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The Laurent Series Around a Singularity

A Taylor series fails the instant a function blows up, but a Laurent series keeps going — it just lets the powers turn negative. Those negative powers are the fingerprint of a singularity, and reading them is how we will tell removable from pole from essential.

Where Taylor's theorem runs out of room

On the power-series rung you earned a remarkable fact: a holomorphic function equals its own Taylor series sum a_n (z - z_0)^n on a disk, and that disk grows until it bumps into trouble. The rule you met was the radius of convergence reaching the nearest singularity: the Taylor disk stops exactly at the closest point where the function misbehaves. But what happens at that troublesome point, and just outside it? A Taylor series, made only of non-negative powers (z - z_0)^n, stays finite and tame at z_0 — so it is simply the wrong tool to describe a function that is blowing up there.

Take the cleanest example, f(z) = 1 / z near z_0 = 0. There is no Taylor series for it at the origin: every Taylor series is finite at its center, but 1/z runs off to infinity as z approaches 0. Yet f is perfectly holomorphic everywhere else — it has a single bad point, an isolated singularity at 0, surrounded by an ocean of good behaviour. We would love a series that describes f on that surrounding region without pretending the bad point isn't there. The trick is almost embarrassingly simple: allow negative powers. Then 1/z is already its own series, the single term z^(-1), and we have said everything there is to say.

The Laurent series: powers in both directions

Here is the central object. A Laurent series centered at z_0 is a two-sided sum that runs over all integers n, from minus infinity to plus infinity: sum a_n (z - z_0)^n. Split it down the middle and it tells its own story. The piece with n >= 0, the ordinary power series sum_{n>=0} a_n (z - z_0)^n, is called the analytic part — it is exactly a Taylor-style series and behaves nicely. The piece with n < 0, the part made of negative powers a_{-1}/(z - z_0) + a_{-2}/(z - z_0)^2 + ..., is called the principal part, and it is where all the misbehaviour lives.

f(z) = ... + a_(-2)/(z-z_0)^2 + a_(-1)/(z-z_0) + a_0 + a_1 (z-z_0) + a_2 (z-z_0)^2 + ...
       \___________________  principal part  __________________/ \____ analytic part ____/
                 (negative powers: the singular behaviour)         (non-negative powers)
A Laurent series reads from both sides of the center. The principal part (negative powers) carries the singularity; the analytic part (non-negative powers) is an ordinary Taylor-style series.

The annulus: where a Laurent series lives

Laurent's theorem says: if f is holomorphic on an annulus — a ring-shaped region r < |z - z_0| < R between two circles around z_0 — then on that whole annulus f equals a unique Laurent series. This is convergence in an annulus, and it is the natural home of the whole idea. The inner radius r can shrink all the way to 0 (a punctured disk with just the center removed) and the outer radius R can swell out to infinity; the theorem holds for any ring on which f stays holomorphic.

The picture to hold is this: the analytic part converges out from the center and covers everything inside the outer circle |z - z_0| < R; the principal part converges in from infinity and covers everything outside the inner circle |z - z_0| > r. Only on the ring where both converge does the full two-sided series make sense — and on that ring it equals f. One subtle but vital warning: the SAME function f can have DIFFERENT Laurent series on different annuli around the same center. The expansion that is valid on a small ring next to z_0 need not match the one valid on a far-out ring, because between them might sit another singularity that splits the plane into separate rings.

Getting a Laurent series the easy way

In practice nobody hunts down Laurent coefficients with that integral. The honest working method is expansion via the geometric series and the Taylor series you already know. You algebraically massage the function until it is built from pieces you can expand, taking care to expand each piece in the right region (close in versus far out), and the negative powers fall out on their own. Let us walk one all the way through, the function f(z) = 1 / (z (z - 1)), expanded around z_0 = 0 on the punctured disk 0 < |z| < 1.

  1. Split into simple pieces by partial fractions: 1 / (z (z - 1)) = -1/z + 1/(z - 1). The first piece, -1/z, is already a pure negative power — there is nothing more to do with it.
  2. Rewrite the second piece to expose a geometric series: 1/(z - 1) = -1/(1 - z) = -(1 + z + z^2 + z^3 + ...), valid because we are on |z| < 1, so the ratio z has |z| < 1 and the geometric series converges.
  3. Add the two expansions: f(z) = -1/z - 1 - z - z^2 - z^3 - ... . The lone term -1/z is the entire principal part; everything from -1 onward is the analytic part.
  4. Read off the coefficients: a_{-1} = -1, and a_n = -1 for every n >= 0. The principal part has exactly one term, so the singularity at 0 is mild — a fact the next guide will name precisely.

Notice the second singularity, at z = 1, sitting just outside our disk. It is what forced the outer radius to be R = 1: our annulus is the ring 0 < |z| < 1, squeezed between the bad point at 0 (inner radius shrunk to 0) and the bad point at 1 (outer radius). If instead you wanted the expansion on the outer region |z| > 1, you would expand 1/(z - 1) the other way, as (1/z)(1 / (1 - 1/z)) = 1/z + 1/z^2 + ..., and you would get a completely different Laurent series — the same function, a different ring, a different expansion, exactly as warned above.

Reading the principal part: a preview of the classification

Now the payoff. Once you have the Laurent series of f around an isolated singularity z_0, the principal part is a complete report card on how badly f behaves there. Just count its negative-power terms — count the Laurent coefficients a_n with n < 0 that are not zero. There are exactly three possibilities, and they are genuinely different worlds, not shades of one thing.

If the principal part is empty — no negative powers at all — the singularity was a fake, a hole that can be filled in; we call it removable, like the seam in sin(z)/z at 0 that smooths over to the value 1. If the principal part has finitely many terms, ending at a_{-m}/(z - z_0)^m, the singularity is a pole of order m, the honest infinity of 1/(z - z_0)^m. And if the principal part has infinitely many negative-power terms, the singularity is essential — the wild case, like e^(1/z) at 0, whose principal part never stops. The next four guides take these one at a time: telling the three apart, the order of a zero and a pole, meromorphic functions that allow only poles, and the genuine wildness of the essential case.