Laurent Series and Isolated Singularities

⚡ Key Takeaways

A Laurent series expands a complex function $f(z)$ that is holomorphic on an annulus into a doubly-infinite power series including negative powers of $(z-z_0)$ . The coefficient of the $(z-z_0)^{-1}$ term, called the residue, encodes the function's behavior near an isolated singularity — a point where $f$ fails to be holomorphic but is holomorphic on a punctured neighborhood. Singularities classify into three types (removable, pole, essential) by the structure of the negative-power "principal part." Developed by Pierre Alphonse Laurent (1843) and refined by Weierstrass and Cauchy; foundational to complex analysis and contour integration.

💡 The Core Insight

Holomorphy on a disk forces a Taylor series. Holomorphy on an annulus — a disk with a hole — relaxes this just enough to admit negative powers. The decomposition $f = (\text{regular part}) + (\text{principal part})$ splits $f$ into a piece holomorphic inside the inner radius and a piece holomorphic outside the outer radius. The principal part is a fingerprint of the singularity, and a single number from it (the residue) governs integration.

DefinitionAnnulus and Isolated Singularity

An open annulus centered at $z_0$ is the set

1

\quad A(z_0; r, R) = \{\, z \in \mathbb{C} : r < |z - z_0| < R \,\}, \qquad 0 \le r < R \le \infty.

Derivation hint: It is the intersection of the exterior of the disk of radius $r$ with the interior of the disk of radius $R$ ; $r=0$ gives a punctured disk, $R=\infty$ an unbounded exterior. A point $z_0$ is an isolated singularity of $f$ if there exists $\rho > 0$ such that $f$ is holomorphic on the punctured disk $A(z_0; 0, \rho) = \{0 < |z-z_0| < \rho\}$ but is not holomorphic (or not defined) at $z_0$ itself. Here $z_0$ symbolizes the center; $r,R$ the radii; the condition $|z-z_0|>r$ excludes inner trouble.

DefinitionLaurent Series

The Laurent series of $f$ about $z_0$ is the doubly-infinite series

2

\quad f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n = \underbrace{\sum_{n=0}^{\infty} a_n (z-z_0)^n}_{\text{regular (analytic) part}} + \underbrace{\sum_{n=1}^{\infty} a_{-n} (z-z_0)^{-n}}_{\text{principal part}}.

Derivation hint: Split the bilateral sum at $n=0$ ; the nonnegative powers form a Taylor-like series (the regular part), the negative powers form the principal part carrying singular behavior. The coefficients are given by the contour integral

3

\quad a_n = \frac{1}{2\pi i} \oint_{\gamma} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}\, d\zeta, \qquad n \in \mathbb{Z},

where $\gamma$ is any positively oriented simple closed contour lying inside the annulus and encircling $z_0$ once. Here $\zeta$ is the integration variable, $n+1$ the exponent shift from the geometric-series kernel, and $2\pi i$ the normalizing constant from Cauchy's theory. Derivation hint: Multiply [2] by $(\zeta-z_0)^{-m-1}$ and integrate term-by-term; only the $n=m$ term survives because $\oint (\zeta-z_0)^{k}\,d\zeta = 2\pi i$ if $k=-1$ and $0$ otherwise.

TheoremLaurent's Theorem — Existence and Uniqueness

If $f$ is holomorphic on the annulus $A(z_0; r, R)$ , then $f$ admits a unique Laurent expansion [2] converging absolutely on $A(z_0; r, R)$ and uniformly on every compact subannulus $r < \rho_1 \le |z-z_0| \le \rho_2 < R$ , with coefficients given by [3]. The choice of contour $\gamma$ (within the annulus, enclosing $z_0$ ) does not affect $a_n$ .

DefinitionClassification of Isolated Singularities

Let $z_0$ be an isolated singularity with principal part $\sum_{n=1}^\infty a_{-n}(z-z_0)^{-n}$ . Then $z_0$ is:

Removable if $a_{-n} = 0$ for all $n \ge 1$ (no principal part). Then $\lim_{z\to z_0} f(z)$ exists finitely and $f$ extends holomorphically.
A pole of order $m$ if $a_{-m} \ne 0$ and $a_{-n} = 0$ for all $n > m$ (finite principal part). A pole of order $1$ is simple.
An essential singularity if infinitely many $a_{-n} \ne 0$ (infinite principal part).

The residue of $f$ at $z_0$ is

4

\quad \operatorname{Res}_{z=z_0} f = a_{-1} = \frac{1}{2\pi i}\oint_\gamma f(\zeta)\,d\zeta.

Derivation hint: Set $n=-1$ in [3]; the kernel exponent $-(-1)-1 = 0$ leaves the integrand $f(\zeta)$ raw.

TheoremResidue Theorem

If $f$ is holomorphic on and inside a positively oriented simple closed contour $\gamma$ except for finitely many isolated singularities $z_1, \dots, z_k$ inside $\gamma$ , then

5

\quad \oint_\gamma f(z)\, dz = 2\pi i \sum_{j=1}^{k} \operatorname{Res}_{z=z_j} f.

Derivation hint: Surround each $z_j$ by a tiny circle; the integral over $\gamma$ equals the sum over the small circles by deforming contours in the holomorphic region, and each small-circle integral is $2\pi i \,\operatorname{Res}$ by [4].

ProofExistence of the Laurent Expansion

Step 1 — Two Cauchy contours. Fix $z$ in the annulus and choose radii with $r < \rho_1 < |z - z_0| < \rho_2 < R$ . Let $C_1, C_2$ be the circles $|\zeta - z_0| = \rho_1, \rho_2$ . Since $f$ is holomorphic on the closed subannulus, Cauchy's integral formula adapted to the doubly-connected region gives

6

\quad f(z) = \frac{1}{2\pi i}\oint_{C_2}\frac{f(\zeta)}{\zeta - z}\,d\zeta \;-\; \frac{1}{2\pi i}\oint_{C_1}\frac{f(\zeta)}{\zeta - z}\,d\zeta.

Derivation hint: Connect $C_1$ and $C_2$ by a cut to form a single simple closed contour bounding a region where $f$ is holomorphic; Cauchy's formula applies and the cut contributions cancel, leaving outer minus inner.

Step 2 — Expand the outer integral. On $C_2$ , $|\zeta - z_0| = \rho_2 > |z-z_0|$ , so

7

\quad \frac{1}{\zeta - z} = \frac{1}{(\zeta-z_0) - (z-z_0)} = \sum_{n=0}^{\infty} \frac{(z-z_0)^n}{(\zeta - z_0)^{n+1}}, \qquad \left|\frac{z-z_0}{\zeta-z_0}\right| < 1.

Derivation hint: Factor out $(\zeta-z_0)$ and apply the geometric series $\frac{1}{1-w} = \sum w^n$ with $w = (z-z_0)/(\zeta-z_0)$ , which converges since $|w|<1$ . Integrating term-by-term (justified by uniform convergence) over $C_2$ yields $\sum_{n\ge 0} a_n (z-z_0)^n$ with $a_n$ as in [3].

Step 3 — Expand the inner integral. On $C_1$ , $|\zeta - z_0| = \rho_1 < |z-z_0|$ , so we factor out $-(z-z_0)$ instead:

8

\quad \frac{-1}{\zeta - z} = \frac{1}{(z-z_0) - (\zeta-z_0)} = \sum_{m=1}^{\infty} \frac{(\zeta-z_0)^{m-1}}{(z - z_0)^{m}}, \qquad \left|\frac{\zeta-z_0}{z-z_0}\right| < 1.

Derivation hint: Now the larger quantity is $(z-z_0)$ ; factor it out and expand $\frac{1}{1-w}$ with $w=(\zeta-z_0)/(z-z_0)$ , generating negative powers of $(z-z_0)$ . Integrating over $C_1$ gives $\sum_{m\ge 1} a_{-m}(z-z_0)^{-m}$ with $a_{-m} = \frac{1}{2\pi i}\oint_{C_1} f(\zeta)(\zeta-z_0)^{m-1}\,d\zeta$ , matching [3] for $n = -m$ .

Step 4 — Combine and unify the contour. Adding the two parts gives [2]. Because the integrands in [3] are holomorphic in $\zeta$ on the whole annulus, Cauchy's theorem lets us replace $C_1$ and $C_2$ by any single contour $\gamma$ encircling $z_0$ inside the annulus. ∎

∎

ProofUniqueness

Suppose $f(z) = \sum_n b_n (z-z_0)^n$ converges on the annulus. Multiply by $(z-z_0)^{-m-1}$ and integrate over a circle $\gamma$ centered at $z_0$ . Uniform convergence permits term-by-term integration, and the orthogonality relation

9

\quad \oint_\gamma (z-z_0)^{k}\,dz = \begin{cases} 2\pi i, & k = -1,\\ 0, & k \ne -1, \end{cases}

isolates $b_m = \frac{1}{2\pi i}\oint_\gamma f(z)(z-z_0)^{-m-1}\,dz = a_m$ . Hence the coefficients are forced. ∎ Derivation hint: Parametrize $z = z_0 + \rho e^{i\theta}$ ; then $\oint (z-z_0)^k dz = i\rho^{k+1}\int_0^{2\pi} e^{i(k+1)\theta}d\theta$ , which vanishes unless $k+1=0$ .

∎

mathematician · Pierre Alphonse Laurent | 1813–1854

A French military engineer and officer in the Corps des Ponts et Chaussées, Laurent submitted his theorem on the expansion of complex functions in annular regions to the Académie des Sciences in 1843, as part of a memoir on the calculus of variations and analysis. The work was never formally published in his lifetime (Cauchy reviewed it favorably but it missed the Grand Prix deadline). Laurent died young at 41; his expansion was published posthumously. Karl Weierstrass had independently obtained an equivalent result in an unpublished 1841 paper. Laurent's name stuck, attached to one of the central tools of complex analysis.

🧠 Intuition

Think of a Taylor series as describing a function from a single clean center point — it needs the function to be smooth all the way through that center. A Laurent series instead describes the function from a ring surrounding a possibly-broken center. The negative powers $(z-z_0)^{-n}$ blow up as $z \to z_0$ ; they are precisely the language needed to record how the function explodes (or oscillates wildly) near the bad point. The order of the principal part measures the severity of the blow-up:

No negative terms → the "singularity" was an illusion; the function is secretly fine (removable).
Finitely many negative terms → the function diverges like $1/(z-z_0)^m$ , a controlled, polynomial-rate explosion (pole).
Infinitely many negative terms → uncontrollable chaos; $e^{1/z}$ near $0$ is the prototype, packing every behavior into any neighborhood (essential).

The residue $a_{-1}$ is special because of equation [9]: it is the only power of $(z-z_0)$ whose integral around a loop survives. Everything else integrates to zero. So when you integrate around a singularity, the function "remembers" only its $a_{-1}$ coefficient.

Figure

The annulus of convergence and classification of singularities by principal part

ExampleA Pole of Order 2 — Coefficients and Residue

Given: $f(z) = \dfrac{e^{z}}{z^{2}}$ , find the Laurent series about $z_0 = 0$ and the residue.

Step 1 — Expand the numerator. The entire function $e^z$ has the Taylor series

10

\quad e^z = \sum_{k=0}^{\infty}\frac{z^k}{k!} = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots

Derivation hint: Standard exponential series, convergent for all $z$ .

Step 2 — Divide by $z^2$ . Shift every exponent down by 2:

11

\quad \frac{e^z}{z^2} = \frac{1}{z^2} + \frac{1}{z} + \frac{1}{2!} + \frac{z}{3!} + \frac{z^2}{4!} + \cdots

Derivation hint: Multiply [10] term-by-term by $z^{-2}$ ; the lowest term becomes $z^{-2}$ .

Step 3 — Identify the structure. The principal part is $z^{-2} + z^{-1}$ . The lowest power is $-2$ with nonzero coefficient $a_{-2}=1$ , so $z_0=0$ is a pole of order 2. The residue is the coefficient of $z^{-1}$ :

12

\quad \operatorname{Res}_{z=0}\frac{e^z}{z^2} = a_{-1} = 1.

Derivation hint: Read off $a_{-1}$ directly from [11].

Check via the pole formula: $\operatorname{Res} = \lim_{z\to 0}\frac{d}{dz}\!\left[z^2\cdot \frac{e^z}{z^2}\right] = \lim_{z\to 0} e^z = 1.$ ✓

\boxed{\;\frac{e^z}{z^2} = z^{-2}+z^{-1}+\tfrac12 + \tfrac{z}{6}+\cdots,\qquad \operatorname{Res}_{z=0}=1\;}

ExampleSame Function, Different Annuli — Computing a Contour Integral

Given: Compute $\displaystyle \oint_{|z|=3}\frac{1}{z(z-1)}\,dz$ using Laurent/residues, then verify the two distinct Laurent expansions about $z=0$ .

Step 1 — Partial fractions.

13

\quad \frac{1}{z(z-1)} = \frac{-1}{z} + \frac{1}{z-1}.

Derivation hint: $\frac{1}{z(z-1)} = \frac{A}{z}+\frac{B}{z-1}$ ; solve $A=-1$ , $B=1$ .

Step 2 — Two annuli about $z=0$ . Singularities sit at $0$ and $1$ , splitting the plane into $0<|z|<1$ and $|z|>1$ . Inner annulus $0<|z|<1$ : expand $\frac{1}{z-1} = -\frac{1}{1-z} = -\sum_{n\ge 0} z^n$ , so

14

\quad \frac{1}{z(z-1)} = -\frac{1}{z} - 1 - z - z^2 - \cdots \quad (0<|z|<1).

Derivation hint: Use $|z|<1$ to expand the geometric series in $z$ ; combine with the $-1/z$ from [13]. *Outer annulus $|z|>1$ :* now $|1/z|<1$ , so $\frac{1}{z-1} = \frac{1}{z}\cdot\frac{1}{1-1/z} = \sum_{n\ge 1} z^{-n}$ , giving

15

\quad \frac{1}{z(z-1)} = -\frac{1}{z} + \frac1z + \frac{1}{z^2}+\cdots = \frac{1}{z^2}+\frac{1}{z^3}+\cdots \quad (|z|>1).

Derivation hint: Expand $\tfrac{1}{1-1/z}=\sum (1/z)^k$ valid for $|z|>1$ ; the two $\pm 1/z$ terms cancel.

Step 3 — Integrate over $|z|=3$ . This contour encloses both singularities. Residues:

16

\quad \operatorname{Res}_{z=0} = -1, \qquad \operatorname{Res}_{z=1} = \lim_{z\to1}\frac{1}{z} = 1.

Derivation hint: For simple poles, $\operatorname{Res}_{z=p}\frac{g}{h}=g(p)/h'(p)$ , or read residues from [13]. By the Residue Theorem [5]:

17

\quad \oint_{|z|=3}\frac{dz}{z(z-1)} = 2\pi i\,(-1 + 1) = 0.

Derivation hint: Sum the residues inside the contour and multiply by $2\pi i$ .

\boxed{\;\oint_{|z|=3}\frac{dz}{z(z-1)} = 0\;}

Note: a single function can have different Laurent series ([14] vs [15]) depending on which annulus you sit in — the expansion is unique per annulus, not globally.

⚠️ A function does NOT have one Laurent series — it has one per annulus

The uniqueness in Laurent's theorem is relative to a fixed annulus. The same $f$ can possess radically different expansions on $0<|z-z_0|<R_1$ versus $R_2<|z-z_0|<R_3$ , as Example 2 shows. Always specify the annulus before writing coefficients. A common error is to compute a residue from the "wrong" expansion: the residue at $z_0$ comes only from the Laurent series valid on a punctured disk $0<|z-z_0|<\rho$ — the one with the smallest inner radius reaching $z_0$ .

⚠️ The order of a pole is not "how big the denominator looks"

Cancellation matters. For $f(z) = \frac{\sin z}{z^2}$ , the denominator suggests order 2, but $\sin z = z - z^3/6 + \cdots$ , so $\frac{\sin z}{z^2} = \frac1z - \frac{z}{6}+\cdots$ — a simple pole (order 1), not order 2. Always expand the numerator before declaring the order. Likewise, $\frac{\sin z}{z}$ has a removable singularity at $0$ , not a pole.

CounterexampleAn essential singularity defeats the limit test

For removable singularities and poles, $\lim_{z\to z_0}|f(z)|$ behaves predictably (finite, or $+\infty$ respectively). One might hope all isolated singularities are so tame. But consider $f(z)=e^{1/z}$ at $z_0=0$ :

18

\quad e^{1/z} = \sum_{n=0}^{\infty}\frac{1}{n!\,z^n} = 1 + \frac{1}{z}+\frac{1}{2z^2}+\frac{1}{6z^3}+\cdots

Derivation hint: Substitute $w=1/z$ into the exponential series $e^w=\sum w^n/n!$ . The principal part is infinite, so $0$ is essential. Approaching along the positive real axis, $f\to +\infty$ ; along the negative real axis, $f\to 0$ ; along the imaginary axis, $f$ oscillates on the unit circle. The limit does not exist in any sense, and by the Casorati–Weierstrass theorem the image of every punctured neighborhood of $0$ is dense in $\mathbb{C}$ . Picard's Great Theorem sharpens this: $f$ attains every complex value (with at most one exception — here $0$ ) infinitely often in any neighborhood of $0$ .

connection

Laurent series sit at the crossroads of several core areas. They generalize the Taylor Series (which is the special case with no principal part, valid on a full disk). The residue extracted from the principal part powers the Residue Theorem, the workhorse of Contour Integration. The classification of singularities feeds directly into the study of Meromorphic Functions (functions whose only singularities are poles) and the theory of the Riemann Sphere, where one analyzes the point at infinity by the Laurent series in $1/z$ . The coefficient formula [3] is a direct descendant of the Cauchy Integral Formula, and the orthogonality relation [9] underlying uniqueness is essentially a Fourier Series in disguise — substituting $z = z_0 + \rho e^{i\theta}$ turns the Laurent coefficients into Fourier coefficients on a circle.

🔬 Where this appears

⏳ History

- 1814–1825: Augustin-Louis Cauchy develops the theory of complex integration, proving the Cauchy integral theorem and integral formula — the bedrock from which Laurent's coefficient formula [3] descends.

- 1826: Cauchy introduces the calcul des résidus ("calculus of residues"), recognizing the special role of the

a_{-1}

coefficient in integration.

- 1841: Karl Weierstrass writes (but does not publish) a paper containing an expansion equivalent to the Laurent series, in his work on power series and analytic continuation.

- 1843: Pierre Alphonse Laurent submits his memoir to the Académie des Sciences in Paris, proving the annular expansion theorem. Cauchy reports favorably but the work misses the prize deadline and is not formally published.

- 1844: Cauchy publishes a note describing Laurent's result, securing the attribution that gives the series its name.

- 1868: Felice Casorati, and independently Weierstrass, establish that near an essential singularity the function values are dense — the Casorati–Weierstrass theorem.

- 1879: Charles Émile Picard proves his Great Theorem, dramatically strengthening Casorati–Weierstrass: near an essential singularity,

f

takes every complex value (save at most one) infinitely often.

- Late 19th c.: Bernhard Riemann's geometric viewpoint (the Riemann sphere, classification of points at infinity) and the systematic theory of meromorphic functions fold Laurent expansions into the mainstream of function theory.

- 20th c.–present: Laurent series become standard machinery in renormalization (QFT), control theory (Z-transform), and analytic number theory.

open-problem · Boundaries of the theory

Laurent's theorem is fully settled for isolated singularities, but the boundary cases are instructive and some neighboring questions remain subtle:

Non-isolated singularities fall outside the theory entirely. For $f(z) = \frac{1}{\sin(1/z)}$ , the points $z = 1/(n\pi)$ are poles accumulating at $0$ ; the origin is a singularity that is not isolated, so no Laurent series about $0$ exists on any punctured disk. Classifying such accumulation points requires cluster-set theory rather than principal parts.
Natural boundaries. Some functions (e.g. lacunary series like $\sum z^{2^n}$ ) cannot be analytically continued past the unit circle at all — every boundary point is singular. Predicting which power series have natural boundaries connects to deep questions in analytic continuation and the Fabry gap theorem; precise structural characterizations remain an active research direction.
Effective Picard exceptions. Picard's theorem guarantees at most one omitted value near an essential singularity, but determining the exceptional value and the distribution of solutions $f(z) = w$ leads into Nevanlinna value-distribution theory, where sharp quantitative versions are still being refined.

Laurent Series and Isolated Singularities

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also