I am confused how the existence of a complex anti derivative on an open set implies the path integral property. The proof of this is shown here and it is very simple.
Is this not a trivial counter example?
$$U = \{ z \in \Bbb C : |z| > 0.5 \space \land |z| <1.5 \} $$ $$g(z):=1/z$$
I believe $g(z)$ has an anti-derivative on all of $U$ but $\int_\gamma g(z)dz = 0$ on $\gamma$ of a unit circle.
This would make more sense if the requirement was $U$ is simply connected, but that requirement does not seem to be used in the proof. What am I missing?
We can define $G(z) = \operatorname{Arg}(z)$ on all of $U$ but it will be discontinuous, and therefore its derivative is not defined everywhere. Then $G$ is not an antiderivative of $g$ on all of $U$.