# Cauchy Goursat, why triangles?

Cauchy Goursat: Let $f$ be analytic in a simply connected domain $D$.If $C$ is a simple closed contour that lies in $D$ , then $$\int_C f(z) dz = 0.$$

I've been reading a lot of proofs on this theorem and all of them treats the contour $C$ as a triangle at first, but doesn't explain why it is sufficient to only show for triangles. Is it because every other simply connected curve is homotopic to a triangle?

Also, why are we allowed to assume, without any loss of generality, that one of the interior triangles is bigger than the other $3$?

• nitpicking: is it Gourst or Goursat? – abel Jan 15 '15 at 3:32
• @abel, it is Goursat. – Hawk Jan 15 '15 at 3:34
• from what i have seen it seems that you can break any region can be partitioned into triangles. and the sum of the integrals is the integral over the boundary. – abel Jan 15 '15 at 3:45
• A rectifiable curve in some open set $U$ can be uniformly approximated by a polygonal curve. Look at the proof of the homotopy version of Cauchy's theorem. – copper.hat Jan 15 '15 at 4:02
• @Hawk: The proof it for triangles first since it is easy to divide a triangle in 4 triangles of half size. (it also works for rectangles). After that, you can divide a polygon into triangles ( it wouldn't work in general to divide it into rectangles...). – Orest Bucicovschi Jan 15 '15 at 6:05

## 1 Answer

First you prove cauchy-goursat for closed triangles and $f$ holomorph in an open set $U \subseteq \mathbb{C}$

Then you prove that if in adition $U$ is convex, then $f$ has a primitive in $U$.

Then you prove cauchy theorem for $U$ in convex set: you just use ths Leibniz-Newton formula in any closed path and it gives zero.

This way you get the result without having to triangulate the closed curve.