Laurent Series Expansion Centered on z=1

[ChemEng1]

Homework Statement

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.

Homework Equations

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? I've tried but haven't been able to find anything that works.

 

[ahsanxr]

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

 

[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}]

 

[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.

 

[ChemEng1]

I'm not too sure about the n+1 on top of the -1.

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 also not too sure whether there should be a factorial there somewhere.

I'm pretty sure the coefficient of the (z-1) term is a multiple of 2 in this case.

 

[ahsanxr]

I'm pretty sure the coefficient of the (z-1) term is a multiple of 2 in this case.

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

 

[ChemEng1]

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

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

 

[ahsanxr]

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

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).

 

[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.