Find the two Laurent series for $f(z)=\frac{1}{z^3-z^4}$ that involve powers of z, and state their domains of convergence.

Question

I'm working on the following problem: Find the two Laurent series for $f(z)=\frac{1}{z^3-z^4}$ that involve powers of z, and state their domains of convergence. I'm not sure how to go about it, any help is appreciated!

Answer 1

Since it is not specified, I suggest finding expansions about $z = 0$. Since Laurent series converge on "annuli" in the complex plane centered at the point of expansion, we look for annuli centered at the origin. Note that $f$ has singularities at $z = 0, 1$. Thus there are two "annuli" on which $f$ is holomorphic: $\{0 < |z| < 1\}$ and $\{1 < |z|\}$. Note that in the former, using the fact that $|z| < 1$, we can expand a geometric series: $$f(z) = \frac{1/z^3}{1 - z} = \frac{1}{z^3}\sum_{k = 0}^\infty z^k = \sum_{k = -3}^\infty z^k$$ which gives the Laurent series.

This trick involving geometric series can also be used for the case when $|z| > 1$. As a hint, note that $|z| > 1$ implies $|1/z| < 1$, so try to expand $f$ using a geometric series with $1/z$ as the ratio.