Show that $f(z) = z^2 \overline z$ is not analytical anywhere Show that $f(z) = z^2 \overline z$ is not analytical anywhere, i.e. $f(z) = z^2 \overline z$ is non holomorphic anywhere.
I think I am supposed to use Cauchy-Riemann equations but not sure how to. 

Hint 
$$f(z)=z^2 \bar{z}=z(z \bar{z})=z |z|^2$$
Writing $z=x+iy$ you get
$$f(x+iy)=(x+iy)(x^2+y^2) $$
Now use the CR Riemann Equations, you will find out that $f$ is differentiable only at one point.
Alternate solution: If $f(z)$ is holomorphic on some domain $D$, then $\frac{f(z)}{z}$ is holomoorphic on the domain $D \setminus \{ 0 \}$. But  $\frac{f(z)}{z}=|z|^2$ only takes real values... 
