# Residue of a function

1. Feb 19, 2007

### 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

Last edited: Feb 19, 2007
2. Feb 19, 2007

### marcusl

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

3. Feb 19, 2007

### DrKareem

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

4. Feb 19, 2007

### HallsofIvy

Staff Emeritus
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.

5. Feb 19, 2007

### 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).

6. Feb 19, 2007

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

7. Feb 20, 2007

### 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.