Find the Laurent Series for (z^2-4)/(z-1) with z_0=1 | Homework Solution

[TheFerruccio]

Homework Statement

Find the Laurent series for the function that converges at 0 < \left|z-z_0\right| < R

Homework Equations

\frac{z^2-4}{z-1}
z_0 = 1

The Attempt at a Solution

I think this is going to be a finite sum. I think it could be wrong, though, because it certainly differs with the answer in the book.

\frac{z^2-4}{z-1} = \frac{z^2}{z-1}-\frac{4}{z-1}

I tried to put the first part in terms of z-1...

\frac{z^2}{z-1} = \frac{z-1+1}{z-1}\frac{z-1+1}{z-1} =

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

= \left(1+\frac{1}{z-1}\right)\left(1+\frac{1}{z-1}\right) = 1+\frac{2}{z-1}+\frac{1}{\left(z-1\right)^2}

I recombined with the last term...

1+\frac{2}{z-1}+\frac{1}{\left(z-1\right)^2}-\frac{4}{z-1} = 1-\frac{2}{z-1}+\frac{1}{\left(z-1\right)^2}

This is the final answer that I get. However, the book says that it is

-\frac{3}{z-1}+2+(z-1)

I am absolutely stumped and I've been racking my brain over it for hours.

 

[micromass]

You found the Laurent series at $1$. I think they want you to find the Laurent series at $0$.

 

[TheFerruccio]

You found the Laurent series at $1$. I think they want you to find the Laurent series at $0$.

They may have wanted me to find the Laurent series at 1. Sorry, I forgot to add to the problem...

z_0 = 1

 

[micromass]

OK, so I'll take a closer look.

I see right away that this is not a correct thing to do:

\frac{z^2}{z-1} = \frac{z-1+1}{z-1}\frac{z-1+1}{z-1} =

 

[TheFerruccio]

OK, so I'll take a closer look.

I see right away that this is not a correct thing to do:

I figured it was the algebra that was holding me back here. Is it because it creates a singularity at 1 or something? Why is this wrong? I do not see it.

 

[micromass]

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

This does not equal $\frac{z^2}{z-1}$.

 

[TheFerruccio]

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

This does not equal $\frac{z^2}{z-1}$.

Holy crap, oh geeze. That's embarrassing. Thanks for pointing that out. Let me try this again.

 

[TheFerruccio]

Alright, well, realizing that that technique is wrong has completely erased the paths that I see to solving this problem. Any hints as to an appropriate next step to take? I've gone the method of splitting z^2-4 into (z-2)(z+2) but I do not know how to go from there.

 

[micromass]

Try to write

\frac{z^2 - 4}{z-1} = \frac{(z-1)^2 + P(z)}{z-1}

and try to find the polynomial $P(z)$.

 

[TheFerruccio]

Try to write

\frac{z^2 - 4}{z-1} = \frac{(z-1)^2 + P(z)}{z-1}

and try to find the polynomial $P(z)$.

Oh, that's clever. Thanks for suggesting this method. This immediately gave me the path I needed to do.

\frac{z^2-4}{z-1} = \frac{\left(z-1\right)^2+P(z)}{z-1}=\frac{{(z-1)}^2+2z-5}{z-1} = \frac{{(z-1)}^2+2(z-1)-3}{z-1} = (z-1)+2-\frac{3}{z-1} which is exactly what I was looking for. I will keep this technique in my toolbox. Thanks for the suggestion ;)

Oh, and it converges for all R > 0.