I am reading Brown and Churchill's Introductory complex analysis book, which states the Cauchy-Goursat theorem as follows:
If a function $f$ is analytic at all points interior to and on a simple closed contour $C$, then $$ \int_{C} f(z) dz =0 $$ I understand this result and its proof just fine. However, there is a second theorem which claims the following:
If a function $f$ is analytic throughout a simply connected domain $D$, then $$ \int_{C} f(z) dz =0 $$ for every closed contour $C$ lying in $D$. So in this second theorem we don't $C$ to be simple due to the fact that $D$ is simply connected. However, if $f$ is analytic at all points interior to and on a closed contour $C$, isn't the interior of the closed contour $C$ a simply-connected domain by default?
To be more precise, if I rewrite the first theorem as: If a function $f$ is analytic at all points interior to and on a closed contour $C$, then $$ \int_{C} f(z) dz =0 $$ is this not true? I don't see why we require $C$ to be simple, because even if $C$ intersects itself, we can just treat our non-simple closed contour as a union of simple closed contours, and those integrals are all $0$.
Best Answer
I think the main issue here is that the `interior' of a closed curve is not necessarily as easy to define as you might think, especially when you start to consider some very pathological curves in the plane. That's why most authors focus on simple closed curves, for which the Jordan curve theorem gives a nice characterization of the interior of a curve (however this is not trivial to prove, as much as it sounds like an obvious statement). If a curve is entirely enclosed within a simply connected region, then one does not have to worry about interior or exterior at all, which is what the second theorem states. You can think of the first theorem as just a special case of the second, but usually the Cauchy-Goursat theorem is mentioned first as simple closed curves are more familiar in a sense than simply connected regions.