<s>[INST] How to show that $f(z)=\frac{z^{5}}{|z|^4}$ satisfies the Cauchy Riemann equations at $z=0$ but not differentable at $z=0$? $$
f(z) = \left\{
        \begin{array}{ll}
            \frac{z^{5}}{|z|^4} & \quad z \neq 0 \\
            0 & \quad z= 0
        \end{array}
    \right.
$$
I am trying to solve this past exam question. Similar question was asked in Show that $f(z)=\frac{z^5}{|z|^4}$ but has not been answered.
My attempt:
Let $z=x+iy$ then we can separate the real and imaginary parts of $f(z)$ as
$$
f(z)=u+iv=\left(\frac{x^5-10 x^3 y^2+5 x y^4}{\left(x^2+y^2\right)^2}\right)+i\left(\frac{5 x^4 y-10 x^2
   y^3+y^5}{\left(x^2+y^2\right)^2}\right)
$$ 
$$
\frac{\partial u}{\partial x}=\frac{5 x^4-30 x^2 y^2+5 y^4}{\left(x^2+y^2\right)^2}-\frac{4 x
   \left(x^5-10 x^3 y^2+5 x y^4\right)}{\left(x^2+y^2\right)^3}
$$
$$
\frac{\partial v}{\partial y}=\frac{5 x^4-30
   x^2 y^2+5 y^4}{\left(x^2+y^2\right)^2}-\frac{4 y \left(5 x^4 y-10 x^2
   y^3+y^5\right)}{\left(x^2+y^2\right)^3}
$$
$$
\frac{\partial u}{\partial y}=\frac{20 x y^3-20 x^3 y}{\left(x^2+y^2\right)^2}-\frac{4 y \left(x^5-10
   x^3 y^2+5 x y^4\right)}{\left(x^2+y^2\right)^3}
$$
$$
\frac{\partial v}{\partial x}=\frac{20 x^3 y-20 x
   y^3}{\left(x^2+y^2\right)^2}-\frac{4 x \left(5 x^4 y-10 x^2
   y^3+y^5\right)}{\left(x^2+y^2\right)^3}
$$
I am not sure what to do next. If I substitute $z=0\rightarrow x=0,y=0$ in the equations above then they become infinite because the bottom term $(x^{2}+y^{2})$ becomes zero. So how can I show that $f(z)$ satisfies the Cauchy Riemann equations at $z=0$. Also, how can I show that $f(z)$ is not differentiable at $z=0$?
Thanks.
 [/INST] It must satisfy the CR condition in $0$, because they only involve partial derivatives in the $x-$ and $y-$ directions, and the function coincides with $f(z)=z$ both on $x$ and $y$-axis. (Note that $|ix|^4=x^4=(ix)^4$ for real $x$.)
</s>