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!}$

Put $z=r.e^{i\theta}$ for polar coordinates then $f(r.e^{i\theta})=r^2(cos\theta-isin\theta)$.
You can see that the real part $u(r,\theta)=r^2.cos\theta$ and imaginary part $v(r,\theta)=-r^2.sin\theta$ satisfy CR-equations 
$ru_r=v_\theta, u_\theta=-rv_r$ iff
$3r^2.cos\theta=0$ and $3r^2.sin\theta=0$ i.e. $r=0$
Since CR-equations merely hold at the point $r=0$ (origin) and not in any open neighborhood of it hence $f$ is differentiable at $z=0$ but not analytic.
