Let $\Omega \subseteq \mathbb{C}$ be open, non-empty and path-connected, i.e. that every two points can be connected by a path that is continuously differentiable in pieces. Furthermore let ¨ f:$\Omega \rightarrow \mathbb{C}$ continuous. Show that the following statements are then equivalent:
(1)f has an antiderivative
(2)For every closed path $\gamma$ in $\Omega$ we have $\int_{\gamma}f(z)dz=0$
I had only an idea for (1)->(2)
:
f has an antiderivative means there is a F: $\Omega \rightarrow \mathbb{C}$, F'(z)=f(z) and $\int_{\gamma}f(z)dz=F(\gamma(b))-F(\gamma(a))$. because $\gamma\ is\ closed, a=b \
and\ it follows \int_{\gamma}f(z)dz=F(\gamma(a))-F(\gamma(a))$=0