Need help finding a Laurent Series

[richyw]

Homework Statement

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

Homework Equations

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

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!

 

[micromass]

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?

 

[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}}

 

[micromass]

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}}

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|&lt;1

or $|z-1|<2$.

Ok, now what about

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

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

 

[richyw]

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

 

[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}

 

[micromass]

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

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|&lt;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.

 

[micromass]

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}

That seems ok.