Very easy and quick residue calculus question.

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    Calculus Residue
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SUMMARY

The forum discussion centers on evaluating the integral \(\oint_{\left|z\right|=3/2} \frac{e^{\frac{1}{z-1}}}{z} dz\) using the residue theorem. The correct evaluation yields an answer of \(2 \pi i\), as the singularity at \(z=0\) is not considered within this specific domain. In contrast, when the domain is \(|z|=1/2\), the singularity at \(z=0\) contributes to the result, yielding \(2\pi i/e\). The discussion emphasizes the importance of correctly identifying and evaluating singularities in complex analysis.

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Homework Statement



\oint_{\left|z\right|=3/2} \frac{e^{\frac{1}{z-1}}}{z} dz<br />

Homework Equations



Using residue theorem, since there are two singularities withing the domain, evaluate residues at each singularity, and multiply by 2\pi i

The Attempt at a Solution



Here is the problem. The answer is 2 \pi i instead of 2 \pi i + \frac{2\pi i}{e}
I don't understand why singularity at zero is ignored.
FYI, if the domain it |z| = 1/2, the answer is 2pi*i/e.
 
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They didn't ignore the singularity at 0. The singularity at z=1 is not what you think it is. Try putting w=z-1 and expanding the power series around w=0. You'll find an infinite number of 1/w terms. Sum them.
 

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