a Laurent series
/ loh-RAHN /
A Taylor series only knows how to climb upward: it adds positive powers (z - z_0)^0, (z - z_0)^1, (z - z_0)^2, and so on, so it can describe a function that is perfectly well-behaved at z_0. But what if your function blows up at z_0, like 1/(z - z_0)? Then you need to be allowed to count downward too — to use 1/(z - z_0), 1/(z - z_0)^2, and the like. A Laurent series is exactly that: a power series that runs over ALL integer powers, negative as well as positive.
Written out, a Laurent series centered at z_0 is the sum over all integers n, from minus infinity to plus infinity, of a_n (z - z_0)^n. It splits naturally into two halves: the terms with n less than 0 (the negative powers, called the principal part) carry all the information about how the function misbehaves at z_0, and the terms with n greater than or equal to 0 (the ordinary Taylor-like part, called the analytic part) describe the tame behaviour. Pierre Alphonse Laurent's theorem says that any function holomorphic on an annulus (a ring r < |z - z_0| < R) has exactly one such expansion that converges there.
The reason Laurent series matter so much is that they let you read off, just from the list of coefficients, precisely what kind of singularity a function has at z_0 — removable, a pole, or essential — and the single coefficient a_{-1} (the residue) is the key to evaluating contour integrals. A Laurent series is the microscope you point at a singular point to see its true shape.
For f(z) = e^z / z near z_0 = 0, write e^z = 1 + z + z^2/2! + ..., then divide by z: e^z / z = 1/z + 1 + z/2! + z^2/3! + .... The single negative-power term 1/z is the principal part; everything else is the analytic part; the coefficient of 1/z is a_{-1} = 1.
A Laurent expansion in action: split a known Taylor series and divide, reading off the negative powers.
A Laurent series is not unique on its own — the SAME function has DIFFERENT Laurent series on different annuli around z_0; you must say which annulus you mean, and only the one on the innermost punctured disk reveals the singularity at z_0.