# Homework Help: Laurent Series Expansion Centered on z=1

1. Jul 16, 2012

### ChemEng1

1. The problem statement, all variables and given/known data
Find a Laurent Series of f(z)=$\frac{1}{(2z-1)(z-3)}$ about the point z=1 in the annular domain $\frac{1}{2}<|z-1|<2$.

2. Relevant equations

3. The attempt at a solution
By partial fraction decomposition, $f(z)=\frac{1}{(2z-1)(z-3)}=\frac{1}{5}\frac{1}{z-3}-\frac{2}{5}\frac{1}{2z-1}$.

When $\frac{1}{2}<|z-1|<2, \frac{(z-1)}{2}<1$.

Hence: $\frac{1}{z-3}=-\frac{1}{3-z}=-\frac{1}{3-z+1-1}=-\frac{1}{2-(z-1)}=-\frac{1}{2}\frac{1}{1-\frac{(z-1)}{2}}$

Therefore: $\frac{1}{z-3}=-\sum_{n=0}^{\infty}\frac{(z-1)^{n}}{2^{n+1}}$

And: $f(z)=-\frac{1}{5}\sum_{n=0}^{\infty}\frac{(z-1)^{n}}{2^{n+1}}-\frac{2}{5(2z-1)}.$

Is this complete? Is there something I could similarly do with the other term? Ive tried but havent been able to find anything that works.

Last edited: Jul 16, 2012
2. Jul 16, 2012

### ahsanxr

I would try explicitly calculating a few terms of the taylor series of the second term and try seeing a pattern.

3. Jul 16, 2012

### ChemEng1

It's this, right? $\frac{1}{2z-1}=\sum^{\infty}_{n=0}(-1)^{n+1}(z-1)^{n}*2^{n}$

So the expansion would become: $f(z)=-\frac{1}{5}\sum^{\infty}_{n=0}(z-1)^{n}[(-2)^{n+1}-2^{-n-1}]$

4. Jul 16, 2012

### ahsanxr

I'm not too sure about the n+1 on top of the -1. I'm also not too sure whether there should be a factorial there somewhere.

5. Jul 16, 2012

### ChemEng1

You're right about the -1. It should be raise to n, not n+1. I mistakingly thought the index started at 1 instead of 0.
I'm pretty sure the coefficient of the (z-1) term is a multiple of 2 in this case.

6. Jul 16, 2012

### ahsanxr

You're correct, sorry for that. I was forgetting to divide by n! in the formula for a_n of a taylor series.

7. Jul 16, 2012

### ChemEng1

No biggie. So is the last expression the answer then?

8. Jul 16, 2012

### ahsanxr

I still think there are a couple of sign and index errors. Try going through your algebra again to find them (or to prove me wrong).

9. Jul 16, 2012

### ChemEng1

I think I figured the other fraction out:

$\frac{1}{2}<|z−1|<2,\frac{1}{2(z-1)}<1.$

Hence: $\frac{1}{2z-1}=\frac{1}{2z-1-1+1}=\frac{1}{2(z-1)+1}=\frac{1}{2(z-1)}\frac{1}{1+\frac{1}{2(z-1)}}=\frac{1}{2(z-1)}\frac{1}{1-(-\frac{1}{2(z-1)})}=\frac{1}{2(z-1)}\sum^{\infty}_{n=0}(-1)^{n}[2(z-1)]^{-n}=\sum^{\infty}_{n=0}(-1)^{n}[2(z-1)]^{-n-1}$

That looks much better. I hope.