# Singularities: Complicated function

1. Oct 25, 2009

### libelec

1. The problem statement, all variables and given/known data
Find all the singularities of $$f(z) = \frac{{{e^{\frac{1}{{z - 1}}}}}}{{{e^{\frac{1}{z}}} - 1}}$$ in the extended complex field, classify them and find Res(f, 0) and Res(f, infinity)

2. Relevant equations

Res(f, z0) = a-1 in the Laurent series around that z0

$${e^z} = \sum\limits_{n = 0}^\infty {\frac{{{z^n}}}{{n!}}}$$

3. The attempt at a solution

It's kind of a difficult function to find singularities from.

First I tried to look at the numerator. I found that 1 is an essential singularity, because the Laurent series of $${{e^{\frac{1}{{z - 1}}}}}$$ around 1 is $$\sum\limits_{n = o}^\infty {\frac{1}{{n!{{(z - 1)}^n}}}}$$, which is a series formed by negative powers of (z-1) only. So the series has infinite negative powers of (z-1)

Then I tried to look at the denominator. I know zero has to be a singularity of some sort, but I can't find a way to classify it or to justify that.

Finally, for the infinity, I make the variable change w = 1/z, then I evaluate $$\frac{{{e^{\frac{w}{{1 - w}}}}}}{{{e^w} - 1}}$$ when w=0. I find that zero is a simple pole of f(1/w) (because 1/f(1/w) has a simple zero in w=0), then infinity a simple pole of f(z).

How do I find what kind of singularity is z=0?

Thanks

2. Oct 26, 2009

### libelec

Anyone?

3. Oct 26, 2009

Nobody?