^2\bigr)=-(b+i)z^2
\end{cases}
\end{equation}
which is a linear system of equations in four variables $a$, $b$, $\alpha$, and $\beta$. The equation system~\eqref{system-eqs} has the unique solution~\eqref{examp-solution-eq}. By Theorem~\ref{anal-criteria-thm}, we reveal that, under the condition~\eqref{examp-solution-eq}, the function $f(z)$ in~\eqref{examp-f(z)-eq} is holomorphic and may be easily rewritten in terms of $z$, rather than $x$ and $y$, as
\begin{equation*}
f(z)=(1-i)z^2
\end{equation*}
on the whole complex plane $\mathbb{C}$.
\par
The arguments tell us that, to justify a holomorphic function $f(z)$, it is sufficient to solve the equation system~\eqref{f(z)=u-v-0-0-eq}.

\subsection{Example}
It is general knowledge that the logarithmic function
\begin{equation*}
f(z)=\ln z=\ln|z|+i\arg z
\end{equation*}
is holomorphic on the cut plane $\mathcal{A}=\mathbb{C}\setminus(-\infty,0]$.
\par
Taking any point $z_0=x_0+y_0\in\mathcal{A}$ and replacing in~\eqref{F(z)=u-v-0-0-eq} $f(z)$ by the logarithmic function $\ln z$ gives
\begin{equation*}
\begin{cases}
\ln z=\ln|z-iy_0+iy_0|+i\arg(z-iy_0+iy_0)=\ln z;\\
\ln z=\ln|x_0+i\times i(x_0-z)|+i\arg(x_0+i\times i(x_0-z))=\ln z
\end{cases}
\end{equation*}
which is valid on the cut plane $\mathcal{A}$. This implies, by Theorem~\ref{anal-criteria-gen-thm}, that the logarithmic function $\ln z$ is holomorphic on the cut plane $\mathcal{A}$.
\par
Since the imaginary part of $\ln z$ is $v(x,y)=\arg z=\arg(x+iy)$ on $\mathcal{A}$, by Theorem~\ref{anal-criteria-gen-thm}, we further obtain that, for any $z_0=x_0+iy_0\in\mathcal{A}$, the functions
\begin{equation*}
v(z-iy_0,y_0)=\arg(z-iy_0+iy_0)=\arg z
\end{equation*}
and
\begin{equation*}
v(x_0,i(x_0-z))=\arg(x_0-(x_0-z))=\arg z
\end{equation*}
are holomorphic on $\mathcal{A}$. Simply speaking, the argument $\arg z$ is holomorphic on $\mathcal{A}$.

\subsection{Example}
Let
\begin{equation}\label{sqrt-plus-pm-eq}
f(z)=\sqrt{1+z}\,+\sqrt{1-z}\,
\end{equation}
on
\begin{equation}\label{domain-(infty,pm1)-eq}
D=\mathbb{C}\setminus(-\infty,-1]\cup[1,\infty).
\end{equation}
It is clearly a single-valued complex function. See~\cite[p.~73, Exercise~1]{Ahlfors-CA-2nd-1966}.
\par
A direct computation gives
\begin{gather*}
\begin{aligned}
f(x+iy)&=u(x,y)+iv(x,y)\\
&=\sqrt[4]{(x-1)^2+y^2}\, \cos\frac{\arg (1-x-iy)}{2}\\
&\quad+\sqrt[4]{(x+1)^2+y^2}\,\cos\frac{\arg (1+x+iy)}{2}\\
&\quad+i \biggl[\sqrt[4]{(x-1)^2+y^2}\, \sin\frac{\arg (1-x-iy)}{2}\\
&\quad+\sqrt[4]{(x+1)^2+y^2}\, \sin\frac{\arg (1+x+iy)}{2}\biggr],
\end{aligned}\\
u(z,0)=\sqrt[4]{(z-1)^2}\, \cos\frac{\arg (1-z)}{2}+\sqrt[4]{(z+1)^2}\,\cos\frac{\arg (z+1)}{2},\\
v(z,0)=\sqrt[4]{(z-1)^2}\, \sin\frac{\arg (1-z)}{2}+\sqrt[4]{(z+1)^2}\, \sin\frac{\arg(z+1)}{2},\\
u(0,-iz)=\sqrt[4]{1-z^2}\, \biggl[\cos\frac{\arg (1-z)}{2}+\cos\frac{\arg (1+z)}{2}\biggr],\\
v(0,-iz)=\sqrt[4]{1-z^2}\, \biggl[\sin\frac{\arg(1-z)}{2}+\sin\frac{\arg (1+z)}{2}\biggr],
\end{gather*}
and the validity of the equation system~\eqref{f(z)=u-v-0-0-eq} applying to the function~\eqref{sqrt-plus-pm-eq} on $D$. Consequently, by virtue of Theorem~\ref{anal-criteria-thm}, we see that the function~\eqref{sqrt-plus-pm-eq} is holomorphic on the domain~\eqref{domain-(infty,pm1)-eq}.

\subsection{Example}
Suppose $f(z)=u(x,y)+iv(x,y)$ and its conjugate $\overline{f(z)}=u(x,y)-iv(x,y)$ are all holomorphic on a domain $D$ and let $z_0=x_0+iy_0\in D$. By virtue of Theorem~\ref{anal-criteria-gen-thm}, we see that the functions in~\eqref{u-v-u-v-z=0eq} are holomorphic and that the equation systems~\eqref{F(z)=u-v-0-0-eq} and
\begin{equation*}
\begin{cases}
\overline{f(z)}=u(z-iy_0,y_0)-iv(z-iy_0,y_0);\\
\overline{f(z)}=u(x_0,i(x_0-z))-iv(x_0,i(x_0-z))
\end{cases}
\end{equation*}
are valid. Hence, it follows that
\begin{align*}
u(x,y)-iv(x,y)&=\overline{u(z-iy_0,y_0)}-i\overline{v(z-iy_0,y_0)}\\
&=u(z-iy_0,y_0)-iv(z-iy_0,y_0)\\
&=\overline{u(x_0,i(x_0-z))}-i\overline{v(x_0,i(x_0-z))}\\
&=u(x_0,i(x_0-z))-iv(x_0,i(x_0-z)).
\end{align*}
As a result, the functions in~\eqref{u-v-u-v-z=0eq} are real constants on $D$, and so the complex function $f(z)$ and its conjugate $\overline{f(z)}$ are two constants.


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\end{thebibliography}

\end{document}