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How would I use partial fractions to expand the following equation?

$\frac{1}{z^3-z}$

I have tried changing re-writing that as:

$\frac{1}{z(z-1)(z+1)}$

But I have a problem finding the numerators of the fraction when written as:

$\frac{a}{z}+\frac{b}{z+1}+\frac{c}{z-1}$

Amir Omidi
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2 Answers2

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Note

$$a(z^2-1)+bz(z-1)+cz(z+1)=1$$

Setting $z=0$ we get $a=-1$, setting $z=1$ we get $c={1\over 2}$, setting $z=-1$ we get $b={1\over 2}$.

Adam Hughes
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  • Why do we set z = 0? – Amir Omidi Feb 25 '15 at 19:57
  • @AmirOmidi becuase it makes the $b$ and $c$ terms disappear (anything times $0$ is $0$), so we can solve for $a$ quickly and easily without any worries about $b$ or $c$. – Adam Hughes Feb 25 '15 at 19:58
  • Ah Ok! Thanks this suddenly makes a lot more sense! – Amir Omidi Feb 25 '15 at 20:00
  • Note: for more complicated partial fractions - ones with more than just an integer on the right side - you'll need to do more work than this. – Dan Uznanski Feb 25 '15 at 20:05
  • @DanUznanski not really, it mostly depends on the complicatedness of the polynomial, not the original denominator. If I only change the RHS, I can just scale the answers for RHS=1. It's the LHS that can really complicate things. – Adam Hughes Feb 25 '15 at 20:20
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If we add the three fractions we get in the numerator $a(z+1)(z-1) + bz(z-1) + cz(z+1) = (a+b+c)z^2 + (-b+c)z - a$.

Now set this equal to 1.

Henno Brandsma
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