How can $z^\alpha$ for $\alpha>0$ be a conformal mapping?

Question

We know that $z^a = r^\alpha e^{i\alpha\theta}$ so that this function maps the wedge $\{z: \theta_1<Arg(z) <\theta_2\}$ to the wedge $\{z:\alpha\theta_1<Arg(z^\alpha)<\alpha\theta_2\}$. For the mapping to be conformal it seems to me we need $\theta_2-\theta_1=\alpha\theta_2-\alpha\theta_1$ which would imply $\alpha=1$. However, I've read that the only requirement necessary for the map to be conformal is $\alpha\theta_2-\alpha\theta_1\leq 2\pi$.

Answer 1

You are thinking about angle preservation at $0$; this is where the lines $\arg z = \theta$ meet. It is correct that the function $z^\alpha$ is not conformal at $0$, unless $\alpha=0$. But the claim is that it is conformal in the domain $\{z: \theta_1<\operatorname{Arg}(z) <\theta_2\}$, and this domain does not include $0$.

The argument of $0$ is undefined, so any formula that specifies what the argument of $z$ should be automatically excludes $z=0$.

Answer 2

A complex map is conformal on an open set $U$ if and only if it is holomorphic and its derivative is nowhere vanishing on $U$.

These are easy enough to check for your map; the thing about your map is defining it. We have $z^\alpha=\exp(\alpha\,\text{Ln}(z))$, so you must choose a suitable branch of the complex logarithm. The idea is that $\theta_1$ and $\theta_2$ must be such that the domain does not cross the branch cut.