Recipe

Expand a function as a Laurent series in a given annulus

Use when: you're asked for the Laurent series of $f$ centered at $z_0$ , valid specifically on one named annulus between two of $f$ 's singularities

Goal: the two-sided series $\sum c_n(z-z_0)^n$ that actually converges to $f$ on that one annulus, not some other one centered at the same point

Taught in Laurent's theorem

Quick recipe

Locate every singularity and measure its distance from $z_0$
Split f into partial fractions, one term per singularity
Pick inside or outside expansion for each term
Expand each term and add the two series together

1
Find every point where $f$ fails to be analytic. Record each one's distance from the center $z_0$ . These distances cut the plane into the largest annuli that Laurent's theorem allows.

Example

For $f(z)=\dfrac{1}{(z+1)(z-2)}$ centered at $z_0=0$ : the singularities sit at distance $1$ and distance $2$ from $0$ , giving three regions, $|z|<1$ , $1<|z|<2$ , and $|z|>2$ .

Why: a different annulus around the same center gives a different Laurent series. So you must name the target region before any algebra.
2
Write $f$ as a sum of simple fractions. Each one has exactly one of the singularities from Step 1 in its denominator.

Example

$\dfrac{1}{(z+1)(z-2)}=-\dfrac13\cdot\dfrac{1}{z+1}+\dfrac13\cdot\dfrac{1}{z-2}$ .

Why: the geometric-series trick works on one linear factor at a time.
3
For a term $\dfrac{1}{z-a}$ , compare $|z|$ on the target annulus with $|a|$ . Then pick the geometric series that converges there.

Example

On $1<|z|<2$ : $z$ is farther from $0$ than the singularity at $-1$ , so that term takes the outside form. But $z$ is still closer than the singularity at $2$ , so that term takes the inside form.
- $|z|<|a|$ on the target annulus — use $\dfrac{1}{z-a}=-\dfrac1a\sum_{n=0}^\infty\dfrac{z^n}{a^n}$ , all non-negative powers.
- $|z|>|a|$ on the target annulus — use $\dfrac{1}{z-a}=\dfrac1z\sum_{n=0}^\infty\dfrac{a^n}{z^n}$ , all negative powers.
Why: a geometric series only converges when its ratio has modulus below $1$ . Which quantity plays the ratio flips at each singularity's own distance.
4
Write out each chosen expansion. Then add them into one two-sided series in $(z-z_0)$ .

Example

$-\dfrac13\cdot\dfrac1{z+1}=-\dfrac1{3z}+\dfrac1{3z^2}-\cdots$ and $\dfrac13\cdot\dfrac1{z-2}=-\dfrac16-\dfrac{z}{12}-\cdots$ , so on $1<|z|<2$ : $f(z)=\cdots+\dfrac1{3z^2}-\dfrac1{3z}-\dfrac16-\dfrac{z}{12}-\cdots$ .

Why: the two pieces add term by term to give $f$ on the one annulus where both are valid.

Common mistakes

Giving a Laurent series with no annulus named, as if a function had only one such series.
Using an inside expansion for a term when the target annulus lies outside that singularity, or the reverse.
Forgetting to redo Step 3 for every singularity. Each one gets its own inside or outside choice.
Assuming the coefficients from one annulus still work on a different annulus of the same function.