Let $$f(z)=\begin{cases}(z\overline{z}^{-1})^2&z\neq 0\\1&z=0\end{cases}.$$
I need to show that the Cauchy-Riemann equations hold for $f$ in $0$ but $f$ is not (complex) differentiable in $0$. I have shown that $f$ is not continous in $0$, so indeed not differentiable.
Because of $$f(z)=\frac{z^4}{|z^4|}$$ we see $$f|_\mathbb{R}=f|_{i\mathbb{R}}=1=f(0).$$ But with $u=\text{Re}(z),~v=\text{Im}(z)$, how can we see $$\partial_xu(0)=\partial_yv(0)\text{ and }\partial_yu(0)=-\partial_xv(0)?$$
Ok so the first thing is that $u$ is not $Re(z)$. It is the real part of $f$. $u(z)=Re(f(z))$. The same for $v$. Now we can define $u,v$ as two functions $u,v:\mathbb{R}^2 \to \mathbb{R}$. So $u(x,y) = Re(f(x,y)) = \frac{Re((x+iy)^4)}{|x+iy|^4}=\frac{x^4-6x^2y^2+y^4}{|x+iy|^4}$ for $(x,y) \in \mathbb{R}^2\setminus0$ and $u(0,0) = 1$, $v(x,y) = Im(f(x,y)) = \frac{Im((x+iy)^4)}{|x+iy|^4}=\frac{4x^3y-4xy^3}{|x+iy|^4}$ for $(x,y) \in \mathbb{R}^2\setminus0$ and $v(0,0) = 0$.
Then we can derive $u,v$ with respect to $x,y$. The easiest way to do this, is using the definition of derivatives.
$\frac{\partial u}{\partial x}(0,0)=lim_{h \to 0} \frac{u(h,0) - u(0,0)}{h} = lim_{h \to 0} \frac{\frac{h^4}{h^4}-1}{h}=0$
$\frac{\partial v}{\partial y}(0,0)=lim_{h \to 0} \frac{v(0,h) - v(0,0)}{h} = lim_{h \to 0} \frac{0}{h}=0$
$\frac{\partial u}{\partial y}(0,0)=lim_{h \to 0} \frac{u(0,h) - u(0,0)}{h} = lim_{h \to 0} \frac{\frac{h^4}{h^4}-1}{h}=0$
$\frac{\partial v}{\partial x}(0,0)=lim_{h \to 0} \frac{v(h,0) - v(0,0)}{h} = lim_{h \to 0} \frac{0}{h}=0$
Now we have shown that the Cauchy Riemann equations hold.
