# Extending Cauchy Goursat from triangles to simply connected domains

In this page Cauchy Integral Theorem, the three Cauchy integral theorems are presented. The first is about simply connected domains and the second is a generalisation of the 1st that includes removable singularities. The third is the Cauchy Goursat theorem regarding triangles or rectangles. At one point (just before the 3rd theorem) it writes:

"Goursat's argument makes use of rectangular contour (many authors use triangles though), but the extension to an arbitrary simply-connected domain is relatively straight-forward"

How is that extension straight-forward? What is the proof of that? In other words how can I prove the 1st theorem for the 3rd (using the triangle version of Cauchy Goursat prefferably) and then how do I prove that the 1st theorem implies the 2nd? ($3\Rightarrow 1\Rightarrow 2$). It would be preferrable if someone linked a pdf file containg the proofs I am asking for. Also note that I am looking for elementary self contained answers that don't invole a lot of topological notions but basic topology. Thank you for your answers.

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$(3) \Rightarrow (1)$: Proof Cauchy-Goursat-Theorem (Theorem 15.2)
$(1) \Rightarrow (2)$: It's a consequence of the residue theorem - probably you'll find a proof (using Cauchy's Theorem) in every book about analytical functions (or here: Proof 1/Proof 2 (short one))
Thanks for the $(3)\Rightarrow (1)$ proof, it is exactly what I asked. Now the $(1)\Rightarrow (2)$ theorem is necessary (to my knowledge) to prove Cauchy's Integral Fomula and thus the residue theorem... –  Nameless Nov 10 '12 at 17:03
Ah, okay, you want to do it the other way round. In this case, $(1) \Rightarrow (2)$ is a simple consequence of Riemann's theorem (since the function is bounded in a neighborhood of the singularities, the singularities are removable.) –  saz Nov 10 '12 at 17:26
Wait a minute. In the proof of $(3)\Rightarrow (1)$ the 3rd step (the one I am asking for) is pretty much omitted... –  Nameless Nov 10 '12 at 17:49
That's true, but I doubt that there is a more detailed proof (especially since there are a lot of other (in my oppinion easier) ways to proof this theorem). But anyway: You can cover im $\gamma$ by arbritrary small squares and obtain a polygonal contour $\Gamma$ arbritary close to $\gamma$ (I tried to draw a picture ). Since $f$ is continuous it works fine. –  saz Nov 10 '12 at 18:20