Laurent Series
A Taylor series
needs a point of holomorphy at its centre. But the most interesting points in complex analysis
are the bad ones — where a function blows up. Near such an isolated singularity there is
no Taylor series, yet there is still a series, one that allows negative powers:
the Laurent series.
f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n = \underbrace{\cdots + \frac{a_{-2}}{(z - z_0)^2} + \frac{a_{-1}}{z - z_0}}_{\text{principal part}} + \underbrace{a_0 + a_1 (z - z_0) + \cdots}_{\text{analytic part}}.
The terms with negative index form the principal part — they encode the
singular behaviour — while the non-negative terms are an ordinary power series. A Laurent series
is valid not on a disk but on an annulus
r < |z - z_0| < R: a ring that excises the troublesome centre.
Why an annulus
Step 1 — separate the two parts. The analytic part is a power series in
(z - z_0); like any
complex power
series it converges inside a disk
|z - z_0| < R.
Step 2 — the principal part is a power series in
1/(z - z_0). Writing
w = 1/(z - z_0), the negative-power tail
\sum_{n\ge 1} a_{-n} w^n converges inside a disk in
w, i.e. for |w| < 1/r, which means
outside a circle: |z - z_0| > r.
Step 3 — intersect the two regions. Both parts converge together exactly on
the overlap — the annulus
r < |z - z_0| < R.
When the centre is the only singularity inside, r can shrink to
0 and the annulus becomes a punctured disk
0 < |z - z_0| < R.
Two worked Laurent series
An essential singularity. Substitute
1/z into the exponential series. Every term is a negative power, so
the principal part is infinite:
e^{1/z} = \sum_{n=0}^{\infty} \frac{1}{n!\,z^n} = 1 + \frac{1}{z} + \frac{1}{2!\,z^2} + \frac{1}{3!\,z^3} + \cdots, \qquad 0 < |z| < \infty.
The coefficient of 1/z is
a_{-1} = 1, of
1/z^2 is a_{-2} = 1/2!, and so on.
Different annuli, different series. The function
\frac{1}{z(z - 1)} has singularities at
0 and 1. Partial fractions give
\frac{1}{z(z - 1)} = \frac{1}{z - 1} - \frac{1}{z}.
On the annulus 0 < |z| < 1 expand
\frac{1}{z - 1} = -\sum_{n\ge 0} z^n, so
\frac{1}{z(z-1)} = -\frac{1}{z} - 1 - z - z^2 - \cdots, \qquad 0 < |z| < 1.
The same function has a completely different Laurent expansion on
|z| > 1 — the annulus chosen decides the series.
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A function holomorphic on an annulus
r < |z - z_0| < R has a unique Laurent expansion
f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n there.
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The negative-power terms are the principal part; the coefficients are
a_n = \dfrac{1}{2\pi i}\oint_\gamma \dfrac{f(w)}{(w - z_0)^{n+1}}\,dw.
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Different annuli around the same point give different series, e.g.
1/[z(z-1)] on 0 < |z| < 1 versus
|z| > 1.
A Taylor series only ever describes a function where it is well-behaved; it goes silent at the
singularities that make a function interesting. The Laurent series is the tool built precisely
for those points. Its principal part is a fingerprint of the singularity — a single
1/z term, a few negative powers, or infinitely many — and reading
that fingerprint is how we will classify singularities and, ultimately, evaluate integrals by
residues.
Trace the annulus of convergence
The ring between the inner radius r and outer radius
R (both adjustable) is the region where the Laurent series about
z_0 = 0 converges. Move the test point
z and the readout reports |z| and whether
it sits inside the ring, too close to the centre, or beyond the outer edge.