I am asked to compute the Laurent series expansion of $\frac{1}{z(z-1)(z+1)}$ in the annulus $0 < |z| < 1$.
The only examples I have seen of computing a Laurent series expansion is by rewriting the above function to look like a geometric series using partial fraction decomposition. In this particular example, we have,
$$ \frac{1}{z(z-1)(z+1)} = - \frac{1}{z} + \frac{1}{2(z-1)} + \frac{1}{2(z+1)}$$
Rewriting this,
$$ - \frac{1}{z} + \frac{1}{2z} \frac{1}{1- \frac{1}{z}} + \frac{1}{2z} \frac{1}{1 - (-\frac{1}{z})}$$
$$= - \frac{1}{z} + \frac{1}{2z} \sum \left ( \frac{1}{z} \right)^n + \frac{1}{2z} \sum \left (- \frac{1}{z} \right)^n$$
However, the summations only converge for $|z| > 1$ and thus will not converge inside of the given annulus.
What do I do in this case?
Since $z^2-1=(z-1)(z+1)$, $$ \frac{1}{z (z-1)(z+1)}=-\frac{1}{z(1-z^2)}$$ and $$ \frac{1}{1-z^2}=1+z^2+z^4+\cdots $$ for $0 <|z|<1$.