$z^\alpha$ is defined and holomorphic on $\mathbb{C}^{\times}$ only for $\alpha$ an integer.

Question

I think I found a solution for this for all $\alpha$: for $\alpha$ rational and not an integer one gets an impossible endomorphism of $\pi_1(\mathbb{C}^{\times})=\mathbb{Z}$, and for $\alpha$ irrational one can show that $z^\alpha$ basically induces the same impossible map as a $z^\frac{p}{q}$ when $\frac{p}{q}$ is a sufficiently good rational approximation of $\alpha$. However I'm afraid I'm missing something because I find the the irrational case to be a bit too complicated for the exercise sheet I was given. Is there a better, simpler solution ? Thanks

Answer 1

If $f: \Bbb C^\times \to \Bbb C$ is holomorphic and locally $$ f(z) = z^\alpha = e^{\alpha \log z} $$ for some holomorphic branch of the logarithm then $$ \frac{f'(z)}{f(z)} = \frac{\alpha}{z} $$ for all $z \in \Bbb C^\times$. In particular, $$ \frac{1}{2\pi i} \int_\gamma \frac{f'(z)}{f(z)} \, dz = \frac{1}{2\pi i} \int_\gamma \frac{\alpha}{z} \, dz = \alpha $$ where $\gamma(t) = e^{i t}$ for $0 \le t \le 2 \pi$ is a closed curve surrounding the origin once.

The left-hand side is the winding number of $f \circ \gamma$ with respect to zero, and that is always an integer (see for example Winding number (demonstration)).