A statement of Cauchy's theorem is that:

Let $\gamma$ be a closed, null homotopic contour on an open,bounded and simply connected subset $\Omega$ of $\mathbb{C}$. If $f:\Omega\mapsto\mathbb{C}$ is holomorphic, then $\int_{\gamma}f(z)dz=0$

I get that $\gamma$ would have residues if $\Omega$ wasn't simply connected and the integral wouldn't be equal to zero, but what is the importance of the null homotopic, closed $\gamma$ assumption and the open,bounded $\Omega$ assumption?



1 Answer 1


The fact that $\Omega$ is simply connected is useless here.

On the other hand, it should be clear that $\gamma$ must be closed. Otherwise, suppose that $\Omega=\mathbb C$ and the $\gamma\colon[0,1]\longrightarrow\mathbb C$ is simply $\gamma(t)=t$. Then, if $F$ is a primitive of $f$, $\int_\gamma f(z)\,\mathrm dz=F(1)-F(0)$, which is not $0$ in general.

Usually we assume that $\Omega$ is is open because that's part of the standard definition of holomorphic function.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .