Index in complex analysis is an integer : Intuition? In complex analysis course, we prove that given a closed path $\gamma$ and $a\notin \gamma^*$ the following number:
$$
\frac{1}{2i\pi}\int_{\gamma}\frac{dz}{z-a} 
$$
is an integer. The integral can be written as $\frac{1}{2i\pi}\int_{t_0}^{t_1}\frac{\lambda'(t)}{\lambda(t)-a}dt$, the proof is more real analysis than complex analysis. I am not asking on the proof, witch is pretty easy to learn. 

The fact that the index is an integer surprise me a lot, is there any intuition to get the idea ? Or perhaps, any combinatorics explanation ?

 A: When I learned complex analysis, this integral was called the winding number, and it counts how many times $\gamma(t)$ "goes around" the point $a$ on the way round the curve -- with sign, so going counterclockwise around $a$ counts as $+1$ and going clockwise counts as $-1$.
This is most easy to explain in the simple case that $a=0$ and $\gamma(t_0)=\gamma(t_1)=1$. In that case our integral is
$$ \int_\gamma \frac1z dz $$
Now, if $\frac1z$ had a global antiderivative (which would be a logarithm) this would be easy; we would then have
$$ \int_{\gamma[t_0...t]} \frac1z dz = \log \gamma(t)$$
and in particular $\int_\gamma \frac1z dz=\log\gamma(t_1)=0$.
But really the complex logarithm is multi-valued, so if $\gamma$ goes around the origin and comes back to $1$, we'll end up in a different branch of the logarithm. The different branches of the logarithm differ by multiples of $2i\pi$ (because that is the period of the exponential function which is the inverse of the logarithm), so if we divide the value of the integral by $2i\pi$, we get how many times we have needed to "move to the neighboring branch" of the logarithm while following the curve.
A: The integral in question is the "winding number", a measure of the "number of times $\gamma$ travels around $a$ in the counterclockwise direction".
The function $f(z) = 1/(z - a)$ has "primitive" (or "antiderivative") $F(z) = \log(z - a)$. This primitive is multiple-valued, with two values differing by $2\pi ni$ for some integer $n$. (This is a direct consequence of the fact that $\exp(z + 2\pi ni) = \exp(z)$ for all complex $z$.) Integrating $dz/(z - a)$ over a closed curve therefore gives an integral multiple of $2\pi i$, the integer measuring the number of times $\gamma$ crosses a cut for the logarithm.
As a simple confirmation, if $n$ is an integer, $r > 0$ is real, and $\gamma(t) = a + re^{2\pi nit}$, with $0 \leq t \leq 1$, is "the circle of radius $r$ around $a$, traced $n$ times", then
$$
\frac{1}{2\pi i}\int_{\gamma} \frac{dz}{z - a} = \frac{1}{2\pi i} \int_{0}^{1} \frac{2\pi ni\, re^{2\pi nit}\, dt}{re^{2\pi nit}} = n.
$$
