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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))

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    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](http://www.fkuehn.de/download/ct.png) ). Since $f$ is continuous it works fine.2012-11-10