Laurent expansion by geometric series and partial fractions

The integral formula for Laurent coefficients is honest but slow. In practice almost nobody computes those integrals; instead they reach for a single familiar tool — the geometric series 1/(1 - w) = 1 + w + w^2 + w^3 + ..., valid for |w| < 1 — and bend the function into a shape where it applies. Combined with partial fractions, this turns finding a Laurent series into a piece of clever algebra.

The method for a rational function goes in plain steps. First, split f into partial fractions, so it becomes a sum of simple pieces like A/(z - a). Second, for each piece decide which annulus you are in, because that fixes how to expand. To expand 1/(z - a) around z_0 = 0 in the region |z| < |a| (inside), write it as -1/(a(1 - z/a)) = -(1/a)(1 + z/a + z^2/a^2 + ...) — a geometric series in the small quantity z/a, giving non-negative powers. To expand the SAME piece in the region |z| > |a| (outside), factor out z instead: 1/(z(1 - a/z)) = (1/z)(1 + a/z + a^2/z^2 + ...) — now a geometric series in a/z, giving negative powers. The rule of thumb: always arrange the ratio inside the series to have modulus less than 1, then expand. Third, add the pieces back together and collect powers.

This is the workhorse of every residue calculation and every series-based singularity classification, and it generalizes: divide a known Taylor series to handle things like e^z / z^2 or (cos z)/z^3, or substitute w = 1/z into a known series to expand around infinity or to see an essential singularity (as for e^(1/z)). Master the trick of choosing the right factorization for the annulus you want, and you can write down Laurent series by hand without ever touching the integral formula.