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The residue theorem

Let $\Omega\subseteq \mathbb{C}$ open, $f$ meromorphic on $\Omega$ and $A$ be the set of the poles of $f$. If $\Gamma$ is a cycle in $\Omega\backslash A$ with $\mathrm{ind}_{\Gamma}(\alpha)=0$ for $\alpha\in \mathbb{C}\backslash \Omega$, we have: $$ \frac{1}{2\pi i}\int_{\Gamma}f(z)\mathrm{d}z=\sum_{a\in A}{\rm Res}(f,\ a)\cdot \mathrm{ind}_{\Gamma}(a) $$

Usually, there is a sketch of the proof on wikipedia. But, there isn't one for this theorem. Can someone point me or explain me an outline of the proof to this one? Just the most important steps.

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1 Answer 1

up vote 6 down vote accepted

Basically, you slice your contour in bits, each containing one pole which is possible, because poles are isolated. Then integral over $\alpha$ is split into the sum of integrals over the parts. Note that integrals along the cuts cancel out . By Cauchy theorem or otherwise show that it is equivalent to integrating around small circles surrounding the poles. Now expand $f$ in Laurent series and note that only the terms of the power -1 survive, since

$$\int_C\frac{dz}{(z-a)^n}=0,\quad n\ne 1$$ $$\int_C\frac{dz}{z-a}=2\pi i\cdot \operatorname{ind}_C(a)$$

That's a very rough indication of the proof.

enter image description here

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+1) You might wish to reflect the image though, making the arrows go in the other direction. –  AD. Jun 7 '12 at 18:55
    
yes, sure i can never figure out my lefts and rights –  Valentin Jun 7 '12 at 19:10
    
Thank you very much! –  Chris Jun 7 '12 at 19:39

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