Finding the Laurent series of a function in a given annulus

Question

Find the Laurent series, valid in $ \{z \in \Bbb C : 1 < |z| < 3 \}$ for:

$$f(z) = \frac{(9z+1)}{(z^2-9)(z^3+1)} $$

Partial fractions seems too messy which makes me think I'm missing something?

Answer 1

By πr8's comment, $$f(z) = \frac{(9z+1)}{(z^2-9)(z^3+1)} = \frac{1}{z^2-9} - \frac{z}{z^3+1}.$$ Moreover, for $|z|<3$, $$\frac{1}{z^2-9}=\frac{-1/9}{1-(z/3)^2}=-\frac{1}{9}\sum_{k=0}^{\infty}(z/3)^{2k}=-\sum_{k=0}^{\infty}\frac{z^{2k}}{9^{k+1}},$$ and, for $|z|>1$, $$\frac{z}{z^3+1}=\frac{z/z^3}{1-(-1/z^3)}= \frac{1}{z^2}\sum_{k=0}^{\infty}(-1/z^3)^{k}= \sum_{k=0}^{\infty}\frac{(-1)^k}{z^{3k+2}}.$$ Hence the Laurent series in $1<|z|<3$ is $$-\sum_{k=0}^{\infty}\frac{(-1)^k}{z^{3k+2}}-\sum_{k=0}^{\infty}\frac{z^{2k}}{9^{k+1}}.$$