Is $ f(z)= |z|.\bar z $ analytic? I want to know whether it is analytic - and if so, to find $ f´(z)$
What I do: 
I use the polar form.
$ z= x + iy \\  |z|=r  \\ \bar z= (cos (\phi) + isin(\phi))$
then: $ f(z)= r^2(cos (\phi) + isin(\phi)) = r^2cos (\phi) + i r^2sin(\phi)$
Cauchy-Riemann:
$\begin{cases} \\
\frac{\partial u}{\partial r} \ = \frac{1}{ r}\frac{\partial v}{\partial \phi} \\[2ex] \frac{\partial u}{\partial r} \ = \frac{1}{ r}\frac{\partial v}{\partial \phi}\\
\end{cases} \quad  \Rightarrow \quad 
\begin{cases}\\
2r\cos \phi= -r\cos\phi \\[2ex] -2r\sin \phi=r\sin\phi\\ \end{cases} $ 
$\text{It is differentiable when $ \ 0\le\phi\le 2\pi \ , \ r=0 \ ?$ }$ 
$ \text{ Any help would be greatly appreciated. Thanks!}$

Note that $f(x+iy)=x\sqrt{x^2+y^2}-i y\sqrt{x^2+y^2}=u+iv.$
So $u_x=\frac{2x^2+y^2}{\sqrt{x^2+y^2}}$ and $v_y=-\frac{2y^2+x^2}{\sqrt{x^2+y^2}}.$ 
From C-R equations:
$u_x=v_y$ is possible only for $x=o, y=0$ and this implies f is nowhere analytic. 
