Null homolog cycle

Question

Let $\gamma$ be a cycle on open set $A$. Suppose that for all analytic functions $f:A\to \mathbb{C}$ we have that $\int\limits_\gamma f(z)dz=0$. Does it follow that $\gamma$ is null-homolog?

Answer 1

Yes. If $\gamma$ is not homologous to zero, then there exists $z_0 \in \mathbb{C} \setminus A$ such that the winding number of $\gamma$ about $z_0$ is nonzero. Now, take the function $f(z) = 1/(z - z_0)$. Then $f$ is analytic on $A$ and the residue theorem tells us that $\int_{\gamma} f(z) \, dz \neq 0$.