Let $G \subset \mathbb{R}^2$ and let $F = (F_1,F_2)$ be a smooth (which means it is $C^1)$ vector field in $G$. $F$ satisfies $$\int_\gamma F_1dx + F_2dy = \int_\gamma F_2dx - F_1dy = 0$$ for every closed and smooth path $\gamma$.
I need to prove that there exists a harmonic function $u: G \to \mathbb{R}$ such that $F = \nabla u$.
I wasn't sure what to do, so I just started trying different things. I think there is a theorem that if for every closed path $\gamma$ it satisfies $\int_\gamma \omega = 0$, then $\omega$ is an exact differential form.
If I write $\omega_1 = F_1dx + F_2dy$ and $\omega_2 = F_2dx - F_1dy$ then both are exact and there exist $u_1,u_2: G \to \mathbb{R}$ such that $\omega_1 = du_1$ and $\omega_2 = du_2$
From here I am not so sure what to do. I think I can also use Green's theorem here, but I am not sure how it helps.
Help would be appreciated.
By Stokes with the first integral, for every closed disk $D$, $$\int_D{d\omega_1}=0.$$
From this follows that $\omega_1$ is closed. Since it is a $1$-form, and provided $G$ is simply connected, $\omega_1$ is exact and there exists some $u$ such that $\omega_1=du$, ie $F=\nabla u$.
For the same reason, $\omega_2$ is closed, and you can easily check that $-d\omega_2=(\nabla \cdot F)dx \wedge dy$, thus $0=\nabla \cdot F=\Delta u$.