Satisfies Cauchy-Riemann equations but is not holomorphic at $0$ I am trying to solve the following problem:

Consider the function defined by $f(x+iy)=\sqrt{|x||y|}$ whenever
$x,y\in \mathbb{R}$. Show that $f$ satisfies the C-R equations at the
origin yet $f$ is not holomorphic at $0$.

Firstly, what would be $u$ and $v$ in this problem? Since it is a mapping from $\mathbb{C}\to\mathbb{R}$, would it be that $f(x+iy)=\sqrt{|x||y|}+i0$
where $u(x,y)=Re(f(z))=\sqrt{|x||y|}$ and $v(x,y)=Im(f(z))=0$?
Secondly, I have attempted to compute the partial derivatives of $u$ w.r.t $x$ and $y$ as follows:
$$\frac{\partial u}{\partial x}=\frac{\partial \sqrt{|x||y|}}{\partial x}=\frac{{\sqrt{|y|}\text{sgn}(x)}}{\sqrt{|x|}}$$
$$\frac{\partial u}{\partial y}=\frac{\partial \sqrt{|x||y|}}{\partial y}=\frac{{\sqrt{|x|}\text{sgn}(y)}}{\sqrt{|y|}}$$
However, I am confused because sgn(x) is not defined for $0$ so I am unable to determine what the value of the partial derivatvies at the origin. I have seen some people compute this direcrly from definition and get zero as an answer, but why am I unable to compute this from the partial derivatives I have calculated?
 A: By definition,
$$
\begin{align}
\frac{\partial u}{\partial x}(0,0)&=
\lim_{x\to0}\frac{u(x,0)-u(0,0)}{x-0}\\
&=\lim_{x\to0}\frac{0-0}{x-0}\\
&=0
\end{align}
$$
and similarly $\frac{\partial u}{\partial y}(0,0)=0$
Computing the partial derivatives at a general point $(x,y)$ is unnecessary and misleading. To answer your question in a comment, if you wanted to compute $\frac{\partial u}{\partial x}(0,y)$ at some $y\neq0$, you'd get
$$
\begin{align}
\frac{\partial u}{\partial x}(0,y)&=
\frac{u(x,y)-u(0,y)}{x-0}\\
&=\lim_{x\to0}\frac{\sqrt{|xy|}-0}{x-0}\\
&=\sqrt{|y|}\lim_{x\to0}\frac{\sqrt{|x|}}{x}
\end{align}
$$ which doesn't exist.
A: Well since
\begin{eqnarray*}
f:\Omega \subseteq \mathbf{C}&\longrightarrow& \mathbf{C}\\
(x,y)&\longmapsto& f(x,y)=\left(\sqrt{|xy|},0\right)
\end{eqnarray*}
so we can see that

*

*$u(x,y)=\sqrt{|xy|}$.

*$v(x,y)=0$
Now the Cauchy-Riemann state that if $f$ is holomorphic function on $\Omega$ so
$$({\rm C-R}):\begin{cases}\displaystyle \frac{\partial u}{\partial x}(a,b)=\frac{\partial v}{\partial y}(a,b)\\\displaystyle \frac{\partial u}{\partial y}(a,b)=-\frac{\partial v}{\partial x}(a,b) \end{cases}, \quad (a,b)\in \Omega. \quad (*)$$
Setting $(a,b)=(0,0)$ so we have that
$$\frac{\partial v}{\partial y}(0,0)=0, \quad \frac{\partial u}{\partial x}(0,0)=0,\quad \frac{\partial u}{\partial y}(0,0)=0, \quad \text{and} \quad -\frac{\partial v}{\partial x}(0,0)=0$$
Therefore $(*)$ holds at $(0,0)$, but $f$ is not holomorphic in $(0,0)$ (just use the definition in $(0,0)$).
