# Need help finding a Laurent Series

1. Apr 6, 2014

### richyw

1. The problem statement, all variables and given/known data

Let $f(z) = \frac{1}{z^2-1}$. Find Laurent Series valid for the following regions.

• 0<|z−1|<2
• 2<|z−1|<∞
• 0<|z|<1

2. Relevant equations

$$\frac{1}{1-z}=\sum^{\infty}_{n=0}z^n,\: |z|<1$$
$$f(z)=\sum^{\infty}_{n=0}a_n(z-z_0)^n+\sum^{\infty}_{n=1}b_n(z-z_0)^{-n}$$

3. The attempt at a solution

I really have no idea what to do, especially for the first two regions. I have written the function as
$$f(z)=\frac{1}{(z+1)(z-1)}$$ and then attempted to find the laurent series for$\frac{1}{z-1}$ & $\frac{1}{z-1}$, then shifted it to 1<|z|<3 and multiply the two sums together, but I think this the wrong way to do it. It's been a couple days and I still can't figure this out!

2. Apr 6, 2014

### micromass

Staff Emeritus
Let's start with the first one. You need to find the Laurent series in the region $0<|z-1|<2$. This means that you need to find a series of the form

$$\sum_{k=-\infty}^{+\infty} a_k (z-1)^k$$

which converges on the right region.

Can you first try to express

$$\frac{1}{z+1}$$

as such a series? What do you get? Hint:

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

Can't you just multiply the series with the factor $\frac{1}{z-1}$ to get the final answer?

3. Apr 6, 2014

### richyw

i'm still quite confused. I don't really remember from calculus how to get series expansions that aren't around z=0.

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

4. Apr 6, 2014

### micromass

Staff Emeritus
Yep, that is exactly right. But you should always mention for which $z$ this series expansion is valid.

Basically, you have the series

$$\frac{1}{1-a} = \sum a^n$$

but this is only valid for $|a| < 1$. You applied this series expansion to $a = -(z-1)/2$, which is fine, but that only works for

$$|(z-1)/2|<1$$

or $|z-1|<2$.

$$\frac{1}{(z-1)(z+1)}$$

Just multiply the series you have with the factor $1/(z-1)$ and you're finished!

5. Apr 6, 2014

### richyw

so just $$f(z)=\sum_{n=0}^{\infty}\frac{(-1)^{n}(z-1)^{n-1}}{2^{n+1}},\: |z-1|<2$$

6. Apr 6, 2014

### richyw

is this correct?

for 2 < |x-1|< ∞ I got
$$\sum^\infty_{n=0}\frac{2^n(-1)^n}{(z-1)^{n+2}}$$
and for 0 <|z| < 1 I got
$$\sum^\infty_{n=0}-z^{2n}$$

7. Apr 6, 2014

### micromass

Staff Emeritus
Fine, but here you also want $0<|z-1|$ since it isn't defined in $1$.

So, now about $2<|z-1|<+\infty$. You can't apply the previous series decomposition anymore. But this time you do know that

$$\left|\frac{2}{z-1}\right|<1$$

So that means that this time you can use that

$$\frac{1}{1 - 2/(z-1)}= \sum \left(\frac{2}{z-1}\right)^n$$

so you must use this somehow.

8. Apr 6, 2014

### micromass

Staff Emeritus
That seems ok.