I need help for this question: Find the Laurent series for $f(z) = \frac{z^3}{3-2z}$ in $A = \{2<|z|<3\} $
I am not sure how to start.. My idea is that, since the annulus is $\{2<|z|<3\}$, I tried to decompose the $f(z)$ to fractions with denominator $({1-\frac{2}{z}})$ and $(1-\frac{z}{3})$ However, I can't seem to fit them back into $f(z)$
Or can I find the Laurent series for $f(z)$ in $B = \{\frac{3}{2}<|z|<\infty\}$ and conclude that since the required annulus is a subset of $B$, then the laurent series I got for $B$ is valid for $A$ as well? (by uniqueness of laurent series)
Hint: In your case the annulus is $2<|z|<3$. So you can consider $2<|z|$ as $|2/z|<1$ and similarly $|z|<3$ as $|z/3|<1$ and then proceed. Hopes this helps.