# Partial Fractions in Laurent Series Expansion

## Homework Statement

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

Partial fraction

## The Attempt at a Solution

R1 = 0 < z < 1
R2 = 1 < z < 2
R3 = z > 2

f = $\frac{1}{z(z-1)(z-2)}$ = $\frac{1}{z}$ * ($\frac{A}{z-1}$ + $\frac{B}{z-2}$)
Where A = -1 , B = 1.
f = $\frac{1}{z}$ * ($\frac{1}{z-2}$ + $\frac{1}{1-z}$)

My question is, why do you have to use partial fractions?

Can't you just leave the initial multiplication of three terms and expand them individually;
$\frac{1}{z-1}$ (dividing 1/z to numerator and denominator at R2,R3)
and
$\frac{1}{z-2}$ (dividing 1/z to numerator and denominator at R3)

so for example,

for R2,
f = $\frac{1}{z}$ * $\frac{1}{z-2}$ * (-$\frac{1}{z}$*$\frac{1}{1-z^{-1}}$)

You can expand 1/(1-z) and 1/(z-2) using taylor series.

Should you always use partial fractions?
Or does expanding without using partial fractions still work?
]

## The Attempt at a Solution

Related Calculus and Beyond Homework Help News on Phys.org
STEMucator
Homework Helper
I believe your partial fraction expansion is incorrect, you should check it again.

##\frac{A}{z} + \frac{B}{z-1} + \frac{C}{z-2}##

$\frac{A}{z}$ this part is not necessarily needed for partial fractions because
it doesn't require any expansions..

STEMucator
Homework Helper
$\frac{A}{z}$ this part is not necessarily needed for partial fractions because
it doesn't require any expansions..
You don't really even need partial fractions at all. You could write it like this:

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

Then write the terms as their respective series, for example:

##\frac{1}{z} = \sum_{n=0}^{∞} (-1)^n (-1+z)^n ## where ##|1-z|<1##

I believe that's what you were asking anyway.

Cool, I get it now. You answered my question.

Thanks Zondrina!