Confusion about a particular differential form being exact?

Consider the differential form:

$$\omega=-\frac y{x^2 + y^2}dx + \frac x{x^2 + y^2}dy$$

I believe that this form is not exact. i.e: There does not exist a $\psi$ such that $d\psi=\omega$.

In particular, this is supposed to be true since the line integral along a loop, $$\int_{C}\omega = 2\pi \neq 0$$

But, what about $$\psi=-\arctan(\frac x{y})$$

According to Maple,

As you can see, Maple says that $d\psi=\omega$. So what's the deal here? Is $\omega$ exact or not?

Thank You!

• The relevant concept is "exact on a given domain". Note that $\arctan(x/y)$ is defined in two half planes that don't contain the circle $C$. – Gribouillis Nov 20 '17 at 8:17
• arctan is not a "single-valued" function. With the usual definitions it will have discontinuities on the punctured plane. – Lord Shark the Unknown Nov 20 '17 at 8:18
• We see $$\omega=-\frac y{x^2 + y^2}dx + \frac x{x^2 + y^2}dy=\dfrac{xdy-ydx}{x^2 + y^2}dy=d\arctan\dfrac{y}{x}$$ this relation holds where $(x,y)\neq(0,0)$. – Nosrati Nov 20 '17 at 8:28

Your $\psi$ is just defined for $y\neq 0$. Therefore $d\psi=\omega$ holds on $A=\{(x,y)\in\mathbb R^2~:~ y>0\}$ and on $B=\{(x,y)\in\mathbb R^2~:~y<0\}$. You can conclude that for any closed curve $\gamma$ in $A$ or in $B$ you get $\int_\gamma \omega =0$. But your loop doesn't belong to $A$ or $B$. Moreover, you can conclude from $\int_C\omega\neq 0$ for a loop $C$ that there isn't any $\psi$ such that $d\psi=\omega$ on $\mathbb R^2\setminus\{(0,0)\}$.
The differential form $\omega$ is exact in some region where $\arctan(y/x)$, or more generally the phase $\operatorname{atan}(x,y)$, makes sense. For example, it is exact in the domain you get when you throw away the non-positive half of the $x$-axes.
However, it is not exact in the punctured plane $\Bbb{R}^2\setminus\{0,0\}$. That non-vanishing path integral around the closed loop proves that.