If $\gamma$ is a closed path, then $\int_{\gamma}\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\in2\pi\mathbb{Z}$ As the title says:
Let $\gamma$ be a smooth closed path that avoids $(0,0)$. I need to show that the line integral: $$\int_{\gamma}\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$ is a multiple of $2\pi$.
Note that $\gamma$ is allowed to self-intersect itself.
Intuitively this value should represent the "number of rotations around $(0,0)$", as they are counted with orientation, but this intuition is all I have.
Any ideas?
 A: Take $\gamma : [0,1] \to \mathbb C$ a complex valued function, so that the quotient $\gamma'/\gamma$ is exactly what you obtain when replacing $\gamma$ into
$$\frac{dz}{z} = \frac{x-iy}{x^2+y^2}(dx+i dy)$$
and multiplying out. Thus, consider the function $h(t) = \int_0^t \frac{\gamma'(s)}{\gamma(s)}ds$. Set $H(t) =\gamma(t) \exp(-h(t))$, and observe that $H'(t) = 0$. This means $H(t)$ is constant, and then that $H(1) = H(0)$. All in all, we have $\gamma(1) \exp (-h(1)) = \gamma(0) \exp (-h(0))$
which gives that $\exp h(1) = 1$ or, what is the same, that $h(1) = 2\pi i n$ with $n\in \mathbb Z$.
Remark: this elementary argument is to be found in Apostol's Mathematical Analysis. 
A: Let $M = (0,\infty)\times\mathbb{R}$ and $\Phi : M \to \mathbb{R}^2 \setminus\{\mathbf{0}\}$ be the covering map $\Phi(r, \theta) = (r\cos\theta, r\sin\theta)$. Then we easily check that
$$ \omega = \frac{x dy - y dx}{x^2 + y^2} \qquad \Rightarrow \qquad \Phi^* \omega = d\theta $$
Now assume that $\gamma : [0, 1] \to \mathbb{R}^2 \setminus \{\mathbf{0}\}$ is a smooth closed curve and let $\tilde{\gamma} : [0, 1] \to M$ be the lifting of $\gamma$, i.e., $\gamma = \Phi\circ\tilde{\gamma}$. Then $\tilde{\gamma}(1)$ and $\tilde{\gamma}(0)$ is mapped to the same point on $\mathbb{R}^2 \setminus\{\mathbf{0}\}$ and hence they differ by $(0, 2\pi n)$ for some $n \in \mathbb{Z}$. So
$$ \int_{\gamma} \omega = \int_{\tilde{\gamma}} \Phi^*\omega = \int_{\tilde{\gamma}} d\theta = 2\pi n. $$
A: We want to apply Green's Theorem but discontinuity at the origin poses a problem, we will avoid this.
I have drawn an arbitrary curve $c_1 \bigcup c_2$. Then I have added a circle that avoids the origin $c_4$. I have added a line connecting the arbitrary curve and the circle. We pass through this line twice, going in different directions so the effect of this line is nothing.
Together I call the curves $\sum$.
Integrating over $\sum$ is the same as integrating over $c_1 \bigcup c_2$ and $c_4$. 

Integrating over $\sum$ gives zero by Greens Theorem. So it must be that,
$$0=\oint_{c_4}+\oint_{c_1 \bigcup c_2}=\oint_{\sum}$$
So,
$$\oint_{c_1 \bigcup c_2}=-\oint_{c_4}$$
Compute minus the integral over $c_4$, a circle of radius $r$, and we are done. Your result follows: If the curve $c$ intersect itself, divide it into closed non-self-intersecting curves $k_1,k_2,...k_n$. Then $\oint_{k_i}=\pm 2\pi$ with sign depending on orientation. The sum is thus in $2\pi\mathbb{Z}$.
A: Write $x=rsin(u), y=rsin(u)$, the form  $\alpha={{-y}\over{x^2+y^2}}dx+{{x}\over{x^2+y^2}}dy=du$ then take a circle $C$ contained in $\gamma$ and apply Stokes formula, since $d\alpha=0$, you obtain $\int_C\alpha=\int_{\gamma}\alpha=2\pi$.
