# Complex Variables: Poles

1. Sep 9, 2009

### Winzer

1. The problem statement, all variables and given/known data
Determine the nature of the singularity at z=0

2. Relevant equations
$f(z)=\frac{1}{cos(z)}+\frac{1}{z}$

3. The attempt at a solution
by expanding into series:

$f(z)=\Sigma_{n=0}^{\infty} \frac{(2n)! (-1)^n}{x^{2n}} + \Sigma_{n=0}^{\infty} (-1)^n (z-1)^n$
Now $\frac{1}{z}$ has no principle part, [tex] b_m=0[/itex].
This leaves the only principle part from cos. $b_m=(2m)! (-1)^m$. There are infinite bm
so the behaviour is an essential singularity.

I don't feel too confident about this answer. I feel I have overlooked a step.

2. Sep 9, 2009

### Dick

Ouch. If you are looking for a singularity at z=0, why are you expanding 1/z around z=1? And cos(0)=1, it's not a singularity of 1/cos(z) at all. And you can't invert a power series by inverting each term in the power series. 1/(a+b) is not equal to 1/a+1/b.