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Compute $\int_{|z|=2} \frac{dz}{z^2-1}$ for the positive sense of the circle.

Please don't use knowledge after Cauchy Theorem and Cauchy Integration Formula.Thank you!

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    substitute $z=2e^{i\phi}$ and integrate from $\phi=0$ to $2\pi$. Fairly straight forward2012-04-09

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Substituting $z=2e^{i\phi}$ results in (using $dz=2i e^{i\phi}\, d\phi$) $$ \int_{|z|=2} \frac{dz}{z^2-1} = \int_0^{2\pi}\frac{2 i e^{i\phi}}{4e^{2i\phi}-1}\,d\phi $$ $$ = \int_0^{\pi}\frac{2 i e^{i\phi}}{4e^{2i\phi}-1}\,d\phi + \int_{\pi}^{2\pi}\frac{2 i e^{i\phi}}{4e^{2i\phi}-1}\,d\phi $$ $$ = \int_0^{\pi}\frac{2 i e^{i\phi}}{4e^{2i\phi}-1}\,d\phi + \int_{0}^{\pi}\frac{2 i e^{i(\phi+\pi)}}{4e^{2i(\phi+\pi)}-1}\,d\phi $$ $$ = \int_0^{\pi}\frac{2 i e^{i\phi}}{4e^{2i\phi}-1}\,d\phi + \int_{0}^{\pi}\frac{-2 i e^{i\phi}}{4e^{2i\phi}-1}\,d\phi $$ $$ = 0 $$

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    It doesn't make much of a difference but I interpreted "in the positive sense of the circle" as "clockwise", so I thought $\gamma(t) = 2e^{-i t}$.2012-04-09
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    @Matt Usually the positive sense in mathematics has the enclosed area to the left (ie counter clockwise), so I interpreted that as $e^{i\phi}$.2012-04-09
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    But of course you are right. it does not make a difference2012-04-09
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    Ah, thanks! I didn't know that.2012-04-09
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    But don't integrate around the two singularities $\pm 1$? Sorry, I just started with this complex integration buisness...2012-04-09
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    @draks There are two singularities at $\pm 1$ and both are within the circle around which we integrate. But that must not concern us when doing the integral explicitly. With cauchys theorem we could have only looked at those two singularities - but then again with cauchy it is even simpler to see this curve as the boundary of everything but the disc that includes these two singularities ;)2012-04-09
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    hm, I'm still a little confused. [Here](http://en.wikipedia.org/wiki/Cauchy%27s_integral_formula#Proof_sketch) they say, that $\oint_C \frac{1}{z-a} \,dz=2\pi i$. Where's the difference? That they don't evaluate the *integral explicitly*?2012-04-09
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    @draks the difference is the second pole. you could evaluate the given integral with cauchys theorem and would get the sum over the residuals of all singularities in the enclosed are. In this case that would be something like $2\pi i-2\pi i=0$ (maybe with a slightly different factor).2012-04-09
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    @draks since $\frac{1}{z^2-1}=1/2( \frac{1}{z-1)-\frac{1}{z+1} )$,so there are two poles2012-04-09
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    Ah, I hoped that it's such... thanks (+1 finally)2012-04-09