Let $f(z): \mathbb{C}\rightarrow \mathbb{C}$ be a complex function such that $f(z) = \begin{Bmatrix} \frac{z^5}{|z|^4},z\neq 0\\ 0, z=0 \end{Bmatrix}$.
Prove that Cauchy-Riemann equations(in x and y version) are met in $(0,0)$ but the function is not differentiable in $(0,0).$
In order to avoid using the binomial theorem on the expression $(x+yi)^5$ and getting a general expression for $u(x,y),v(x,y)$. I did as followed -
$u_x(0,0) = \lim_{x\rightarrow 0} \frac{u(x,0)-u(0,0)}{x}$
an expression for $u(x,0)$ would be -
$f(x+0i) = u(x,0)+v(x,0)i$ = $\frac{(x+0i)^5}{x^4}$ = $x$.
so $u_{x}(0,0)$ is just 1.
and getting $u_{y}(0,0),v_{x}(0,0),v_{y}(0,0)$ would be in a similiar way.
Is that legal mathematically? or a logical mistake was fallen somewhere? beacuse there are a lot of arguments around my fellows in class.
Thanks in advance.