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

[DrKareem]

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

 

[marcusl]

Perfect, except you need z^(-2) in the 3rd term above (i'm sure you made a typo).

 

[DrKareem]

Oh yes it is a typo. Thanks for the input.

 

[HallsofIvy]

Except that z= 0 is NOT a pole for e^1/z. A point is a "pole of order n" for function f(z) is zⁿf(z) is analytic but no z^kf(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 e^1/z has all negative integers as powers, z= 0 is an essential singularity, not a pole.

 

[arunma]

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).

 

[DeadWolfe]

Halls is correct. The residue is defined only for a pole of finite order, which you do not have.

 

[DrKareem]

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.