Taylor and Laurent Series

⚡ Key Takeaways

Taylor and Laurent series represent analytic functions as power series. A function holomorphic on a disk equals its Taylor series there; a function holomorphic on an annulus equals its Laurent series (including negative powers). Used to classify singularities, compute residues, and approximate functions. Developed by Brook Taylor (1715), Colin Maclaurin, and Pierre Alphonse Laurent (1843).

💡 The Core Insight

Holomorphy is equivalent to local power-series representability. On a disk you need only non-negative powers (Taylor); on an annulus around an isolated singularity you also need negative powers (Laurent). The type of singularity is read directly off how many negative-power terms appear: none (removable), finitely many (pole), infinitely many (essential).

DefinitionTaylor Series

Let $f$ be holomorphic on an open disk $D(z_0, R) = \{z : |z - z_0| < R\}$ . Its Taylor series about $z_0$ is

1

\quad f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n, \qquad a_n = \frac{f^{(n)}(z_0)}{n!}.

Derivation hint: Differentiate the series $n$ times term-by-term and evaluate at $z=z_0$ ; all terms vanish except the $n$ -th, giving $f^{(n)}(z_0) = n!\,a_n$ . Here $z, z_0, a_n \in \mathbb{C}$ , $f^{(n)}$ is the $n$ -th complex derivative, and $n! = 1\cdot 2 \cdots n$ with $0! = 1$ . When $z_0 = 0$ this is called the Maclaurin series.

DefinitionLaurent Series

Let $f$ be holomorphic on an annulus $A = \{z : r < |z - z_0| < R\}$ with $0 \le r < R \le \infty$ . Its Laurent series is

2

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

Derivation hint: Use Cauchy's integral formula over two concentric circles bounding the annulus, expanding the inner kernel as a geometric series in $(z-z_0)^{-1}$ and the outer in $(z-z_0)$ . The coefficients are given by

3

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

Derivation hint: Multiply [2] by $(\zeta-z_0)^{-m-1}$ , integrate over a circle $\gamma$ in the annulus, and apply $\oint (\zeta-z_0)^{k}\,d\zeta = 2\pi i\,\delta_{k,-1}$ . where $\gamma$ is any positively-oriented circle $|z - z_0| = \rho$ with $r < \rho < R$ . The integer $\delta_{k,-1}$ in the hint is the Kronecker delta.

TheoremCauchy–Taylor Theorem

If $f$ is holomorphic on $D(z_0, R)$ , then the Taylor series [1] converges to $f(z)$ for every $z$ with $|z - z_0| < R$ , and

4

\quad a_n = \frac{f^{(n)}(z_0)}{n!} = \frac{1}{2\pi i}\oint_{|\zeta - z_0| = \rho} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}\, d\zeta, \quad 0 < \rho < R.

Derivation hint: This is [3] restricted to $n \ge 0$ ; for $n < 0$ holomorphy on the full disk makes the integrand holomorphic, so those coefficients vanish by Cauchy's theorem.

TheoremLaurent's Theorem

Every function holomorphic on an annulus $A$ admits a unique Laurent expansion [2] there, converging absolutely on $A$ and uniformly on compact subsets. The radius of convergence of the analytic part is $\ge R$ and the principal part converges for $|z - z_0| > r$ .

ProofCauchy Integral Formula to Power Series

Steps deriving the Taylor expansion:

Step 1. Fix $z$ with $|z - z_0| < \rho < R$ . By Cauchy's integral formula,

5

\quad f(z) = \frac{1}{2\pi i} \oint_{|\zeta - z_0| = \rho} \frac{f(\zeta)}{\zeta - z}\, d\zeta.

Derivation hint: Standard Cauchy formula; $z$ lies inside the circle $|\zeta - z_0| = \rho$ .

Step 2. Rewrite the kernel by factoring:

6

\quad \frac{1}{\zeta - z} = \frac{1}{(\zeta - z_0) - (z - z_0)} = \frac{1}{\zeta - z_0}\cdot \frac{1}{1 - \frac{z - z_0}{\zeta - z_0}}.

Derivation hint: Add and subtract $z_0$ in the denominator; factor out $(\zeta - z_0)$ .

Step 3. Since $\left|\frac{z - z_0}{\zeta - z_0}\right| = \frac{|z - z_0|}{\rho} < 1$ , expand as a geometric series:

7

\quad \frac{1}{\zeta - z} = \sum_{n=0}^{\infty} \frac{(z - z_0)^n}{(\zeta - z_0)^{n+1}}.

Derivation hint: Use $\frac{1}{1-w} = \sum w^n$ with $w = \frac{z-z_0}{\zeta-z_0}$ ; convergence is uniform in $\zeta$ on the circle.

Step 4. Substitute [7] into [5] and exchange sum and integral (justified by uniform convergence):

8

\quad f(z) = \sum_{n=0}^{\infty} \left( \frac{1}{2\pi i} \oint \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}\, d\zeta \right) (z - z_0)^n.

Derivation hint: Interchange $\sum$ and $\oint$ legitimately because the series converges uniformly on the compact contour.

Step 5. The bracketed coefficient is exactly $a_n$ of [4], completing the proof. For Laurent's theorem, one applies this argument twice — once on the outer boundary (giving non-negative powers) and once on the inner boundary with the kernel expanded in $(z-z_0)^{-1}$ (giving the principal part). $\blacksquare$

∎

mathematician · Pierre Alphonse Laurent | 1813–1854

A French military engineer and officer in the Corps of Engineers, Laurent worked on harbor construction at Le Havre. In 1843 he submitted a memoir to the French Academy generalizing Taylor's theorem to the annulus, introducing what we now call the Laurent series. Cauchy, who reviewed it, noted he had a related unpublished result, but acknowledged Laurent's contribution. The paper was not published in his lifetime; Laurent died at 41. His name endures on one of the most-used tools in complex analysis.

🧠 Intuition

Think of a power series as a "fingerprint" of a function recorded at a single point. In real analysis the fingerprint can lie — $e^{-1/x^2}$ has every derivative zero at $0$ yet is nonzero nearby. In complex analysis, holomorphy is so rigid that the fingerprint always tells the truth: the disk version is the Taylor series, faithful out to the nearest singularity. The Laurent series is what you get when there's a singularity at the center. You stand in a ring around the trouble spot. The negative powers (the principal part) describe how fast the function blows up as you approach the hole; the non-negative powers describe the smooth background. The coefficient $c_{-1}$ — the residue — is special because it is the only term whose contour integral survives, making it the bridge to integration.

Figure

Regions of validity: a Taylor disk versus a Laurent annulus around an isolated singularity

ExampleMaclaurin series of the exponential

Given: $f(z) = e^z$ , expand about $z_0 = 0$ .

Step 1. All derivatives: $f^{(n)}(z) = e^z$ , so $f^{(n)}(0) = 1$ for all $n$ .

Step 2. Coefficients $a_n = f^{(n)}(0)/n! = 1/n!$ .

Step 3. Series:

9

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

Derivation hint: Insert $a_n = 1/n!$ into [1].

Step 4. Radius of convergence: $e^z$ is entire (no singularities), so $R = \infty$ — valid for all $z \in \mathbb{C}$ .

\boxed{\,e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}, \quad R = \infty\,}

ExampleLaurent series and a residue

Given: $f(z) = \dfrac{1}{z(z-1)}$ , find the Laurent expansion in the annulus $0 < |z| < 1$ and the residue at $z=0$ .

Step 1. Partial fractions:

10

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

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

Step 2. For $|z| < 1$ , expand $\frac{1}{z-1} = -\frac{1}{1-z} = -\sum_{n=0}^\infty z^n$ .

Step 3. Combine:

11

\quad f(z) = -\frac{1}{z} - \sum_{n=0}^{\infty} z^n = -\frac{1}{z} - 1 - z - z^2 - \cdots, \quad 0 < |z| < 1.

Derivation hint: Add the principal part $-1/z$ from [10] to the geometric expansion.

Step 4. The principal part is the single term $-1/z$ ; the coefficient of $z^{-1}$ is the residue.

\boxed{\,\operatorname{Res}_{z=0} f = c_{-1} = -1\,}

Check: $z=0$ is a simple pole; $\operatorname{Res}_{z=0} = \lim_{z\to 0} z\,f(z) = \lim_{z\to 0}\frac{1}{z-1} = -1$ . ✓

⚠️ Common mistake

The Laurent series is not unique unless you fix the annulus. The same function can have different Laurent series in different annuli centered at the same point. For $f(z) = \frac{1}{z(z-1)}$ the expansion in $0<|z|<1$ (above) differs from the one valid in $|z|>1$ . Always state the region. Uniqueness holds only within a fixed annulus.

⚠️ Common mistake

Negative powers do NOT automatically mean a pole. A pole of order $m$ has finitely many negative-power terms (highest being $(z-z_0)^{-m}$ ). If the principal part has infinitely many nonzero terms, the singularity is essential, not a pole. Confusing these misreads the function's behavior entirely.

CounterexampleReal analyticity is strictly weaker than complex differentiability

Consider the real function

12

\quad g(x) = \begin{cases} e^{-1/x^2}, & x \ne 0 \\ 0, & x = 0 \end{cases}.

Derivation hint: Show by induction that $g^{(n)}(0)=0$ using L'Hôpital on $e^{-1/x^2}/x^k$ . Every derivative at $0$ is zero, so its Maclaurin series is identically $0$ — yet $g(x) \ne 0$ for $x \ne 0$ . The Taylor series converges everywhere but represents the function nowhere except the origin. This pathology is impossible for a holomorphic complex function: if a complex $f$ has all derivatives zero at $z_0$ , then $f \equiv 0$ on the whole connected domain. Complex differentiability is genuinely stronger.

connection

Taylor and Laurent series are the engine behind the Residue Theorem: only the coefficient $c_{-1}$ survives contour integration, turning integrals into algebra. The classification of Isolated Singularities (removable / pole / essential) is defined by the structure of the principal part. The existence of Taylor expansions is the content of Cauchy's Integral Formula differentiated repeatedly. Convergence radius is governed by the Radius of Convergence formula $1/R = \limsup |a_n|^{1/n}$ , and the nearest singularity in the plane caps $R$ — explaining why the real series for $1/(1+x^2)$ diverges past $|x|=1$ despite the function being smooth on all of $\mathbb{R}$ . They connect to Analytic Continuation, where overlapping power-series representations extend a function's domain.

🔬 Where this appears

⏳ History

- 1668 — Nicolaus Mercator publishes the series for

\ln(1+x)

, an early power-series result.

- 1715 — Brook Taylor states the general expansion theorem in Methodus Incrementorum Directa et Inversa, though without convergence concerns.

- 1742 — Colin Maclaurin popularizes the special case

z_0 = 0

in his Treatise of Fluxions; the name "Maclaurin series" follows.

- 1797 — Lagrange attempts to base all of calculus on power series in Théorie des fonctions analytiques, coining "analytic function."

- 1821 — Cauchy's Cours d'analyse sets rigorous foundations for convergence and limits.

- 1825 — Cauchy develops the integral theorem and formula, from which the Taylor coefficients [4] emerge rigorously.

- 1843 — Pierre Alphonse Laurent submits his memoir generalizing the expansion to annuli (published posthumously, 1863).

- 1851 — Riemann's dissertation deepens the geometric theory of holomorphic functions.

- 1876 — Weierstrass and Casorati clarify the behavior at essential singularities (the Casorati–Weierstrass theorem), completing the singularity classification.

open-problem

While the undergraduate theory is complete, the boundary behavior of power series remains subtle. A Taylor series convergent on $|z-z_0| < R$ may converge, diverge, or behave wildly on the boundary circle itself, and characterizing this is delicate (Abel's theorem, Tauberian theorems give partial answers). For lacunary series like $\sum z^{2^n}$ , the boundary circle is a natural boundary — analytic continuation across it is impossible everywhere. More broadly, deciding whether a given power series admits continuation, and locating its singularities from coefficient asymptotics, connects to deep questions in analytic number theory (e.g. the location of zeros via Dirichlet series, related to the Riemann Hypothesis). The interplay between coefficient growth and singularity structure is an active research theme.

Formal Statement

Why It Holds

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also