Help with Laurent Series for Convergence in Different Regions

[MaxManus]

Homework Statement

Could anyone help me with Laurent series? I do not understand it at all even though the book has several examples.

And here is one with my comments

Find the Laurent series of
\frac{1}{(z-1)(z-2)}

a in the region abs(z) < 1
b in the region 1< abs(z) < 2
c in the region abs(z) > 2

The Attempt at a Solution

They are using a theorem which says that any convergent series of the form
\sum_{j= - \infty}^{\infty} c_j (z-z_0)^j

Using partial fractions

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

Now we proceed differently in each region to derive the convergent series

a For abs(z) < 1

6) \frac{1}{z-2} = - \frac{1}{2} \frac{1}{1 -z/2} = - \frac{1}{2} \sum_{j=0}^{\infty} = -\sum_{j=0}^{\infty} \frac{z^j}{2^j+1}

The equation 6) I am not sure about, why did the rewrite it in the second from the left and how did they see that this could be written as the series in the third from the left. It is the just taylor series around zeros? and do they use zero because it is the center of the disk? And do the use the taylor series because the circle does not include the number 2?

they do the same for:
7) \frac{1}{z-1} = - \frac{1}{1-z} = -\sum_{j=0}^{\infty} z^j

And then it is of course just to subtract the two equations and you get
\sum_{j=0}^{\infty} (-\frac{1}{2^{j+1}} + 1)z^j

b for 1 < abs(z) < 2, equation 6 is still valid, but we have

8)
\frac{1}{z-1} = \frac{1}{z} \frac{1}{1-1/z} = \frac{1}{z} \sum _{j=0}^{\infty} \frac{1}{z^j} = \sum_{j=0}^{\infty} \frac{1}{z^{j+1}}

why was 6) still valid and 7) had to be replaced with 8)?

For abs(z) > 2, Equation 8 is still valid, why?
and
\frac{1}{z-2} = \frac{1}{z} \frac{1}{1-2/z} = \frac{1}{z} \sum_{j=0}^{\infty} (\frac{2}{z})^j = \sum_{j=0}^{\infty} \frac{2^j}{z^{j+1}}

Edit: Not sure why the two last latex equations fail, they work in
http://www.codecogs.com/latex/eqneditor.php

 

[vela]

One basic series you need to know is

\frac{1}{1-z} = 1+z+z^2+z^3+\cdots

It's just the Taylor series about z₀=0. Using convergence tests, you can show it's only valid when |z|<1.

The general approach to the problem is to rewrite the terms in such a way that it looks like that series above, where the denominator is 1-something and where the modulus of the something is less than 1.

 

[MaxManus]

Thanks, but why did they get different Laurent series for different regions

 

[vela]

For instance, when 1<|z|<2, you can't use

\frac{1}{1-z} = 1+z+z^2+z^3+\cdots

because the series won't converge since you don't have |z|<1. However, in that region, you have |1/z|<1, so if you can expand in powers of 1/z instead, you'll have a series that converges.

 

[gomunkul51]

But(!) you can use:

1<|z|

<br /> \frac{1}{1-\frac{1}{z}} = 1+\frac{1}{z}}+\frac{1}{z^2}} +\cdots<br />

...Basically this and what vela wrote are the two things you have to know :)

 

[MaxManus]

Thanks for all the help. All my exercises does in fact end up with \frac{1}{1-z}