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?
Best Answer
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.