Laurent Series

The Laurent expansion is the computational backbone of the residue theorem, which reduces contour integrals to reading off a single coefficient ($a_{-1}$) from a local expansion — this is how physicists and engineers evaluate real integrals like $\int_{-\infty}^\infty \frac{dx}{1+x^4}$ via complex contours. It underlies the classification of isolated singularities (removable, pole, essential) purely by counting negative-power terms, which in turn controls the behavior of meromorphic functions used throughout analytic number theory (e.g. the Laurent expansion of the Riemann zeta function at its pole $s=1$) and in the study of generating functions in combinatorics. In physics, Laurent expansions around poles of propagators and Green's functions extract physically meaningful quantities (masses, coupling residues) in quantum field theory and statistical mechanics. In engineering, the z-transform of a discrete-time signal is literally a Laurent series in $z^{-1}$, and questions of stability and causality of digital filters translate directly into questions about the annulus of convergence of that series. In numerical analysis, Laurent/Fourier-type expansions on annuli underlie fast algorithms for evaluating functions on circles and for constructing conformal maps.

Ask the tutor: How does the number of nonzero negative-power Laurent coefficients determine whether a singularity is removable, a pole, or essential?

Think of a Taylor series as a lens that only sees the interior of a disk: it is built from data at a single point $z_0$ and it necessarily converges out to the nearest singularity. A Laurent series instead is built for an annular neighborhood, a ring-shaped region that can have singularities both at its center and beyond its outer edge. The negative-power terms are exactly what is needed to describe behavior as $z \to z_0$, when $z_0$ itself is a hole in the domain, while the nonnegative-power terms describe ordinary analytic behavior filling out toward the outer radius $R$. A very useful mental model: the analytic part $\sum_{n\ge 0} a_n(z-z_0)^n$ converges on the whole disk $|z-z_0|<R$ and represents a function holomorphic there; the principal part $\sum_{n\ge1} a_{-n}(z-z_0)^{-n}$, upon substituting $w = 1/(z-z_0)$, becomes an ordinary power series $\sum_{n\ge 1} a_{-n}w^n$ in $w$, convergent for $|w| < 1/r$, i.e. for $|z-z_0|>r$. So the principal part converges on the exterior of a disk of radius $r$. A Laurent series therefore converges exactly on the overlap of an "inside" region and an "outside" region — the annulus $r<|z-z_0|<R$ — and the function it represents is genuinely built from two different analytic pieces glued along that ring.

Concretely, if $f(z) = \sum_{n\ge 0}(z-z_0)^n$ has radius of convergence $1$ and $g(w)=\sum_{n\ge1}\left(\tfrac{1}{3}\right)^n w^n$ has radius of convergence $3$, then setting $w=1/(z-z_0)$ turns $g$ into a series converging for $|z-z_0|>1/3$. The combined Laurent series $f+g$ converges exactly on the annulus $1/3 < |z-z_0| < 1$ — a concrete, checkable instance of the general picture above.

Example 1: One function, three different annuli. Let

This function is holomorphic on $\mathbb{C}\setminus\{1,3\}$, so it has three natural regions of interest centered at $0$: the disk $|z|<1$, the annulus $1<|z|<3$, and the exterior $|z|>3$. Partial fractions give

On $|z|<1$: both $|z|<1$ and $|z|<3$, so expand both terms as geometric series in $z$:

Hence, valid for $|z|<1$:

This is an ordinary Taylor series (pure analytic part, no negative powers), as expected since $f$ is actually analytic at $0$ with nearest singularity at $z=1$. On $1<|z|<3$: here $|z|>1$ so we must expand $1/(z-1)$ in powers of $1/z$, while $|z|<3$ still allows the geometric expansion of $1/(z-3)$ in $z$:

So, valid for $1<|z|<3$:

Now there genuinely is a principal part: $a_{-1} = -1/2$, $a_{-2}=-1/2$, $a_{-n} = -1/2$ for all $n\ge1$ in fact (check: $-\frac12 z^{-n}$ has coefficient $-1/2$ for every $n\ge1$), and analytic part coefficients $a_n = -\frac{1}{6\cdot 3^n}$ for $n\ge 0$. On $|z|>3$: now both $1/z$ and $3/z$ have modulus $<1$, so expand both terms in powers of $1/z$:

So, valid for $|z|>3$:

Here the analytic part is identically zero — every coefficient $a_n$ with $n\ge 0$ vanishes — because $f(z)\to 0$ like $1/z^2$ as $z\to\infty$, so there is no room for a nonzero constant, linear, etc. term. This single computation is the clearest possible illustration of why the annulus is part of the data: the same function $f$ has three genuinely different Laurent expansions centered at the same point $z_0=0$, one for each of the three natural annuli determined by the singularities at $1$ and $3$.

Example 2: An essential singularity, and extracting the residue. Let $g(z) = z^2 e^{1/z}$, holomorphic on $\mathbb{C}\setminus\{0\}$, and expand around $z_0=0$ on the annulus $0<|z|<\infty$. Using the known Taylor series of the exponential, $e^{w} = \sum_{k=0}^\infty w^k/k!$ for all $w\in\mathbb C$, substitute $w=1/z$:

convergent for all $z\ne 0$ (i.e. $r=0$, $R=\infty$) since the exponential series converges for every $w\in\mathbb C$. Multiplying by $z^2$ termwise (justified since we are multiplying a convergent series by a fixed polynomial, which preserves convergence and simply shifts/scales each term):

where explicitly $a_2=1$, $a_1=1$, $a_0=\frac12$, and for $n\le -1$, $a_n = \dfrac{1}{(2-n)!}$ (check $n=-1$: $\frac{1}{3!}=\frac16$ ✓; $n=-2$: $\frac{1}{4!}=\frac{1}{24}$ ✓). By Laurent's theorem this expansion is unique on $0<|z|<\infty$, so we may read off the residue directly:

No contour integral needed to be computed by hand — the uniqueness clause guarantees that any correct method of producing a two-sided series representing $g$ on this punctured plane yields the same $a_{-1}$, which is exactly the value the defining contour integral $\frac{1}{2\pi i}\oint_{|z|=1} g(z)\,dz$ would produce. This is precisely the mechanism the future Residue Theorem exploits.