# Cauchy integral theorem in a Jordan domain?

With the Goursat's lemma we can prove the local version of Cauchy integral theorem in a disk: if $f$ is holomorphic in an open disk $D\subset \mathbb{C}$, then $f$ has a primitive in $D$. But can we use the same proof if we consider a Jordan domain $int(J)\subset \mathbb{C}$ of the Jordan curve $J \subset \mathbb{C}$?

We know that the set $\mathbb{C}\setminus int(J)$ is connected, and domain $int(J)$ is path-connected with rectilinear paths, so we can use the Goursat's lemma?

## 1 Answer

By the Jordan-Schoenfliess theorem, $\textrm{int}(J)$ is simply connected. As it is also bounded, it is conformally equivalent to the unit disc by the Riemann mapping theorem. As every holomorphic function has a primitive in the unit disc, the same is true for $\textrm{int}(J)$.

• I do know that simply connected domain is the same as Jordan domain. But my question is that can we infer the result with only Jordan curve theorem. It looks very plausible. – Hulkster Mar 17 '18 at 3:39
• @Hulkster All Jordan by itself tells us is that $U=\textrm{int}(J)$ is a bounded open set. For every holomorphic function there to have a primitive, it is necessary that $U$ be simply connected. The usual proof of Goursat uses convexity (or some weaker form like star-shapedness) that implies simply connectedness. – Lord Shark the Unknown Mar 17 '18 at 3:46
• Yes, but we use Goursat's lemma to prove Cauchy theorem in an open disk. Can we use the same method in the case of $int(J)$ ? – Hulkster Mar 17 '18 at 3:51
• @Hulkster That seems unlikely to me. The usual proof of Goursat on the disc relies very much on the geometry of the disc being very simple to describe... – Lord Shark the Unknown Mar 17 '18 at 3:54
• OK. Thanks for answering. – Hulkster Mar 17 '18 at 3:58