Let $f(z)=\begin{cases} \frac{z^5}{\left | z \right |^4} & \text{ if } z\neq 0 \\ 0 & \text{ if } z=0 \end{cases} $
I could show this is continuous on $\mathbb{C}$.
And, I would like to show this satisfies the C-R equation at $z=0$.
However, there are some things confusing me, especially the meaning of 'a function satisfies C-R equation at some specific point.'
Let me share my work.
Since $f$ is continuous on $\mathbb{C}$, we can let $f(z) = u(x,y) + iv(x,y)$ for some continuous $u,v$ on $\mathbb{C}$.
Since $f(0) = 0, \lim_{x\rightarrow 0, y\rightarrow 0}u(x,y)=0$ and $\lim_{x\rightarrow 0, y\rightarrow 0}v(x,y)=0$.
This means $u(0,0) = 0$ and $v(0,0)=0$.
Thus, $\frac{\partial u}{\partial x}=\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y} =0$
Hence, the C-R equation satisfies for $f$ at $z=0$.
Is this how I should approach for this kind of problems?
Or should I let $z = x+yi$ and do calculations? I also tried this way, but it is not that simple as I expected. This is why I want to see if this proof works.