Winding number of differential curve Consider the one-form $\omega$ on $\textbf{R}^2$\ {(0,0)} defined by
$\omega$ = $\frac{xdy-ydx}{x^2+y^2}$
Let K $\subset$$\textbf{R}^2$\ {(0,0)} denote the positive x-axis.
Let $\gamma$ : $[a,b]$ -> $\textbf{R}^2$\ {(0,0)} be a loop which is transverse to K.
Show that the intersection number $A*\gamma \in$ Z satisfies $\frac{1}{2\pi}$ $\int_{\gamma}^{}\omega$=$A*\gamma$

Can anyone help me starting this one?
Thank you.
 A: (1) If $\gamma(t)=(x(t),y(t))=r(t)(\cos\ s(t),\sin\ s(t))$ then on
$\gamma$,
$$ \omega = \frac{xy' -yx' }{x^2+ y^2} dt = s'(t)dt $$
Consider a curve $\gamma$ with $$ s(0)=0,\ 2\pi > s(t)>0\ (0<t<2\pi
),\ s(2\pi)=2\pi,
$$ That is, this curve meets positive $x$-axis at $t=2\pi$ first (except starting), and maybe
$r(0)\neq r(2\pi)$. Then
$$ \int_\gamma \omega = \int_0^{2\pi} s'(t) dt=2\pi $$
Here $s(2\pi)=2\pi$ implies that it winds the $(0,0)$ one times.
Assume that $\gamma$ with $$ s(0)=0,\ 2\pi > s(t)>0\ (0<t<2\pi ),\
s(2\pi)=0,
$$
That is it meets positive $x$-axis but not wind $(0,0)$ :
$$\int_\gamma \omega = \int_0^{2\pi} s'(t) dt= 0 $$
That is, in this case we do not count.
(2) Assume that a closed curve $\gamma$
meets $K$-times with positive $x$-axis and it starts at a point on
positive $x$-axis. If we reparametrize $\gamma $, so we have $$
\gamma : [0,2\pi K ] \rightarrow \mathbb{R}^2 - \{0\} $$ s.t. $
\gamma (2\pi i) $ is in positive $x$-axis and $\gamma(0)=\gamma(2\pi
K)$.
So we have a claim : $\int_{\gamma|_{[2\pi i,2\pi(i+1)]} }\omega =2\pi $ which is proved by (1).
