Need help finding a Laurent Series

[richyw]

Homework Statement

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

• 0<|z−1|<2
• 2<|z−1|<∞
• 0<|z|<1

Homework Equations

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

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
[tex]f(z)=\frac{1}{(z+1)(z-1)}[/tex] and then attempted to find the laurent series for[itex]\frac{1}{z-1}[/itex] & [itex]\frac{1}{z-1}[/itex], 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

[tex]\sum_{k=-\infty}^{+\infty} a_k (z-1)^k[/tex]

which converges on the right region.

Can you first try to express

[tex]\frac{1}{z+1}[/tex]

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

[tex]\frac{1}{z+1} = \frac{1}{2 + (z-1)} = \frac{1}{2} \frac{1}{1 + (z-1)/2}[/tex]

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 [tex]\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}}[/tex]

 

[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 [tex]\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}}[/tex]

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

Basically, you have the series

[tex]\frac{1}{1-a} = \sum a^n[/tex]

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

[tex]|(z-1)/2|<1[/tex]

or ##|z-1|<2##.

Ok, now what about

[tex]\frac{1}{(z-1)(z+1)}[/tex]

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

 

[richyw]

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

 

[richyw]

is this correct?

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

 

[micromass]

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

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

[tex]\left|\frac{2}{z-1}\right|<1[/tex]

So that means that this time you can use that

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

so you must use this somehow.

 

[micromass]

is this correct?

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

That seems ok.