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

Question

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:

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

Answer 1

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 $b={1\over 2}$.

Answer 2

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.