Complex Analysis: Theory and Applications · Problem 7.3.1

The key is recognizing that this function has a singularity at $z=0$ and another at $z=1$, so different regions around the origin will yield different expansions. Factor out $z^3$ from the denominator to get $\frac{1}{z^3(1-z)}$, then use the geometric series $\frac{1}{1-z} = \sum z^n$ for $|z|<1$. For the region $|z|>1$, rewrite the denominator to make the series converge by factoring differently, so you're summing powers that decay as $z$ grows large. Each region gives you a distinct Laurent series with different ranges of validity.

The key is recognizing that this function has a singularity at $z=0$ and another at $z=1$, so different regions around the origin will yield different expansions. Factor out $z^3$ from the denominator to get $\frac{1}{z^3(1-z)}$, then use the geometric series $\frac{1}{1-z} = \sum z^n$ for $|z|<1$. For the region $|z|>1$, rewrite the denominator to make the series converge by factoring differently, so you're summing powers that decay as $z$ grows large. Each region gives you a distinct Laurent series with different ranges of validity.