Why is the 1-form given by the derivative of argument on the punctured plane not exact?

Question

In this wikipedia article the 1-form $$\omega = \frac{-ydx+xdy}{x^2 + y^2}$$ is introduced as the derivative of argument on the punctured plane $\mathbb{R}^2\setminus\{0\}$. It is brought up as an example for a form which is closed but not exact. I can verify that it is closed: \begin{align} \omega =\ &\frac{-y}{x^2 + y^2}dx + \frac{x}{x^2 + y^2}dy\\ \Rightarrow d\omega =\ &\left( \frac{\partial}{\partial x} \frac{x}{x^2 + y^2} - \frac{\partial}{\partial y} \frac{-y}{x^2 + y^2} \right) dx \wedge dy\\ =\ &\left( \frac{\partial}{\partial x} \frac{x}{x^2 + y^2} + \frac{\partial}{\partial y} \frac{y}{x^2 + y^2} \right) dx \wedge dy\\ =\ &\left( \frac{y^2-x^2}{\left(x^2 + y^2\right)^2} + \frac{x^2 - y^2}{\left(x^2 + y^2\right)^2} \right) dx \wedge dy\\ =\ &0 \end{align} However, I am not convinced yet that $\omega$ is not exact. I understand that the argument on the punctured plane does not constitute an actual function, but maybe some other function exists which has $\omega$ as its derivative. I would just like to see a prove that $\omega$ is not exact just by looking at the expression above.

Answer 1

On a simply connected domain, the path integral over $\omega$ is path independent, since on such domains, it is exact (since $d\omega=0$). This means that the argument is the (up to a constant) unique integral of $\omega$. By looking at the half slit domain ($\mathbb R^2$ minus a ray emanating from the origin) it follows from this, that an integral of $\omega$ cannot be continued continuously to the plane minus the origin.