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## Homework Statement

f = [itex]\frac{1}{z(z-1)(z-2)}[/itex]

## Homework Equations

Partial fraction

## The Attempt at a Solution

R1 = 0 < z < 1

R2 = 1 < z < 2

R3 = z > 2

f = [itex]\frac{1}{z(z-1)(z-2)}[/itex] = [itex]\frac{1}{z}[/itex] * ([itex]\frac{A}{z-1}[/itex] + [itex]\frac{B}{z-2}[/itex])

Where A = -1 , B = 1.

f = [itex]\frac{1}{z}[/itex] * ([itex]\frac{1}{z-2}[/itex] + [itex]\frac{1}{1-z}[/itex])

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;

[itex]\frac{1}{z-1}[/itex] (dividing 1/z to numerator and denominator at R2,R3)

and

[itex]\frac{1}{z-2}[/itex] (dividing 1/z to numerator and denominator at R3)

so for example,

for R2,

f = [itex]\frac{1}{z}[/itex] * [itex]\frac{1}{z-2}[/itex] * (-[itex]\frac{1}{z}[/itex]*[itex]\frac{1}{1-z^{-1}}[/itex])

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?

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