Let $u$ be a harmonic function on a connected open set. If $\int_{\gamma}*du = 0$ for any cycle $\gamma$ then $u$ has a harmonic function.
This question arises from an answer to this post
Please do modify this question's assumptions if I am not stating some things correctly.
We know that the converse is true, namely that for any harmonic $u$ on an open connected set that has a harnonic conjugate, $\int_{\gamma}*du = 0$. I am thinking maybe Morera's theorem has to do something with this..
I suggest looking at this from a real calculus of two variables perspective. You are given that $$\int_\gamma -\frac{\partial u}{\partial y} \mathrm{d} x+\frac{\partial u}{\partial x} \mathrm{d} y \equiv 0, $$ for all cycles $\gamma$ in the domain. This allows you to safely define a function $$v(x,y) := \int_{(x_0,y_0)}^{(x,y)} -\frac{\partial u}{\partial y} \mathrm{d} x+\frac{\partial u}{\partial x} \mathrm{d} y . $$ with $(x_0,y_0)$ a fixed point in the domain. The expression for $v(x,y)$ is well defined, since the difference between any two integration paths is a cycle.
