- #1

billiards

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## Homework Statement

Why is the pole at [itex]z=+i[/itex] simple for the function [itex]1/(1+z^{2})[/itex]?

## Homework Equations

The function can be written as:

[itex]h(z)=1/(1+z^{2})=1/[(z-i)(z+i)][/itex]

It can be expanded as a Laurent series:

[itex]h(z)=\sum^{\infty}_{-\infty}a_{n}(z-z_{0})^{n}[/itex]

The function is defined as "simple" if terms for [itex]n<-1[/itex] do not contribute to the Laurent series. That is:

[itex]h(z)=\sum^{\infty}_{-1}a_{n}(z-z_{0})^{n}[/itex]

## The Attempt at a Solution

Been struggling with this one for a few days now. Tried to do a Taylor series expansion of the function but ended up with something like:

[itex]h(z)=1-2z/(1+z^{2})^{2}+...[/itex]

So it looks like z has exponents less than -1, which to my simplistic mind suggests like the Laurent series has terms contributing for n<-1. This suggests to me that there is a problem in my understanding somewhere along the line.

My intuition is telling me to take a different tack, I think the solution will probably have something to do with the fact that we are showing that

*the pole at +i*is simple. So perhaps I can then sub [itex]z_{0}=i[/itex]?

In which case:

[itex]h(z)=\sum^{\infty}_{-\infty}a_{n}(z-i)^{n}=\frac{1}{(z-i)(z+i)}[/itex]

Hmmm.. (I literally just thought of this as typing) .. is this moving in the right direction?