How do you prove that a closed integral around a vector field must be a multiple of $2\pi$?

Question

A closed line integral is always going to be equal to $2\pi n$ where n is some integer:

$$\oint d\mathbf{R}\cdot\mathbf{A}(\mathbf R)=2\pi n$$

Why is this the case?

Answer 1

If ${\bf F}=(P,Q)$ is any old vector field in the plane ${\mathbb R}^2$, and $\gamma$ is a closed curve therein then the value of the line integral $$\int_\gamma {\bf F}\cdot d {\bf z}\tag{1}$$ can be any real number whatsoever. Physicists can interpret this value as work done (or received) when a cart is moved against the force ${\bf F}$ along $\gamma$.

There are some fields ${\bf F}$, called conservative, for which the integral $(1)$ is $=0$ for all closed curves $\gamma$. Such fields are gradients of some potential function $f:\>{\mathbb R}^2\to{\mathbb R}$, i.e., one has ${\bf F}=\nabla f$. A necessary cndition for this to be the case is that ${\rm curl}({\bf F})=Q_x-P_y\equiv0$.

Now there is this special field $${\bf A}(x,y)=\nabla{\rm arg}(x,y)=\left({-y\over x^2+y^2},{x\over x^2+y^2}\right)\ ,$$ which is only defined in the punctured plane ${\mathbb R}^2\setminus\{{\bf 0}\}$. Even though ${\rm curl}({\bf A})\equiv0$ this field is not conservative. The line integral $$\int_\gamma {\bf A}\cdot d {\bf z}\tag{2}$$ for closed curves $\gamma$ that avoid the origin has a geometrical meaning: It is the sum of all increments $\Delta\phi$ of the polar angle $\phi$ along $\gamma$. If the given $\gamma$ goes $n\in{\mathbb Z}$ times counterclockwise around the origin then the value of $(2)$ is $2\pi n$.

Answer 2

Try to write same integral as a contour integral of some complex function. If the corresponding complex valued function has one(may be more) singularity on region inside the curve with residue $-ni$ then the integral is $2\pi i (-in)=2\pi n$. I assumed that the corresponding complex function is exist but I don't know if it is the case always.