Laurent series, deep confusion

[futurebird]

I'm so lost. I can follow the steps in the examples in the book, but all of the examples seem easier than this problem-- and in each case there is a point where you "spot" a similarity to the geometric series or something and then you can just patch it in... But, I just don't know how to make that happen here.

Given the function $$f(z)=\frac{z}{a^{2}-z^{2}}$$ Expand f(z) in a laurent series in the regions: |z| < a and |z| > a.

From another problem I know that:

$$\frac{1}{1-z}=\displaystyle\sum_{n=0}^{\infty}z^{n}$$, |z| < 1 so:

$$\frac{1}{a-z}=\displaystyle\sum_{n=0}^{\infty}z^{n}$$, |z| < a (right?)

Now I want to use this fact so I write:

$$f(z)=\frac{z}{(a+z)(a-z)}=\frac{z}{(a+z)}\left(\displaystyle\sum_{n=0}^{\infty}z^{n}\right)$$

um... and I don't know what to do next. I've looked at the example in the book and they used -1/z to repace z when working with $$\frac{z}{(a+z)}$$.

But, if I do that it changes the boundaries for the whole problem. I tried writing:

$$=\left(\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}}{z^{n+1}}\right)\left(\displaystyle\sum_{n=0}^{\infty}z^{n}\right)$$

for my next step, but I know this is wrong. I can only do that replacement if I change the boundary... I'm sort of confused about how to go about doing these problems. Help?

 

[HallsofIvy]

Why not use "partial fractions"? That is, write
[tex]\frac{z}{a^2- z^2}= \frac{z}{(a-z)(a+z)= \frac{A}{a- z}+ \frac{B}{a+z}[/itex]
Then write the two separate fractions as power series in z using the fact that
$\frac{1}{1- z}= \sum_{n=0}^{\infty}z^n$
and add corresponding terms of the two series.

 

[futurebird]

Okay, I have

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

Then,

$$=\frac{1}{2}\left(\displaystyle\sum_{n=0}^{\infty}z^{n}-\displaystyle\sum_{n=0}^{\infty}(-1)^{n}z^{n}\right)$$

$$=\displaystyle\sum_{n=0}^{\infty}z^{2n+1}$$

for |z| <a

okay, but what about |z|> a?

Something about this makes no sense... where are the negative powers?

 

[Kreizhn]

It is important to consider the cases of $|z|<a \text{ and } |z|>a$ seperately.

Let us first consider $|z|>a$.

We want to be able to use the fact that

$$\frac{1}{1-x} = \displaystyle\sum_{n=0}^{\infty}x^{n}$$

Thus factor a z out of the addition term to get

$$\frac{1}{z\left( \frac{a}{z} +1 \right) }$$

Now since $|z|>a$, then we know that the sum will converge, and so we can proceed as before.

For the record, it is not necessary to use partial fractions for either case.

$$f(z)=\frac{z}{(a^2-z^2)}=\frac{z}{a^2 \left( 1- \frac{z^2}{a^2} \right) }$$

 

[Kreizhn]

You'll note that using

$$f(z)=\frac{z}{(a^2-z^2)}=\frac{z}{a^2 \left( 1- \frac{z^2}{a^2} \right) }$$

is a much faster way of getting your solution. Furthermore, when you get your closed solution, who says that the index variable must be a positive integer?