What happens to the angles at the origin under the mapping $f(z)=z^{\alpha}$ when a) $\alpha>1$ b) for $0<\alpha<1$ ($z\in\mathbb{C}$)?
I know that the angles increases resp. decreases but I am unable to show why.
I've tried rewriting the mapping using the parametrization $z(t)$ where $t_0=z_0$. And since the angle is given by the derivative the angle of the mapping is: $$arg\Big(f'(z(t_0))\Big)=arg\Big( f'(z(t))z'(t)\Big)=arg(f'(z(t)))+arg(z'(t)).$$
At $t=0$ the second term is zero but what about the first term? Is it correct to say:
$$arg(\alpha z^{\alpha -1}(0))=arg(\alpha)+0?$$
If this is true then I don't know why the origin would be anymore interseting than any other point since the same relation (i.e. $arg(\alpha)+arg(z(t)^{\alpha - 1})$ is true whenever.