Differential forms, number of zeros on disk

Question

Let $f$ be a holomorphic function on $U \subset \mathbb{C}$ and $f'(z) \neq 0$, $z \in U$. Then how do I show that all zeros of $f$ are simple and positive, and that if I have a disk $D \subset U$ where there's no zeros in $\partial D$, then $$\#\text{ of zeros of }f\text{ in }D = {1\over{2\pi i}} \int_{\partial D} {{df}\over{f}}.$$ I know how to show this more "powerful" techniques from ecomplex analysis, but is there a way to prove this with only the language of differential forms and real functions?

Answer 1

The fact that the zeros of $f$ are simple and positive comes from $$\det(df) = u_x^2 + u_y^2 = |f'(z)|^2 > 0.$$A straightforward computation shows that$${{df}\over{f}} = {{du + i\,dv}\over{u + iv}} = {{u\,dv + v\,du}\over{u^2 + v^2}} + i{{u\,dv - v\,du}\over{u^2 + v^2}}$$$$={1\over2}d(\log(u^2 + v^2)) + i{{u\,dv - v\,du}\over{u^2 + v^2}}.$$Hence, by Kronecker's formula,$${1\over{2\pi i}} \int_{\partial D} {{df}\over{f}} = {1\over{2\pi}} \int {{u\,dv - v\,du}\over{u^2v^2}} = \# \text{ of zeroes of }f\text{ in }D.$$