Laurent Series Expansion of $\frac{1}{z^2-1}$

[gtfitzpatrick]

Homework Statement

Calculate the laurent series expansion about he points specified, classify the singularity and sate the region of convergence for.

$\frac{1}{z^2 - 1}$ at (i) z=1 (ii) z=-1 (iii)z=0

Homework Equations

The Attempt at a Solution

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

i observe that f(z) is analytic in {z≠ -1,1} so is anyalytic in -1 < $\left|z\right| < 1$

@z=1 $\frac{-1}{2}\frac{1}{1-(-z)} = \frac{-1}{2}\sum (-z)^n$

similarly

@z=-1 $\frac{1}{2}\frac{1}{(z-1)} = \frac{-1}{2}\frac{1}{1-z)} = \frac{-1}{2}\sum (z)^n$

and

@z=0 is it just the sum of both of them? singularities are at 1, -1 and region of convergence -1 < $\left|z\right| < 1$

am i anywhere near?

 

[jackmell]

Homework Statement

Calculate the laurent series expansion about he points specified, classify the singularity and sate the region of convergence for.

$\frac{1}{z^2 - 1}$ at (i) z=1 (ii) z=-1 (iii)z=0

Homework Equations

The Attempt at a Solution

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

i observe that f(z) is analytic in {z≠ -1,1} so is anyalytic in -1 < $\left|z\right| < 1$

@z=1 $\frac{-1}{2}\frac{1}{1-(-z)} = \frac{-1}{2}\sum (-z)^n$

similarly

@z=-1 $\frac{1}{2}\frac{1}{(z-1)} = \frac{-1}{2}\frac{1}{1-z)} = \frac{-1}{2}\sum (z)^n$

and

@z=0 is it just the sum of both of them? singularities are at 1, -1 and region of convergence -1 < $\left|z\right| < 1$

am i anywhere near?

No. When you expand around a point z0, then the series should be in powers of (z-z0). So for the first one, the first partial term is already in powers of z-1. However for the second term:

$$-1/2 \frac{1}{z+1}$$

that's not. But you can do so by forcing a factor of z-1 in the denominaor by:

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

which you can then easily express that as a power series in terms of (z-1) right? Same dif with the other ones.

 

[gtfitzpatrick]

thanks a million jackmell