Winding number integral/index in plane Let $B(x_0,y_0)$ be an open unit disk.  Assume that $F(x,y)=(f_1 (x,y),f_2 (x,y)):\overline{B(x_0,y_0)}\rightarrow \mathbb{R}^2$ is a diffeomorphism. Assume also that $F(x,y) \ne (s_0,t_0) $, for all $(x,y)\in \partial B(x_0,y_0)$.
What is the elementary way to evaluate the following result?
$$\frac{1}{2\pi}\int_{\partial B(x_0,y_0)}\frac{(f_1-s_0) df_2-(f_2-t_0)df_1}{(f_1-s_0) ^2+(f_2-t_0) ^2}=sign \det F'(x_0,y_0)=±1. $$
Can that be shown by using Green's theorem and then change of variables formula? Since $F$ is a diffeomorphism, due to inverse function theorem $\det F'(x,y) \ne 0$, for all $(x,y) \in B(x_0,y_0)$, and by Jordan curve theorem $F(\partial B(x_0,y_0))$ is a Jordan curve. Therefore inner domain is simply connected and we can use Green's theorem. But I'm not sure how to calculate that integral.
 A: Assume that $B$ is a disk with center ${\bf z}_0$ in the $(x,y)$-plane,  that
$$f:\quad \bar B\to{\Bbb R}^2, \qquad (x,y)\to(u,v):=\bigl(f_1(x,y),f_2(x,y)\bigr)$$
is a diffeomorphism, and that ${\bf 0}\notin f(\partial B)$.   Then
$$N:={1\over2\pi}\int_{\partial B}{f_1\>df_2-f_2\>df_1\over f_1^2+f_2^2}=
\left\{\eqalign{{\rm sgn}\bigl(J_f({\bf z}_0)\bigr)&\qquad\bigl({\bf 0}\in f(B)\bigr)\cr
0\qquad&\qquad\bigl({\bf 0}\notin f(B)\bigr) \cr}\right.$$
Proof. Consider the vector field
$${\bf A}(u,v):=\left({-v\over u^2+v^2}, \>{u\over u^2+v^2}\right)=\nabla\arg(u,v)$$
 in the punctured $(u,v)$-plane. I claim that 
 $$N={1\over2\pi}\int_{f(\partial B)}{\bf A}\cdot d{\bf w}\ .\tag{1}$$
 Subproof.  Let $$t\mapsto\bigl(x(t),y(t)\bigr)\qquad(0\leq t\leq T)$$ be a parametrization of $\partial B$. Then
 $$t\mapsto {\bf w}(t):=\bigl(f_1\bigl(x(t),y(t)\bigr), f_2\bigl(x(t),y(t)\bigr)\bigr)\qquad (0\leq t\leq T)$$
 is a parametrization of $f(\partial B)$. It follows that
 $$\eqalign{&\int_{f(\partial B)}{\bf A}\cdot d{\bf w}
 =\int_0^T\left({-f_2\over f_1^2+f_2^2}\bigg|_{{\bf w}(t)}\dot u(t)+ {f_1\over f_1^2+f_2^2}\bigg|_{{\bf w}(t)}\dot v(t) \right)\ dt\cr
 &=\int_0^T \left({-f_2\over f_1^2+f_2^2}\bigl(f_{1.1}\dot x(t)+f_{1.2}\dot y(t)\bigr)+
  {f_1\over f_1^2+f_2^2}\bigl(f_{2.1}\dot x(t)+f_{2.2}\dot y(t)\bigr)\right)\>dt \cr
  &=\int_{\partial B}{f_1\>df_2-f_2\>df_1\over f_1^2+f_2^2}\quad.\qquad\square\cr}$$
 Note that
 $${\rm curl}\>{\bf A}(u,v)\equiv0\ .$$
When ${\bf 0}\notin f(B)$ we can apply Green's theorem in $(1)$ and obtain
$$N={1\over2\pi}\int_{f(\partial B)}{\bf A}\cdot d{\bf w}={1\over2\pi}\int_{f(B)}{\rm curl}\>{\bf A}(u,v)\>{\rm d}(u,v)=0\ .$$
When ${\bf 0}=f({\bf p})\in f(B)$ we  introduce $B':=B\setminus B_\epsilon({\bf p})$ and apply Green's theorem to $f(B')$. We then get
$$N={1\over2\pi}\int_{f(\partial B)}{\bf A}\cdot d{\bf w}={1\over2\pi}\int_{f(\partial B')}{\bf A}\cdot d{\bf w}+{1\over2\pi}\int_{f(\partial B_\epsilon)}{\bf A}\cdot d{\bf w}\ .$$
Since the first term on the right hand side vanishes we have
$$N={1\over2\pi}\int_{f(\partial B)}{\bf A}\cdot d{\bf w}={1\over2\pi}\int_{f(\partial B_\epsilon)}{\bf A}\cdot d{\bf w}\ .$$
Now $f(\partial B_\epsilon)$ is a tiny ellipse around ${\bf 0}$, and a suitable limiting argument should then prove that $N= {\rm sgn}\bigl(J_f({\bf z}_0)\bigr)$ in this case.
