Laurent series of $1/({z^3-z})$

Question

Question: Find the Laurent series of the function $$f(z) = \frac{1}{z^3 - z}$$ at the domain $|z-1|>2$.

Attempt: So we have $$\frac{1}{z(z-1)(z+1)}$$ and we only have to find a Laurent series for $1/(z+1)$ and leave $1/z$ and $1/(z-1)$ how they are right?

Do you put it into partial fractions and then find Laurent series of $1/(z+1)$ or do you put it like $$\frac1{z(z-1)} \cdot \frac1{z+1}$$ and then sub the Laurent series for $1/(z+1)$ in?

Answer 1

Hint: Write $z + 1 = z - 1 + 2 $, then

$$\frac{1}{z+1} = \frac{1}{(z-1)}\frac{1}{1 + \frac{2}{z-1}} = \sum_{n=0}^{\infty} (-1)^{n}\frac{2^n}{(z-1)^{n+1}}$$

where $\frac{2}{|z-1|} < 1 \implies |z-1| > 2$.

Use partial fraction.