Residue Theorem: Finding Residue at ##|z|=2##

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Homework Statement



Find the residue of ##\oint { \frac { sinz }{ 2z-\pi } } dz## where ##\left| z \right| =2##[/B]

Homework Equations



##f\left( z_{ o } \right) =\frac { 1 }{ 2\pi i } \oint { \frac { f\left( w \right) }{ w-z_{ o } } } dw##

The Attempt at a Solution



It seems to me that the answer is ##2\pi i##, but the book gives ##\pi i##.

Not sure what I did wrong, I am pretty confident in my answer.

Chris
 
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For your integral, f(z_0)=\frac 1 2 \sin \frac \pi 2 and z-z_0=z-\frac \pi 2.
 
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I got \pi i as well. You probably forgot a factor of half when calculating the residue.
 
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Where does the 1/2 factor come from?

Thanks,
Chris
 
Never mind, I see that it the bottom needs to have a two factored out to fit the form.

Thanks,
Chris
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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