Laurent series, finding annulus

Question

I am trying to find the annulus, which are interesting for the following function:

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

I already found: $|z|< 2$, $2 < |z| < 3$, $3 < |z| < \infty$. Is this correct so far and are there more?

Answer 1

You can write, for instance, the Laurent series on the annulus $2\lt |z|\lt3$ as a product of two geometric series. One converges on $|z|\gt2$, the other on $|z|\lt3$. Then using, if you like, the Cauchy formula for the product, you can discover the $a_n$ in $\sum_na_nz^n$, for the Laurent series on the overlap.

In general, there are formulae for the inner and outer radius: $r=\limsup_{n\to\infty}|a_{-n}|^{1/n}, 1/R=\limsup_{n\to\infty}|a_n|^{1/n}$.

Or you could check for the distance to nearest singularities. In this case they are at $-2i$ and $3$.

As you noted, you can also do $|z|\lt2$ and $|z|\gt3$.

As for Laurent series about $0$, that'll be it, by uniqueness.

To get residues, you need to do the Laurent series at the pole in question.