Laurent Series

A Taylor series is the language of the well-behaved: it captures an analytic function on a disk where nothing goes wrong. But the functions physics cares about are rarely so polite — propagators blow up at particle masses, Green's functions carry poles, a tunnelling amplitude hides an $e^{-1/g^2}$ that no power series can see. To describe a function near the point where it misbehaves, we abandon the disk and work on an annulus — a ring $r<|z-z_0|<R$ that punctures out the bad point at its centre. On such a ring Laurent's theorem promises what a Taylor series cannot: a unique expansion in both positive and negative powers of $(z-z_0)$ . The construction is nothing more than Cauchy's integral formula applied twice. Surround the evaluation point by an outer and an inner circle. On the outer one the Cauchy kernel expands as a geometric series in $(z-z_0)/(\zeta-z_0)$ and delivers non-negative powers; on the inner one it expands in the reciprocal ratio and delivers negative powers. The two assemble into one doubly-infinite series, and Cauchy's theorem collapses both circles onto a single loop — one uniform formula for every coefficient at once. The negative-power terms — the principal part — are the whole point. They fingerprint the singularity the annulus encircles: empty means the singularity was an illusion (removable), finite means a pole of definite order, infinite means an essential singularity where Picard's theorem takes over. And buried in that principal part is the single most useful number in complex analysis: the coefficient $a_{-1}$ , the residue, which contour integration extracts and the residue theorem turns into the value of an integral. Choosing which annulus to expand in is the skill this article is really about.

Key takeaways

A function $f$ analytic on an annulus $r<|z-z_0|<R$ has a unique two-sided expansion $f(z)=\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n$ converging throughout the ring.
Every coefficient is one contour integral, $a_n=\frac{1}{2\pi i}\oint_C f(\zeta)/(\zeta-z_0)^{n+1}\,\mathrm{d}\zeta$ , over any loop $C$ in the annulus encircling $z_0$ once.
The negative-power principal part classifies the central singularity: empty (removable), finite (pole of order $m$ ), or infinite (essential); the residue is exactly $a_{-1}$ .
One algebraic expression yields different Laurent series in different annuli separated by singularities — the expansion belongs to the annulus, not to the point.
Laurent's theorem is Cauchy's integral formula split over the two boundary circles of the ring, then merged by Cauchy's theorem into a single loop.
The whole machine of residues, generating functions, asymptotics, and the $z$ -transform is Laurent series wearing different hats.
Key idea: split Cauchy's integral formula over the inner and outer boundaries of an annulus — the outer circle gives a Taylor-like series in positive powers, the inner circle gives a geometric series in negative powers, and together they form a unique two-sided expansion whose coefficients are single contour integrals.

The core insight

Split Cauchy's integral formula over the inner and outer boundaries of an annulus: the outer circle yields a Taylor-like series in positive powers, the inner circle yields a geometric series in negative powers, and together they give a unique two-sided expansion whose coefficients are single contour integrals.

Where this lives today

Laurent series are the working representation of singular functions across modern physics and engineering. In quantum field theory the Feynman propagator is analytic in the complex energy plane except at poles whose locations are particle masses and whose residues fix coupling strengths; the analytic-S-matrix programme reads scattering data off pole positions and residues, exactly the $a_{-1}$ of a local Laurent expansion. Dimensional regularization pushes this further: loop integrals are expanded as Laurent series in $\epsilon=4-d$ , and the coefficients of the $1/\epsilon^k$ poles — a genuine principal part — are the ultraviolet divergences renormalization subtracts. In digital signal processing the $z$ -transform is literally a Laurent series in $z^{-1}$ , and its annular region of convergence decides whether the filter is causal and stable. Special-function theory leans on the same object: the Bessel generating function $e^{(x/2)(t-1/t)}=\sum_{n=-\infty}^{\infty}J_n(x)\,t^n$ is a Laurent series in $t$ , so each $J_n(x)$ is a Laurent coefficient. And in semiclassical physics the essential singularity of $e^{-1/g^2}$ in the coupling $g$ — invisible to any power series — is the analytic fingerprint of instantons, the reason perturbation theory alone can never see tunnelling.

The physical picture

Cauchy's formula is a sampling device: to learn $f$ at a point you surround it with a loop and average $f$ over it, weighted by the kernel $1/(\zeta-z)$ . On a disk there is one loop and the answer is a Taylor series. On an annulus there are two boundaries reporting different things. The outer circle sees only the smooth analytic content of $f$ and returns it as positive powers. The inner circle wraps the hidden singularity and returns its strength as negative powers — a memory of how badly $f$ would blow up if you filled the hole back in. The first figure makes that two-contour anatomy concrete, with the single formula both circles collapse into.

Figure

Two concentric contours build the expansion around an annulus centred on the singularity z₀: the outer circle C_R returns the positive powers, the inner circle C_r returns the negative powers, and Cauchy's theorem merges them into one common loop C on which every coefficient is a single contour integral

The negative powers carry the physics. An empty principal part means the singularity was never real and $f$ extends smoothly across $z_0$ ; one stopping at $(z-z_0)^{-m}$ is a pole of order $m$ , a source of definite strength; one running forever is an essential singularity, the signature of non-perturbative behaviour. And because analyticity is demanded only on the ring, the same algebraic $f$ splits the plane into nested annuli, one per gap between singularities, each carrying its own different Laurent series. The second figure shows this for a rational function whose middle-annulus coefficients we compute below.

Figure

The single function f(z) = 1/((z−1)(z−3)) has three distinct Laurent regions about z₀ = 0, carved by its poles at z = 1 and z = 3: the disk (pure Taylor), the middle annulus whose leading coefficient is a₋₁ = −1/2, and the exterior of only negative powers

The core equations

The starting object is the expansion itself. On an annulus of analyticity, $f$ is written as a doubly-infinite power series — the ordinary Taylor part $n\ge0$ plus a principal part $n<0$ that a disk would never allow.

f(z)=\sum_{n=-\infty}^{\infty} a_n\,(z-z_0)^n, \qquad r<|z-z_0|<R \qquad (1)

where: $f$ — function analytic on the open annulus; $z_0$ — centre of the annulus, typically an isolated singularity; $a_n$ — Laurent coefficients, indexed over all integers $n\in\mathbb{Z}$ ; $r,R$ — inner and outer radii with $0\le r<R\le\infty$ . Reading it: the series splits cleanly into $\sum_{n\ge0}a_n(z-z_0)^n$ (analytic part) and $\sum_{n<0}a_n(z-z_0)^n$ (principal part), and both must converge on the ring. Each coefficient is a single contour integral. The remarkable fact — proved in Derivation A — is that one formula covers positive and negative $n$ alike, over any loop in the annulus.

a_n=\frac{1}{2\pi i}\oint_C \frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\,\mathrm{d}\zeta \qquad (2)

where: $C$ — any positively oriented closed contour lying in the annulus and encircling $z_0$ once; $\zeta$ — integration variable on $C$ ; $n$ — any integer, positive, zero, or negative. Reading it: for $n\ge0$ this reduces to a Cauchy derivative formula; for $n<0$ it probes the principal part. The contour may be deformed freely within the annulus because the integrand is analytic there for each fixed $n$ . The single most useful special case is $n=-1$ . Every other power integrates to zero around a closed loop, so the entire contour integral of $f$ collapses onto this one coefficient — the residue.

a_{-1}=\operatorname{Res}_{z=z_0} f(z)=\frac{1}{2\pi i}\oint_C f(z)\,\mathrm{d}z \qquad (3)

where: $a_{-1}$ — coefficient of $(z-z_0)^{-1}$ ; $\operatorname{Res}$ — the residue at the enclosed singularity. Reading it: this is the bridge to the residue theorem — integration reads off one Laurent coefficient, and summing residues evaluates real and complex integrals alike. The engine that manufactures both halves of the series is the elementary geometric sum, applied to the two ways of writing the Cauchy kernel.

\frac{1}{1-w}=\sum_{k=0}^{\infty} w^k,\qquad |w|<1 \qquad (4)

where: $w$ — a placeholder that becomes $(z-z_0)/(\zeta-z_0)$ on the outer circle (giving positive powers) or its reciprocal $(\zeta-z_0)/(z-z_0)$ on the inner circle (giving negative powers). Reading it: the whole theorem is this one identity, chosen so that the ratio in $w$ has modulus below one on the relevant circle. The coefficients cannot grow arbitrarily. Bounding the integral in (2) by the maximum of $|f|$ on a circle gives a Cauchy-type estimate that controls the tail and pins down the type of singularity.

|a_n|\le \frac{M(\rho)}{\rho^{n}},\qquad M(\rho)=\max_{|z-z_0|=\rho}|f(z)| \qquad (5)

where: $M(\rho)$ — maximum modulus of $f$ on the circle of radius $\rho$ , with $r<\rho<R$ ; $n$ — any integer. Reading it: for large positive $n$ the bound forces rapid decay of $a_n$ (radius of convergence); for negative $n$ it constrains the principal part and underlies the classification of singularities. Finally, on a fixed circle the Laurent series is a Fourier series — the same expansion in two dictionaries. Writing $z-z_0=\rho e^{i\theta}$ turns the coefficient integral into a Fourier integral.

a_n=\frac{1}{2\pi\,\rho^{n}}\int_0^{2\pi} f(z_0+\rho e^{i\theta})\,e^{-in\theta}\,\mathrm{d}\theta \qquad (6)

where: $\rho$ — radius of the chosen circle, $r<\rho<R$ ; $\theta$ — angular parameter along it. Reading it: $\rho^n a_n$ are the Fourier coefficients of the boundary values $f(z_0+\rho e^{i\theta})$ , so Laurent series inherit every intuition about spectra.

Derivation ALaurent's theorem from Cauchy's integral formula

The goal is to prove the whole theorem — the existence of the two-sided expansion and the single coefficient formula (2) — starting only from Cauchy's integral formula and the geometric series. The strategy is to enclose the evaluation point between two concentric circles inside the annulus, apply Cauchy's formula to the region between them, and then expand the kernel differently on each circle so that both expansions converge. Fix $z$ with $r<|z-z_0|<R$ , and choose circles $C_r$ of radius $\rho_1$ and $C_R$ of radius $\rho_2$ with $r<\rho_1<|z-z_0|<\rho_2<R$ , so that $z$ sits strictly between them.

Step 1. The boundary of the ring between $C_r$ and $C_R$ is $C_R$ traversed counterclockwise together with $C_r$ traversed clockwise. Cauchy's integral formula on this region gives $f(z)=\frac{1}{2\pi i}\oint_{C_R}\frac{f(\zeta)}{\zeta-z}\,\mathrm{d}\zeta-\frac{1}{2\pi i}\oint_{C_r}\frac{f(\zeta)}{\zeta-z}\,\mathrm{d}\zeta$ . Why: $z$ lies inside the annular region but outside the inner disk, so only the annulus boundary contributes — the inner circle enters with a minus sign because its outward orientation is clockwise relative to the region.

Step 2. On $C_R$ we have $|z-z_0|<|\zeta-z_0|$ , so write $\frac{1}{\zeta-z}=\frac{1}{(\zeta-z_0)}\cdot\frac{1}{1-\frac{z-z_0}{\zeta-z_0}}=\sum_{n=0}^{\infty}\frac{(z-z_0)^n}{(\zeta-z_0)^{n+1}}$ . Why: the ratio $(z-z_0)/(\zeta-z_0)$ has modulus below one on the outer circle, so the geometric series (4) converges uniformly in $\zeta$ and may be integrated term by term.

Step 3. Integrating term by term gives the analytic part $\frac{1}{2\pi i}\oint_{C_R}\frac{f(\zeta)}{\zeta-z}\,\mathrm{d}\zeta=\sum_{n=0}^{\infty}a_n(z-z_0)^n$ with $a_n=\frac{1}{2\pi i}\oint_{C_R}\frac{f(\zeta)\,\mathrm{d}\zeta}{(\zeta-z_0)^{n+1}}$ . Why: uniform convergence lets the sum and integral swap, and each $\zeta$ -integral is exactly the coefficient formula (2) for $n\ge0$ .

Step 4. On $C_r$ we instead have $|\zeta-z_0|<|z-z_0|$ , so expand the reciprocal ratio: $\frac{1}{\zeta-z}=-\frac{1}{(z-z_0)}\cdot\frac{1}{1-\frac{\zeta-z_0}{z-z_0}}=-\sum_{m=0}^{\infty}\frac{(\zeta-z_0)^m}{(z-z_0)^{m+1}}$ . Why: now it is the reciprocal ratio that is small, so we factor out $(z-z_0)$ rather than $(\zeta-z_0)$ — this is the single change of viewpoint that generates negative powers.

Step 5. Integrating the inner term gives $-\frac{1}{2\pi i}\oint_{C_r}\frac{f(\zeta)}{\zeta-z}\,\mathrm{d}\zeta=\sum_{m=0}^{\infty}\Big[\frac{1}{2\pi i}\oint_{C_r}f(\zeta)(\zeta-z_0)^m\,\mathrm{d}\zeta\Big](z-z_0)^{-m-1}$ . Setting $n=-m-1$ recovers the coefficient formula (2) for $n<0$ . Why: relabelling the index shows the inner circle produces precisely the negative-power coefficients, in the very same form as the positive ones.

Step 6. For every fixed $n$ the integrand $f(\zeta)(\zeta-z_0)^{-n-1}$ is analytic throughout the annulus, so Cauchy's theorem lets both $C_R$ and $C_r$ be deformed to one common loop $C$ . The two sums merge into $f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$ with $a_n$ given by (2). Why: contour independence within a region of analyticity is what unifies the two circles into a single formula valid for all $n$ at once. *Why this is a consistency check: as $r\to0$ with $f$ analytic at $z_0$ , the inner integrals with $m\ge0$ vanish (analytic integrand, no enclosed singularity), so $a_n=0$ for $n<0$ and $a_n=f^{(n)}(z_0)/n!$ — Laurent reduces exactly to Taylor.*

Go deeper · Derivation B · 1 minUniqueness of the Laurent coefficients

Derivation BUniqueness of the Laurent coefficients

The goal is to show the expansion is forced: if any two-sided power series represents $f$ on the annulus, its coefficients must be the ones given by (2). This matters because it licenses every shortcut used in practice — geometric series, partial fractions, multiplying known expansions — since whatever series you legitimately produce is guaranteed to be the Laurent series. The tool is the orthogonality of powers on a circle. Suppose $f(z)=\sum_{n=-\infty}^{\infty} b_n (z-z_0)^n$ converges uniformly on a circle $|z-z_0|=\rho$ lying in the annulus.

Step 1. Multiply by $(z-z_0)^{-m-1}$ and integrate around the circle: $\oint\frac{f(z)}{(z-z_0)^{m+1}}\,\mathrm{d}z=\sum_{n}b_n\oint(z-z_0)^{n-m-1}\,\mathrm{d}z$ . Why: uniform convergence on the circle justifies interchanging the sum and the integral, turning one integral into a sum of elementary ones.

Step 2. Evaluate the power integral by parametrising $z-z_0=\rho e^{i\theta}$ : $\oint(z-z_0)^{k}\,\mathrm{d}z=i\rho^{k+1}\int_0^{2\pi}e^{i(k+1)\theta}\,\mathrm{d}\theta=2\pi i\,\delta_{k,-1}$ , which is $2\pi i$ when $k=-1$ and $0$ otherwise. Why: the angular integral of $e^{i(k+1)\theta}$ over a full period vanishes unless the exponent is zero, i.e. unless $k=-1$ — this is the orthogonality of the powers.

Step 3. Only the term with $n-m-1=-1$ , i.e. $n=m$ , survives, giving $\oint\frac{f(z)}{(z-z_0)^{m+1}}\,\mathrm{d}z=2\pi i\,b_m$ , hence $b_m=\frac{1}{2\pi i}\oint\frac{f(z)}{(z-z_0)^{m+1}}\,\mathrm{d}z=a_m$ . Why: the single surviving term isolates one coefficient, and it is identical to the formula (2), so the two series coincide term by term. *Why this is a consistency check: two expansions of the same $f$ on overlapping annuli must agree on the overlap, and the orthogonality integral reproduces $a_{-1}$ as the residue in (3), consistent with Derivation A.*

Go deeper · Derivation C · 1 minThe residue of a pole of order m

Derivation CThe residue of a pole of order m

The goal is a working formula for the coefficient that dominates applications — the residue $a_{-1}$ — at a pole, without expanding the whole series. An essential singularity admits no shortcut and demands full expansion, but a pole of finite order $m$ lets us strip the singularity, differentiate, and read the residue off. The idea is to multiply $f$ by $(z-z_0)^m$ to kill the principal part's tail, leaving an analytic function whose Taylor coefficient we want. Suppose $f$ has a pole of order $m$ at $z_0$ , so $f(z)=\sum_{n=-m}^{\infty}a_n(z-z_0)^n$ with $a_{-m}\ne0$ .

Step 1. Form $g(z)=(z-z_0)^m f(z)=\sum_{n=-m}^{\infty}a_n(z-z_0)^{n+m}=a_{-m}+a_{-m+1}(z-z_0)+\dots+a_{-1}(z-z_0)^{m-1}+\dots$ . Why: multiplying by $(z-z_0)^m$ shifts every exponent up by $m$ , converting the principal part into an ordinary Taylor series — $g$ is analytic at $z_0$ with a removable singularity.

Step 2. The residue $a_{-1}$ is the coefficient of $(z-z_0)^{m-1}$ in $g$ , which by Taylor's theorem is $g^{(m-1)}(z_0)/(m-1)!$ . Why: the $(m-1)$ -th derivative of a power series picks out the $(m-1)$ -th coefficient, and $a_{-1}$ sits at exactly that position after the shift.

Step 3. Therefore $a_{-1}=\operatorname{Res}_{z=z_0}f=\frac{1}{(m-1)!}\lim_{z\to z_0}\frac{\mathrm{d}^{m-1}}{\mathrm{d}z^{m-1}}\big[(z-z_0)^m f(z)\big]$ . For a simple pole ( $m=1$ ) this is just $\lim_{z\to z_0}(z-z_0)f(z)$ . Why: the limit handles the removable singularity cleanly, and the $m=1$ case recovers the familiar simple-pole rule used constantly in contour integration. *Why this is a consistency check: for $f=e^z/z^3$ (pole of order $m=3$ at $0$ ), the formula gives $\frac{1}{2!}\lim_{z\to0}\frac{\mathrm{d}^2}{\mathrm{d}z^2}e^z=\frac12$ , matching the direct expansion $e^z/z^3=z^{-3}+z^{-2}+\tfrac12 z^{-1}+\cdots$ .*

Limiting cases

The structure of the principal part $\sum_{n<0}a_n(z-z_0)^n$ is a complete diagnostic for the singularity at the centre of the annulus. Reading it off is the everyday use of a Laurent series.

Case	Result	Meaning
Removable singularity	$a_n=0$ for all $n<0$	Empty principal part; $f$ extends analytically to $z_0$ , e.g. $\frac{\sin z}{z}=1-\frac{z^2}{6}+\cdots$ at $z=0$
Pole of order $m$	$a_{-m}\ne0,\ a_n=0$ for $n<-m$	Finite principal part with lowest power $(z-z_0)^{-m}$ , e.g. $\frac{e^z}{z^3}$ has a pole of order $3$
Essential singularity	$a_n\ne0$ for infinitely many $n<0$	Principal part never terminates, e.g. $e^{1/z}=\sum_{k\ge0}\frac{1}{k!\,z^k}$ ; Casorati–Weierstrass and Picard apply
Pure Taylor limit	$r=0,\ a_n=\frac{f^{(n)}(z_0)}{n!}$	Inner radius shrinks to zero when $f$ is analytic at $z_0$ ; Laurent collapses to Taylor
Expansion at infinity	series in $z^{-k}$ on $\|z\|>R$	Beyond the outermost singularity; the negative-power series is the large- $z$ asymptotic expansion

Common trapthe misconception that a function has one Laurent series

It does not. A Laurent series belongs to an annulus, not to a function or even to a point — and a single function generally has several, one for each ring of analyticity carved out by its singularities. The textbook trap is $\frac{1}{z-1}$ . Inside the disk $|z|<1$ it expands as $\frac{-1}{1-z}=-\sum_{k\ge0}z^k=-1-z-z^2-\cdots$ , all non-negative powers and no residue at $0$ . Outside, on $|z|>1$ , the very same function expands as $\frac{1}{z}\cdot\frac{1}{1-1/z}=\sum_{k\ge0}z^{-k-1}=z^{-1}+z^{-2}+\cdots$ , all negative powers. These look nothing alike, both are correct, and confusing them silently corrupts every residue you extract. The discipline is unbreakable: name the annulus first — which region contains your contour or your point of interest — and only then expand. The same rule underlies the region-of-convergence bookkeeping of the $z$ -transform, where the annulus decides whether a filter is causal and stable.

Worked examples

1 · Three annuli of a rational function, and its residue (fully worked)

Expand $f(z)=\dfrac{1}{(z-1)(z-3)}$ about $z_0=0$ in the middle annulus $1<|z|<3$ , give the coefficients as numbers, and use $a_{-1}$ to evaluate $\oint_C f\,\mathrm{d}z$ for a contour in that ring.

Step 1. Partial fractions: write $\frac{1}{(z-1)(z-3)}=\frac{A}{z-1}+\frac{B}{z-3}$ with $1=A(z-3)+B(z-1)$ . Setting $z=1$ gives $A=-\tfrac12$ ; setting $z=3$ gives $B=\tfrac12$ . So $f=-\tfrac12\,\frac{1}{z-1}+\tfrac12\,\frac{1}{z-3}$ .

Step 2. Expand each piece so that it converges on $1<|z|<3$ . Since $|z|>1$ , treat $\frac{1}{z-1}=\frac1z\cdot\frac{1}{1-1/z}=\sum_{k=0}^{\infty}z^{-k-1}=\sum_{k\ge1}z^{-k}$ . Since $|z|<3$ , treat $\frac{1}{z-3}=-\frac13\cdot\frac{1}{1-z/3}=-\sum_{k=0}^{\infty}\frac{z^k}{3^{k+1}}$ . Each ratio has modulus below one on the ring, so both series converge there.

Step 3. Combine: $f=-\tfrac12\sum_{k\ge1}z^{-k}-\tfrac12\sum_{k\ge0}\frac{z^k}{3^{k+1}}$ . Reading off coefficients: for $n<0$ , $\ a_n=-\tfrac12$ ; for $n\ge0$ , $\ a_n=-\dfrac{1}{2\cdot3^{\,n+1}}$ . So $a_{-1}=-\tfrac12$ , $a_0=-\tfrac16$ , $a_1=-\tfrac1{18}$ — the values in the second figure.

Step 4. The residue payoff: by (3), $\oint_C f\,\mathrm{d}z=2\pi i\,a_{-1}=2\pi i\left(-\tfrac12\right)=\boxed{-\pi i}$ . Cross-check: a contour in $1<|z|<3$ encircles only the pole at $z=1$ , whose residue is $A=-\tfrac12$ , so the residue theorem gives $2\pi i(-\tfrac12)=-\pi i$ . The $a_{-1}$ of the middle-annulus series and the enclosed residue agree exactly.

2 · A residue at an essential singularity (fully worked)

Evaluate $\displaystyle\oint_{|z|=1} z^2\,e^{1/z}\,\mathrm{d}z$ .

Step 1. The point $z=0$ is an essential singularity, so there is no pole-order shortcut — the Laurent expansion itself is the only route. Use the known series $e^{1/z}=\sum_{k=0}^{\infty}\frac{1}{k!\,z^k}$ , valid on $0<|z|<\infty$ .

Step 2. Multiply by $z^2$ : $z^2 e^{1/z}=\sum_{k=0}^{\infty}\frac{z^{2-k}}{k!}$ . We need the coefficient of $z^{-1}$ , i.e. the term with $2-k=-1$ , so $k=3$ , giving coefficient $\frac{1}{3!}=\frac16$ . Thus $\operatorname{Res}_{z=0}\big(z^2 e^{1/z}\big)=a_{-1}=\tfrac16$ .

Step 3. Apply the residue theorem: $\oint_{|z|=1} z^2 e^{1/z}\,\mathrm{d}z=2\pi i\,a_{-1}=2\pi i\cdot\tfrac16=\boxed{\dfrac{\pi i}{3}}\approx 1.047\,i$ . The whole integral is fixed by a single coefficient buried in an infinite principal part.

3 · Reading pole order and residue with the shortcut

Classify the singularity of $f(z)=\dfrac{e^z}{z^3}$ at $z=0$ and find its residue.

Step 1. Divide the exponential series: $\frac{e^z}{z^3}=\frac{1}{z^3}\left(1+z+\frac{z^2}{2}+\cdots\right)=z^{-3}+z^{-2}+\tfrac12 z^{-1}+\tfrac16+\cdots$ . The principal part stops at $z^{-3}$ with $a_{-3}=1\ne0$ , so this is a pole of order $3$ .

Step 2. The residue is the $z^{-1}$ coefficient, $a_{-1}=\tfrac12$ . Confirm with the order- $m$ formula: $a_{-1}=\frac{1}{2!}\lim_{z\to0}\frac{\mathrm{d}^2}{\mathrm{d}z^2}\big[z^3\cdot\frac{e^z}{z^3}\big]=\frac12\,e^0=\boxed{\tfrac12}$ . Direct expansion and the shortcut agree.

Physics of the portrait

The Laurent series is where the disciplined machinery of nineteenth-century complex analysis — Cauchy's contour integrals, Weierstrass's power series, the taxonomy of singular points — comes together in one construction, named for a French military engineer who proved it in a prize memoir and died young.

Origins · 1 minPierre Alphonse Laurent and how this was discovered

Pierre Alphonse Laurent

1813–1854 · complex analysis and the annular expansion of analytic functions

Pierre Alphonse Laurent was, improbably for a mathematician whose name physicists speak daily, a career officer in the French army's Corps of Engineers. Trained at the École Polytechnique, he spent years directing harbour works at Le Havre, doing mathematics on the side. Around 1843 he submitted a memoir to the Paris Academy of Sciences for its Grand Prix; in it he proved that a function analytic on an annulus admits an expansion in both positive and negative powers — extending Cauchy's theorem, which handled only the disk. The prize deadline had passed, so the memoir was never formally entered, and Cauchy himself, reporting on it, urged publication; it appeared only posthumously. Unknown to Laurent, Karl Weierstrass had proved an equivalent result in 1841 in a paper unpublished until 1894, so the theorem is doubly discovered and singly named. Laurent died in 1854 at forty-two, a chef de bataillon, his reputation resting almost entirely on this one theorem. The construction that carries his name turned the classification of singularities — removable, pole, essential — into an exact statement about which coefficients vanish, and made the residue simply the coefficient $a_{-1}$ .

History

1821

— Cauchy's Cours d'analyse founds rigorous power-series complex analysis

1831

— Cauchy's integral formula expresses coefficients as contour integrals

1841

— Weierstrass proves the two-sided annular expansion, unpublished until 1894

1843

— Laurent presents to the Paris Academy the theorem that now bears his name

1879

— Picard's great theorem sharpens the essential-singularity picture

Going further

Texts. Stone & Goldbart, Mathematics for Physics, §17–18 (complex analysis) · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., §11.6 · Hassani, Mathematical Physics, 2nd ed., Ch. 10 · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 6 · Nakahara, Geometry, Topology and Physics, App. on complex analysis · Ahlfors, Complex Analysis, 3rd ed., Ch. 5 · Whittaker & Watson, A Course of Modern Analysis, §5.6. All formulas are standard results. Key anchor values: the orthogonality integral $\oint(z-z_0)^k\,\mathrm{d}z=2\pi i\,\delta_{k,-1}$ follows by direct parametrisation; the middle-annulus coefficients of $1/[(z-1)(z-3)]$ are $a_{-1}=-\tfrac12$ , $a_0=-\tfrac16$ , $a_1=-\tfrac1{18}$ , giving $\oint f=-\pi i$ ; $\operatorname{Res}_{z=0}z^2 e^{1/z}=\tfrac16$ so $\oint_{|z|=1}z^2 e^{1/z}\,\mathrm{d}z=\pi i/3$ ; $\operatorname{Res}_{z=0}e^z/z^3=\tfrac12$ by both direct expansion and the order- $m$ formula; and the Bessel generating identity $e^{(x/2)(t-1/t)}=\sum J_n(x)t^n$ holds with each $J_n(x)$ a Laurent coefficient. No nonstandard numerical constants appear.

Prerequisites. Cauchy's Integral Formula and Derivatives · Cauchy's Integral Theorem · Taylor Series of Analytic Functions · Complex Integration and Contour Integrals. Leads to. Zeros, Poles, and Isolated Singularities · Residues and the Residue Theorem.

Ask the tutor: Give me any function and an annulus, and I will walk you through building its Laurent series term by term and reading off the residue.

What exactly is an annulus, and why can't I just use a disk?

An annulus is a ring-shaped region $r<|z-z_0|<R$ — everything between two concentric circles, with the centre punched out. You need it precisely because the interesting functions are singular at $z_0$ : a Taylor series lives on a full disk and demands analyticity everywhere inside, including the centre, which a singular function cannot provide. By excising the centre, the annulus lets $f$ be analytic on the ring while still misbehaving at $z_0$ , and the price of that freedom is the appearance of negative powers. When $f$ happens to be analytic at the centre too, the inner radius can shrink to zero, the hole disappears, and the annulus degenerates back into a disk — at which point the Laurent series is just the Taylor series.

Is the Laurent series of a function unique?

Yes and no, and the distinction is a common source of error. On a fixed annulus the expansion is completely unique — Derivation B shows the coefficients are forced by the orthogonality of powers, so any legitimate method must land on the same answer. But a function with several singularities has several annuli, and each carries its own unique series. So "the Laurent series of $f$ " is ambiguous until you name the ring: fix the annulus and there is exactly one expansion; change it and you generally get a completely different one, as $1/(z-1)$ shows inside versus outside the unit circle.

How is the residue related to the coefficient a₋₁?

They are the same number. The residue at $z_0$ is defined as the coefficient $a_{-1}$ of $(z-z_0)^{-1}$ in the Laurent expansion valid on a punctured neighbourhood of $z_0$ . Its importance comes from integration: on a closed contour every power $(z-z_0)^n$ integrates to zero except $n=-1$ , which contributes $2\pi i$ . So $\oint f\,\mathrm{d}z=2\pi i\,a_{-1}$ , meaning the entire contour integral is controlled by this one coefficient. That is the mechanism of the residue theorem, and it is why so much of applied complex analysis is really the art of extracting $a_{-1}$ .

How do I actually compute Laurent coefficients in practice?

Almost never from the contour-integral formula directly — that is for proving theorems. In practice you manipulate known series. For a rational function, split it by partial fractions and expand each piece with the geometric series, keeping every ratio below one in modulus on your chosen annulus (Example 1). For products with $e^{1/z}$ , $\sin(1/z)$ , and the like, multiply the standard Maclaurin series and collect the power you want (Example 2). For a pole of known order, use the shortcut of Derivation C to get the residue without the full series. The contour integral (2) is a last resort, useful mainly when $f$ is given only implicitly or on the boundary.

What really separates a pole from an essential singularity?

Count the negative powers. A pole of order $m$ has a principal part that stops — from $(z-z_0)^{-m}$ up, with $a_{-m}\ne0$ and nothing below — so $|f|\to\infty$ in a controlled, power-law way. An essential singularity has a principal part that never stops: infinitely many nonzero negative coefficients, as in $e^{1/z}=\sum_{k\ge0}z^{-k}/k!$ . The behaviour is wildly different. Near an essential singularity, the Casorati–Weierstrass theorem says $f$ comes arbitrarily close to every complex value in any neighbourhood, and Picard sharpens this to say it actually attains every value, with at most one exception, infinitely often.

Why does one integral formula work for every coefficient, positive and negative?

Because for each fixed $n$ the integrand $f(\zeta)/(\zeta-z_0)^{n+1}$ is analytic on the whole annulus, and Cauchy's theorem makes its contour integral independent of which loop in the annulus you use. In Derivation A the positive coefficients emerge naturally from the outer circle and the negative ones from the inner circle — two different circles, two different geometric expansions. But once the coefficient is written as an integral of an annulus-analytic integrand, the outer and inner circles can be deformed into any common loop $C$ without changing the value. So the split into two circles is only a device for the proof; the final formula collapses onto a single contour valid for all $n$ simultaneously.

In what sense is a Laurent series just a Fourier series?

Restrict the expansion to a single circle $z-z_0=\rho e^{i\theta}$ . Then $(z-z_0)^n=\rho^n e^{in\theta}$ , and $f(z_0+\rho e^{i\theta})=\sum_n (a_n\rho^n)\,e^{in\theta}$ is exactly a Fourier series in the angle $\theta$ with coefficients $a_n\rho^n$ ; equation (6) is its Fourier inversion formula. This is why the two theories rhyme — a Laurent series is to an annulus what a Fourier series is to a circle, both encoding how a function behaves as you go around a loop, and both diagnosing structure through the decay (or non-decay) of their coefficients. The radius $\rho$ is a free dial that reweights the spectrum by $\rho^n$ .