Basic question about winding number in Complex Analysis

Question

I have a question which is very basic. In the proof that the complex integral $$W(\gamma,a)=\frac{1}{2\pi i} \int_{\gamma} \frac{dw}{w-a}$$ is an integer, some proof start out-of-the-blue by saying "to prove this, it is sufficient to show that $$exp\left(\int_{\gamma} \frac{dw}{w-a}\right)=1$$ and I have no idea of why/or this statement relates to the theorem to prove... Am I missing something obvious? If the exponential is 1, then the integral is zero, no? Why should it then be an integer?

Answer 1

$e^{z}=1$ iff $z=2\pi i n$ for some integer $n$.

If $e^{z}=1$ and $z=x+iy$ ($x$ and $y$ real) then $e^{x}(\cos y+i\sin y)=1$. Taking modulus we get $e^{x}=1$ so $x=0$. We then get $\cos y=1$ and $\sin y=0$. This gives $y =2\pi n$ for some integer $n$ so $z=2\pi i n$. Converse should be clear.

Answer 2

It's a consequence of Euler's formula !

$$e^{i\pi}=-1$$ hence $$e^{2i\pi}=1$$ and therefore $$e^{2i\pi.n}=1$$ for any $n$ integer

For this integral this means

$$exp\left(\int_{\gamma} \frac{dw}{w-a}\right)=2\pi i.n$$

that is

$$\frac{1}{2i\pi}exp\left(\int_{\gamma} \frac{dw}{w-a}\right)=n$$