So I want to check if $|z|^2$ is complex differentiable. I did not want to use Cauchy-Riemann ODE since we did not have it in the lecture. What I tried is a bit weird and I know that I did a mistake somewhere, but can't figure out where exactly. I want to use the definition only:
$$ \lim_{z\rightarrow z_0} \frac{f(z)-f(z_0)}{z-z_0} = \lim_{z\rightarrow z_0} \frac{x^2+y^2-x_0^2-y_0^2}{(x+iy)-(x_0+iy_0)} $$
now this looks pretty ugly. So I would try to splitt it:
i) $z_1= x_0+iy$ with $y\rightarrow y_0$ (vertical)
ii) $z_2=x+iy_0$ with $x\rightarrow x_0$ (horizontal)
We get:
i) $$ \lim_{z\rightarrow z_0} \frac{f(z_1)-f(z_0)}{z-z_0} = \lim_{y\rightarrow y_0} \frac{y^2-y_0^2}{i(y-y_0)}=\lim_{y\rightarrow y_0} \frac{y+y_0}{i} =\frac{2y_0}{i}$$
ii) $$ \lim_{z\rightarrow z_0} \frac{f(z_2)-f(z_0)}{z-z_0} = \lim_{x\rightarrow x_0} \frac{x^2-x_0^2}{(x-x_0)}=\lim_{x\rightarrow x_0} x+x_0 =2x_0$$
Well now it has to be $\frac{2y_0}{i}=2x_0$. But this is only the case if... $x_0=y_0=0$? It definitly is the case for $0$ but how can I show that it is only the case for $0$? Is the rest right and if not, where is the mistake?