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

• billiards
You will get-\frac{1}{2}\left(1+(zi)+(zi)^2+(zi)^3+...\right)Now replace (zi) with -i(z-i) and you will have the expansion you need.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. You will get-\frac{1}{2}\left(1+(zi)+(zi)^2+(zi)^3
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?

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...

micromass said:
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?

billiards said:
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!

micromass said:
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

billiards said:
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?

micromass said:
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

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.

## What is the concept of "Pole" at z=+i in h(z)?

The concept of "Pole" in mathematics is a point on a complex plane where a function becomes undefined or infinite. At z=+i, h(z) has a pole because it results in a division by zero, making the function undefined.

## What is the significance of z=+i in h(z)?

Z=+i in h(z) is significant because it is the location of the pole, which represents a singularity in the function. It can also affect the behavior and properties of the function in the surrounding area.

## Why is it important to understand the simplicity of Pole at z=+i for h(z)?

Understanding the simplicity of Pole at z=+i for h(z) is important because it helps in analyzing the behavior and properties of the function in that region. It can also provide insights into the overall behavior of the function and its critical points.

## What are the possible applications of understanding the simplicity of Pole at z=+i for h(z)?

Applications of understanding the simplicity of Pole at z=+i for h(z) include solving complex mathematical problems, analyzing the stability of control systems, and predicting the behavior of functions in various engineering and scientific fields.

## How can one determine the simplicity of Pole at z=+i for h(z)?

The simplicity of Pole at z=+i for h(z) can be determined by analyzing the function's Laurent series expansion around that point. The number of terms in the series will indicate the order of the pole, and the coefficient of the first term will determine the residue and the simplicity of the pole.

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