Differential closed form Im trying to go alone through Fultons, Introduction to algebraic topology. He asks whether there is a function $g$ on a region such that $dg$ is the form:
$$\omega =\dfrac{-ydx+xdy}{x^2+y^2}$$
in some regions. I know you can do it on the upper half plane by considering $-arctan(x/y)$. But Im a bit confused. I know that you can measure the angle, except maybe on a fixed hlaf line, like taking away the negative axis, but on the other hand, the tan is bijective on intervals of lenght $\pi$. So, can you find such a function on the union of the right half plane andthe upper half plane? Thanks 
 A: Yes, you can let $g$ be the function that takes a polar representation with $r>0$ and $-\pi<\theta<\pi$ and returns just $\theta$. This is equal to shifted versions of your arctan function pieced together.
A: Yes you can. You can solve the equation $dg=\omega$ in any open subset $U\subset \mathbb R^2\setminus (0,0)$ which does not "enclose" the origin $(0,0)\in\mathbb R^2.$  Formaly: there exists a continous  $\gamma:[0, \infty)\to \mathbb R^2\setminus U$, such that $\gamma(0)=(0,0)$ and $\|\gamma(t)\|\to\infty$ as $t\to\infty$.  (In your example, you can take say $\gamma(t)=(-t,-t)$). 
Once $U$ satisfies this condition, you can define $g(x)$ by integrating $\omega$ along any path connecting some fixed point $x_0\in U$ to $x$ and avoiding $\gamma$. This definition is independent of the path since $\mathbb R^2\setminus\gamma([0,\infty))$ is simply connected, hence any two such paths can be deformed into each other, leaving the integral defining $g(x)$ unchanged, since $\omega$ is closed.  
A: By the way, this form is notable in that it is closed but not exact on $\mathbb{R}^2 \setminus \{0\}$. (You can tell it is not exact by integrating around the unit circle and getting a nonzero result.)
Be careful of the meaning he intends for the word "region"...I suppose the general meaning is "open connected set".
