Homework Help: A Simple Pole

1. Jul 11, 2011

billiards

1. The problem statement, all variables and given/known data

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

2. Relevant equations

The function can be written as:
$h(z)=1/(1+z^{2})=1/[(z-i)(z+i)]$

It can be expanded as a Laurent series:
$h(z)=\sum^{\infty}_{-\infty}a_{n}(z-z_{0})^{n}$

The function is defined as "simple" if terms for $n<-1$ do not contribute to the Laurent series. That is:
$h(z)=\sum^{\infty}_{-1}a_{n}(z-z_{0})^{n}$

3. 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:

$h(z)=1-2z/(1+z^{2})^{2}+....$

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 $z_{0}=i$?

In which case:
$h(z)=\sum^{\infty}_{-\infty}a_{n}(z-i)^{n}=\frac{1}{(z-i)(z+i)}$

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

2. Jul 11, 2011

micromass

You will need to develop a Laurent series around a certain point. So here you must find the Laurent series around i. This means you must be able to write your function in the form

$$f(z)=...+\frac{a_{-2}}{(z-i)^2}+\frac{a_{-1}}{z-i}+a_0+a_1(z-i)+a_2(z-i)^2+...$$

So, you must write your function $f(z)=\frac{1}{1+z^2}$ in this form. To do this, you must start with splitting your function into partial fractions...

3. Jul 13, 2011

billiards

How do you go about expanding that function into partial fractions?

4. Jul 13, 2011

micromass

You'll need to find constants A and B such that

$$\frac{1}{z^2+1}=\frac{A}{z+i}+\frac{B}{z-i}$$

Look at http://en.wikipedia.org/wiki/Partial_fractions for some nice examples!

5. Jul 13, 2011

billiards

Thank you kind sir.

I believe the partial fraction expansion looks like this:

$\frac{1}{2i(z-i)}-\frac{1}{2i(z+i)}$

Ah yes and if we observe that $z+i=(z-i)+2i$ then.... urrrr ... how do we clean up that $z+i$ term? (I think my brain is not working today!)

Do we just ignore the $z+i$ term and note that the exponent of the $z-i$ is $-1$?

Or can we say

6. Jul 13, 2011

micromass

Seems good, and you probably notices that

$$\frac{1}{2i(z-i)}$$

is in the correct form, but the other term is not. You will have to write the term

$$\frac{1}{2i(z+i)}$$

in a series that involves only terms of the kind $(z-i)$. For this, we will have to use the following very well known series

$$\frac{1}{1-z}=1+z+z^2+z^3+z^4+...$$

Can you use the above series to write

$$\frac{1}{2i(z+i)}$$

in the form of a series?

7. Jul 14, 2011

billiards

Hmmm I can't see how to do it that way.

I tried a Taylor series expansion about z=i of that expression and I think it is wrong but I found that....

$$\frac{1}{2i(z+i)}=-\frac{1}{4}-\frac{i}{8}(z-i)-\frac{1}{4}(z-i)^{2}+.....$$

Now that fits, but I don't fully understand my own reasoning for getting there so I don't feel comfortable. Also I would like to see how using the well know series works.

Cheers

8. Jul 14, 2011

micromass

For the well-known series approach, do

$$\frac{1}{2i(z+i)}=\frac{1}{2iz-2}=-\frac{1}{2}\frac{1}{1-zi}$$

Now apply the well-known series

$$\frac{1}{1-a}=1+a+a^2+a^3+a^4+...$$

with a=zi.