# Showing this integral from complex analysis is an integer without residues

I've already proven that the winding number is an integer, now I want to show that, given the following assumptions:

• The function $f$ is holomorphic on the domain $D$
• $\gamma$ is a piecewise-smooth, closed curve in $D$
• $f$ does not vanish on $\gamma$

It's true that

$$\frac{1}{2\pi i}\int_{\gamma}\frac{f '(\xi)}{f(\xi)}d\xi \in \mathbb{Z}.$$

Here's what I'm thinking, although it's not at all a rigorous proof:

$$\frac{1}{2\pi i}\int_{\gamma}\frac{f '(\xi)}{f(\xi)}d\xi=\frac{1}{2\pi i}\int_{\gamma}\frac{d}{d\xi}\log_{R}(f(\xi))d\xi=\frac{1}{2\pi i}\{\log_{R}(f(\xi_1))-\log_{R}(f(\xi_2))\}$$

where $\log_{R}$ is the logarithm function defined on the riemann surface from the wikipedia page on complex logarithms wiki page

and $\xi_1,\xi_2$ are two points which are the same when considered as points of $\mathbb{C}$ (because $\gamma$ is closed) but which have different arguments, differing by $k2\pi i$ for $k \in \mathbb{Z}$, when considered as points on the Riemann surface $R$.

I think you probably get that idea that I've got. IT could be totally off, but this is what I was led to in thinking about it. Trouble is, we haven't really discussed this Riemann surface, so I doubt it's what is expected.

If $z(t)$ parameterizes $\gamma$, then $f(z(t))$ parametrizes a curve which we can call $f(\gamma)$. Then we have $${1 \over 2\pi i} \int_{\gamma} {f'(z) \over f(z)}\,dz = {1 \over 2\pi i} \int_{f(\gamma)}{dw \over w}$$ The right-hand side of this is exactly $n(f(\gamma), 0)$, which is an integer since winding numbers are all integers.