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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 forward –  example Apr 9 '12 at 10:36

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up vote 4 down vote accepted

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 $\gamma(t) = 2e^{-i t}$. –  Matt N. Apr 9 '12 at 10:48
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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 $e^{i\phi}$. –  example Apr 9 '12 at 10:50
    
But of course you are right. it does not make a difference –  example Apr 9 '12 at 10:51
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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). –  example Apr 9 '12 at 13:55
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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 poles –  Jiangnan Yu Apr 9 '12 at 14:06

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