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 $3$?

  • $\begingroup$ nitpicking: is it Gourst or Goursat? $\endgroup$ – abel Jan 15 '15 at 3:32
  • 2
    $\begingroup$ @abel, it is Goursat. $\endgroup$ – Hawk Jan 15 '15 at 3:34
  • $\begingroup$ 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. $\endgroup$ – abel Jan 15 '15 at 3:45
  • 1
    $\begingroup$ 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. $\endgroup$ – copper.hat Jan 15 '15 at 4:02
  • 1
    $\begingroup$ @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...). $\endgroup$ – orangeskid Jan 15 '15 at 6:05

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.