If you have a holomorphic function $f$ from the unit disc $D$ to a unit square $U$, such that:
1) $f$ has a continuous extension to the boundary of $D$,
2) $f$ restricted to the boundary of $D$ is a bijection between boundary of $D$ and boundary of $U$.
Then WTS $f$ is a biholomorphism between $D$ and $U$.
This was a question my professor mentioned as an example of something that might be on the final tomorrow. My hypothesis may be not quite correct, specifically I'm not sure if we are assuming 2 or if we can conclude it. I think we have to assume it?
Anyway, so I want to do this problem by applying the argument principle to the function $g(z) = f(z) - w$ for any $w \in U$. Showing that \int_C \frac{g'(z)}{g(z)}dz=2\pi i will imply that $g(z)$ has exactly one zero (since $f(z)$ is holomorphic on $D$ so is $g(z)$, so it has no poles and we conclude 1 zero from the argument principle) so therefore $f$ is a bijection and we are done.
So I'm not quite sure how to handle this integral. Using contour integration, we get \int_C \frac{f'(z)}{f(z)-w} dz = \int_0^{2\pi} \frac{f'(e^{i\theta})}{f(e^{i\theta})-w}i\cdot e^{i\theta}d\theta. And from here I want to use a logarithm, but I am getting hung up on branch cuts and where to go from here.
Could anyone walk me through how to complete this integral?
Thank you!