I’m reading some lecture notes in French where they give the following form of Cauchy’s integral theorem:
Let $f$ be a holomorphic function on an open set $U \subset
\mathbb{C}$. Then, for every oriented path $\Gamma$ such that
$\Gamma=\partial K$ where $K\subset U$ is a compact set such that
$K=cl(int(K))$ and $\partial K$ is a finite union of piecewise $C^2$,
disjoint and orientable arcs. Then we have$$\int_{\Gamma} f(z)dz=0.$$
(There’s actually a name in French for this kind of compact set, which I don’t know how to translate to English.)
I was trying to understand how this translates to the usual version of Cauchy’s integral theorem I already know, that is:
If $f$ is holomorphic in a simply connected open set $Ω$, then for any
simply closed contour $C$ in $Ω$, the contour integral is zero:$$\int _{C}f(z)dz=0$$
…as the version on my lecture notes seems overly complicated to me.
In particular, I was wondering if any simple closed contour would satisfy or not the property of being the boundary of such a compact set. If not, what would a non-example look like?
Best Answer
Yes. A simple closed contour is a Jordan curve. The well-known Jordan curve theorem says that every simple closed curve in the plane separates the plane into a bounded interior region and an unbounded exterior region. But more is true: The Jordan-Schoenflies theorem tells us that if $C \subset \mathbb C$ is a simple closed curve, then there is a homeomorphism $h : \mathbb C \to \mathbb C$ such that $h(C)$ is the unit circle $S^1$ in the plane. It is the boundary of the unit disk $D^2$ in the plane, thus $K = h^{-1}(D^2)$ is the desired compact subset having $C$ as its boundary.