Understanding the Simplicity of Pole at z=+i for h(z)

[billiards]

Homework Statement

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

Homework 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}

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?

 

[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...

 

[billiards]

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...

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

 

[micromass]

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

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!

 

[billiards]

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!

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

 

[micromass]

Thank you kind sir.

I believe the partial fraction expansion looks like this:

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

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?

 

[billiards]

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?

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

 

[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.