# Homework Help: Expand function in Laurent series

1. Mar 27, 2014

### skrat

1. The problem statement, all variables and given/known data
Expand $f(z)=\frac{1}{z^2(z-1)}$ in Laurent series for $0<|z-1|<1$. Use binomial series.

2. Relevant equations

3. The attempt at a solution

I am looking at this problem for quite some time now and still I got nothing.

I do however think that this will come in handy $\frac{1}{z^2(z-1)}=\frac{1}{z}-\frac{1}{z^2}-\frac{1}{z-1}$. That's all I have and not even a clue on how to continue ... Please help.

2. Mar 27, 2014

### skrat

Update.

Ok, I managed to find some examples http://sym.lboro.ac.uk/resources/Handout-Laurent.pdf and hopefully at least some of the things written below are ok.

In $f(z)=\frac{1}{z^2(z-1)}$ I replace $w=z-1$ so that my open disk $0<|z-1|=|w|<2$ is centered around point $0$.

So $f(w)=\frac{1}{w(w+1)^2}=\frac{1}{w}-\frac{1}{w+1}-\frac{1}{(w+1)^2}$.

Now it makes sense to google binomial series and find out that $\frac{1}{(1-z)^{\beta +1}}=\sum_{k=0}^{\infty}\binom{k+\beta }{k}z^k$.

Now if I am not completely mistaken than $\frac{1}{w}=\frac{-1}{-(1-1+w)}=-\frac{1}{1-(1+w)}=-\sum_{k=0}^{\infty}(1+w)^k$.

And similar for $-\frac{1}{w+1}$ and $-\frac{1}{(w+1)^2}$, which alltogether gives me $f(w)=-\sum_{k=0}^{\infty}[(1+w)^k+(2+w)^k+(1+(1+w)^2)^k]$.

And finally $f(z)=-\sum_{k=0}^{\infty}[z^k+(z+1)^k+(1+(z)^2)^k]$.

Yes or no?

3. Mar 27, 2014

### vela

Staff Emeritus
No. You want to expand about z=1 or w=0. You expanded parts of the function about w=1. The series you end up with in the end should consist of powers of (z-1).

Also, the disk is given by 0 < |z-1| < 1. If w=z-1, then you get 0 < |w| < 1, not 0 < |w| < 2.

4. Mar 27, 2014

### skrat

Ok, how do I do that?

I agree, it was mistake.

5. Mar 27, 2014

### vela

Staff Emeritus
For example, the term 1/w is already in the form of wn. You don't have to do anything with that term.

Expand $\frac{1}{1+w}$ as a power series in $w$. (You should memorize this series if you don't already know it.) Differentiate that and you'll get the series for ${-\frac{1}{(1+w)^2}}$. Put it all together and you have the Laurent series in terms of $w$.

6. Mar 27, 2014

### skrat

That's exactly how I wanted to do it, but the problem states that I have to use binomial series. So, of course my next question is what does binomial series have to do with all this? Where do I than use it if not here?

7. Mar 27, 2014

### vela

Staff Emeritus
You can use the binomial theorem to expand (1+w)-1 and (1+w)-2.

8. Mar 27, 2014

### skrat

Remember: $f(w)=\frac{1}{w(w+1)^2}=\frac{1}{w}-\frac{1}{w+1}-\frac{1}{(w+1)^2}$

Hopefully this sounds a bit better:

$-\frac{1}{w+1}=-\frac{1}{1-(-w)}=-\sum_{k=0}^{\infty}\binom{k}{k}(-1)^kw^k=\sum_{k=0}^{\infty}(-1)^{k+1}w^k$

and

$-\frac{1}{(w+1)^2}=-\frac{1}{(1-(-w))^2}=-\sum_{k=0}^{\infty}\binom{k+1}{k}(-1)^kw^k=\sum_{k=0}^{\infty}(k+1)(-1)^{k+1}w^k$