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!}$
Best Answer
Hint If $f(z)$ would be analytic on any disk, then so would be $$g(z)=zf(z)=|z|^3$$
Now, it is trivial to show that this function is nowhere analytic. This follows immediately from CR, or via many other simple methods.