What is the residue of exp(1/z) at z=0?

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Discussion Overview

The discussion revolves around finding the residue of the function \(\exp{\frac{1}{z}}\) at \(z=0\). Participants explore the nature of the singularity at this point and the implications for calculating the residue, including the use of Laurent series and definitions of poles.

Discussion Character

  • Debate/contested, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant proposes that the residue can be found using the Laurent series expansion, identifying the coefficient of \(1/z\) as 1.
  • Another participant points out a potential typo in the series expansion, suggesting that the third term should include \(z^{-2}\).
  • A different participant argues that \(z=0\) is not a pole but an essential singularity, as the Laurent series for \(\exp{\frac{1}{z}}\) contains all negative integer powers of \(z\).
  • Some participants agree that the residue is only defined for poles of finite order, which is not applicable in this case.
  • One participant expresses gratitude for the clarification regarding the nature of the singularity and its implications for finding the residue.

Areas of Agreement / Disagreement

Participants generally disagree on whether \(z=0\) is a pole or an essential singularity, with some asserting it is an essential singularity and others supporting the initial claim about the residue calculation. The discussion remains unresolved regarding the correct classification of the singularity.

Contextual Notes

Limitations include the dependence on the definitions of poles and essential singularities, as well as the implications for residue calculations. The discussion does not resolve the mathematical steps regarding the classification of the singularity.

DrKareem
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Hi. I'm trying to find the residue of

\exp{\frac{1}{z}}

at z=0 since it is a pole, so I can integrate the function over the unit circle counterclockwise. I expanded this function in Laurent Series to get

\exp{\frac{1}{z}} = 1 + \frac{1}{1!z} + \frac{1}{2!z^2}+ ...

So in this case the residue is the coefficient of 1/z which is 1. Is this method correct? There is no answer to it in the book...

EDIT: If you guys can find the residue using another method, please teach me.
EDIT 2: Fixed the typo :p
 
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Perfect, except you need z^(-2) in the 3rd term above (i'm sure you made a typo).
 
Oh yes it is a typo. Thanks for the input.
 
Except that z= 0 is NOT a pole for e1/z. A point is a "pole of order n" for function f(z) is znf(z) is analytic but no zkf(z) is analytic for k< n. In particular, the Laurent series for f(z) has no power of z less than -n. Since the Laurent series for e1/z has all negative integers as powers, z= 0 is an essential singularity, not a pole.
 
It's been a couple years since I took complex analysis, but yes, this seems like a perfectly accurate way to find the Laurent Series (and by consequence the residue as well).
 
Halls is correct. The residue is defined only for a pole of finite order, which you do not have.
 
Thank you HallsofIvy for clarifying this matter. This must be the reason why the other methods for finding the residue are not applicable, just like what Deadwolfe suggested. That explains a lot. And thanks for the rest for your input.
 

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