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newmike

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**Contour integral with multiple singularities inside domain without residue theorem??**

## Homework Statement

Evaluate

[tex]\oint\frac{dz}{z^{2}-1}[/tex]

where C is the circle [tex]\left|z\right|[/tex] = 2

## Homework Equations

Just learned contour integrals, so not much.

Ok to use Cauchy's Integral formula (if applicable)

Cannot use reside theorem as we haven't learned that yet

## The Attempt at a Solution

I've tried this a few different ways...

Anyway, it is clear that there are two singularities: z=1 and z=-1, and both of which are inside the contour. Because of this, I don't think I can use Cauchy's Integration formula.

By attempting the integral the long way and writing out partial fractions, I've arrived at:

[tex]\frac{1}{2}\oint \frac{2ie^{it}dt}{2e^{it}-1}-\frac{1}{2}\oint \frac{2ie^{it}dt}{2e^{it}+1}[/tex]

where each integral is integrated over 0 to 2*pi

which is where I'm stick. I see that I can turn this into (using substitution):

[tex]\frac{1}{2}ln(\frac{2e^{it}-1}{2e^{it}+1})[/tex]

evaluated on 0 to 2pi which ends up with zero since it's not analytic. It seems I are ignoring the singularities here...

I have a gut feeling the answer shouldn't be zero... Any suggestions?

Thanks,

-Mike

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