How can I show $f(z)=\sqrt{r}e^{i\frac{\theta}{2}}$ is not differentiable at 0?
Solution 1:
If $f(r\mathrm{e}^{i\vartheta})=\sqrt{r}\mathrm{e}^{i\vartheta/2}$, then $$ f^2(z)=z. $$ Hence, if $f$ is analytic in $\Omega\subset\mathbb C$, then $$ 2f(z)f'(z)=1. $$ If $0\in\Omega$, then $$ 2f(0)f'(0)=1, $$ which is IMPOSSIBLE since $f(0)=0$.
Thus, $f$ can not be analytic at $z=0$.