Is Cauchy's integral formula applicable to this type of integral?

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SUMMARY

The discussion focuses on the applicability of Cauchy's integral formula to the integral \(\oint_{C} \frac{z^2+1}{e^{iz}-1}\) where the contour \(C\) is the unit circle. The integral has a pole at \(z=0\), and the user seeks to express the integrand in the form \(\frac{f(z)}{z}\) to utilize the formula. Ultimately, it is concluded that the residue theorem is necessary for evaluation, as the integral cannot be directly simplified to the required form without further manipulation, such as performing a Laurent series expansion.

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Homework Statement



I am trying to determine if Cauchy's integral formula will work on the following integral, where the contour C is the unit circle traversed in the counterclockwise direction.

\oint_{C}^{}{\frac{z^2+1}{e^{iz}-1}}


Homework Equations


See Cauchy's Integral Formula - http://en.wikipedia.org/wiki/Cauchy_integral_formula"

The Attempt at a Solution



I realize that there is a pole at z=0. I realize that if I could get this integral into the form

\frac{f(z)}{z},

with f(z) being analytic in and on the contour C, then I could use the formula. However, I'm not sure how to get the integrand in that form. Is it even possible to use Cauchy's integral formula on this integral, or do I need to use a different method to evaluate this integral?
 
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Ok, so it appears that I need to use the residue theorem in order to evaluate this integral. I was hoping I could just use the integral formula. I haven't got to study the residue theorem yet in my text. Thanks
 
The residue thorem is a simple use of the integral formula.
write f(z)=[z*f(z)]/z
 
so are these the steps:

do a laurent expansion of denominator
cancel with stuff in the numerator
then the coefficient of the z^{-1} term gives us the residue
multiply this by 2 \pi i to give the integral's value

im not too sure about the first of those two steps?
 

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