# Relationship between Simply Connectd Domains, Cauchy's Theorem, and Jordan curves

For a while, before seeing the formal statement of Cauchy's theorem, I used to think that the following is actually Cauchy's Theorem:

Let $$C$$ be a rectifiable (or piecewise smooth) Jordan curve $$C$$ and let $$\Omega$$ be a domain containing $$C$$ and its interior. If $$f$$ is holomorphic on $$\Omega,$$ then $$\int_C f=0.$$

Now, of course I know that the requirenment is actually for $$\Omega$$ do be simply connected. But intuitively, it does seem like the interior of a Jordan curve is probably connected, and that we may extend the interior a bit to cover the curve itself, and still have a simply connected domain.

So I am wondering, is my old version of Cauchy's Theorem actually true?

You say you know that the requirement for $$\int_C f = 0$$ is that $$\Omega$$ is simply connected. In fact, if $$\Omega$$ is simply connected, then for all holomorphic $$f : \Omega \to \mathbb C$$ and all closed curves $$C$$ in $$\Omega$$ we have $$\int_C f = 0$$. However, this is just a special case of the following:

If $$C$$ is a closed curve in $$\Omega$$ which is null-homotopic in $$\Omega$$, then $$\int_C f = 0$$ for all holomorphic $$f : \Omega \to \mathbb C$$.

Now any Jordan curve in $$\Omega$$ such that $$\Omega$$ contains the interior of $$C$$ is null-homotopic in $$\Omega$$.

In fact, this is a consequence of the Schoenflies Theorem (see e.g. https://en.wikipedia.org/wiki/Schoenflies_problem). This says that there exists a homeomorphism $$h : \mathbb C \to \mathbb C$$ such that $$h(S^1) = C$$ (here $$S^1$$ is the standard circle). Let $$B$$ denote the open unit disk in $$\mathbb C$$, $$B' = \mathbb C \setminus (B \cup S^1)$$, $$U$$ the interior of $$C$$ and $$U' = \mathbb C \setminus (U \cup C)$$. Then $$B, B'$$ are the components of $$\mathbb C \setminus S^1$$ and $$U, U'$$ are the components of $$\mathbb C \setminus C$$. Hence $$h$$ maps $$B$$ either onto $$U$$ or onto $$U'$$. But $$D = h(B \cup S^1) = h(B) \cup C$$ is compact which is possible only when $$h(B) = U$$. So $$D = U \cup C$$ is contained in $$\Omega$$ and, being homeomorphic to the closed unit disk $$D^2 = B \cup S^1$$, it is contractible. Hence $$C$$ is null-homotopic in $$D \subset \Omega$$.

• Thank you very much for your answer! I will have to take some time to digest it, but in the meantime, may I ask you a slightly tangential question? I've noticed that when studying the integrals of functions $f \mathbb{\Omega} \subseteq \mathbb{C} \to \mathbb{C},$ we have the option of taking $2$ different points of view: $1)$ Fix the class of functions, such as to "holomorphic functions," and vary the domains to see on which class of domains the integrals of holomorphic functions along simple closed curves vanish...
– Ovi
Jul 1, 2019 at 11:55
• ...$2)$ Fix a class of domains, and try to find the class of functions whose integral along simple closed curves vanishes in that domain. From my short, $1$-semester study of complex analysis, it seems like if anything, we took the first point of view. However, I have noticed the second point of view, in your answer and in other places. My question is: if the first point of view is encountered when studying "pure" complex analysis, then in which context is the second point of view more common? Is it usually taken, for example, by topologists? ...
– Ovi
Jul 1, 2019 at 11:55
• ...Or am I wrong that in complex analysis we tend to take the first point of view? Thank you very much.
– Ovi
Jul 1, 2019 at 11:55
• @Ovi A comment will not be adequate to answer your "tangential question". I suggest you ask a new question. Jul 1, 2019 at 13:56
• Thanks, I will do that. I fear that it might be considered too broad, but I will give it a try later.
– Ovi
Jul 1, 2019 at 15:48

Here's an interesting, different approach from that of Paul Frost.

There is a slightly stronger version of Cauchy's Theorem (the homological one) which states the following:

Theorem: Let $$\Omega\subset\mathbb{C}$$ be an open set and $$f$$ a holomorphic function in $$\Omega$$. If $$C$$ is a rectifiable closed curve that satisfies $$\operatorname{Ind}(C;a)=0 \text{ for all }a\in\mathbb{C}\setminus\Omega$$ then $$\int_C f = 0$$

In your case, $$C$$ is a Jordan curve, so it splits the complex plane in two connected components: the interior (which is contained in $$\Omega$$) and the exterior (which therefore contains $$\mathbb{C}\setminus\Omega$$). It is easy to see that the winding number ($$\operatorname{Ind}(C;\cdot):\mathbb{C}\setminus C\to\mathbb{Z}$$) is continuous, hence locally constant, in particular it is constant in each connected component of $$\mathbb{C}\setminus C$$. It is also easy to see that if $$D$$ is a disk that contains $$C$$, $$\operatorname{Ind}(C;a)=0$$ $$\forall\ a\in\mathbb{C}\setminus D$$. Putting it all together, we get that $$\operatorname{Ind}(C;a)=0$$ for all $$a$$ in the exterior of $$C$$ (which contains $$\mathbb{C}\setminus\Omega$$). Using the above version of Cauchy's Theorem, we easily get that $$\int_C f = 0$$.