# 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.