Taylor and Laurent series

A function that is perfectly smooth — analytic — near a point behaves like an infinite polynomial there: its Taylor series. But complex functions almost always carry singularities somewhere, and near a singularity no ordinary power series can work, because a power series in (z−a)(z-a) is itself analytic at aa. The fix is the Laurent series, a two-sided expansion valid in a ring-shaped annulus around the trouble spot, with negative powers of (z−a)(z-a) soaking up exactly the bad behaviour a pure Taylor series cannot represent. This article builds both expansions from the same coefficient formula, shows why that formula collapses to the familiar f(n)(a)/n!f^{(n)}(a)/n! only for non-negative powers, and works a rational function in two different annuli to show that the "same" function genuinely has different series depending on where you expand it. The payoff is the coefficient c−1c_{-1}, the residue, which is the entire subject of the next topic.

The core insight

An analytic function equals its Taylor series inside the largest disk free of singularities, and near an isolated singularity it equals a two-sided Laurent series — analytic part plus a principal part of negative powers — valid in an annulus, with both series' coefficients extractable either by the same contour-integral formula or, far more practically, by algebraic manipulation of known Maclaurin series.

The core idea

Taylor series come from analyticity in a disk. If ff is analytic throughout ∣z−a∣<R|z-a| < R, it is represented there by

f(z)=∑n=0∞cn(z−a)n,cn=f(n)(a)n!,f(z) = \sum_{n=0}^{\infty} c_n (z-a)^n, \qquad c_n = \frac{f^{(n)}(a)}{n!},

where: cnc_n are the Taylor coefficients and RR, the radius of convergence, is the distance from aa to the nearest point where ff fails to be analytic. This is not a new idea over Unit 1's real Maclaurin/Taylor series — it is the same coefficient formula, now justified rigorously by Cauchy's Integral Formula rather than by repeated differentiation, and RR is found by the same ratio test used for real series (1/R=lim⁡∣cn+1/cn∣1/R = \lim |c_{n+1}/c_n|), except now the "nearest bad point" can be anywhere in the complex plane, not just on the real line.

Laurent series come from analyticity in an annulus. When ff has an isolated singularity at z=az=a, it cannot have a Taylor series centred there, but it is still analytic in a punctured neighbourhood r<∣z−a∣<Rr < |z-a| < R, and on that annulus

f(z)=∑n=−∞∞cn(z−a)n=∑n=0∞cn(z−a)n⏟analytic part+∑n=1∞c−n(z−a)n⏟principal part.f(z) = \sum_{n=-\infty}^{\infty} c_n (z-a)^n = \underbrace{\sum_{n=0}^{\infty} c_n (z-a)^n}_{\text{analytic part}} + \underbrace{\sum_{n=1}^{\infty} \frac{c_{-n}}{(z-a)^n}}_{\text{principal part}}.

where: the analytic part behaves like an ordinary Taylor series, while the principal part carries every negative power and encodes the singularity itself. The single coefficient c−1c_{-1} is singled out and named the residue of ff at aa — it is the only coefficient that survives integration around a closed loop, which is exactly why it drives the residue calculus that follows this topic.

One formula, two behaviours. Both series share the same coefficient definition, cn=12πi∮Cf(z)(z−a)n+1 dzc_n = \frac{1}{2\pi i}\oint_C \frac{f(z)}{(z-a)^{n+1}}\,dz, for CC any positively-oriented simple closed contour lying in the region of validity and encircling aa once. For n≥0n \ge 0 and ff analytic at aa, this integral collapses to the differentiation shortcut f(n)(a)/n!f^{(n)}(a)/n!; for negative nn, or whenever ff is not even defined at aa, no differentiation shortcut exists and the integral (or, in practice, algebra) is the only route.

Figure
A Taylor disk centred at a point next to a Laurent annulus around a singularity, with the disk boundary reaching the nearest bad point and the annulus sitting between two singularities
Taylor disk versus Laurent annulus Taylor series: single disk a nearest singularity R Laurent series: annulus between two singularities a r R The annulus excludes the centre — Laurent, not Taylor, is forced there
Derivation AWhere the coefficient formula c_n comes from

The goal is to derive, from the Laurent series itself, why multiplying f(z)f(z) by (z−a)−m−1(z-a)^{-m-1} and integrating term by term around a closed loop isolates a single coefficient — turning the assumed series f(z)=∑n=−∞∞cn(z−a)nf(z) = \sum_{n=-\infty}^{\infty} c_n (z-a)^n into the extraction formula cn=12πi∮Cf(z)(z−a)n+1 dzc_n = \frac{1}{2\pi i}\oint_C \frac{f(z)}{(z-a)^{n+1}}\,dz that both the Taylor and the Laurent case rely on, and explaining exactly why that formula reduces to f(n)(a)/n!f^{(n)}(a)/n! only when n≥0n \ge 0.

Step 1. Take the key integral as given: for any integer kk and CC a positively-oriented simple closed loop encircling aa once, ∮C(z−a)k dz=2πi\oint_C (z-a)^k\,dz = 2\pi i if k=−1k = -1, and 00 for every other integer kk (positive or negative). This single fact — provable directly by parametrising z−a=εeiθz - a = \varepsilon e^{i\theta} on a small circle — is the entire engine of the derivation.

Step 2. Multiply the assumed series by (z−a)−m−1(z-a)^{-m-1} for a fixed target integer mm, and integrate term by term around CC (legitimate because the series converges uniformly on CC, which lies inside the annulus of validity): ∮Cf(z)(z−a)−m−1 dz=∑n=−∞∞cn∮C(z−a)n−m−1 dz\oint_C f(z)(z-a)^{-m-1}\,dz = \sum_{n=-\infty}^{\infty} c_n \oint_C (z-a)^{n-m-1}\,dz.

Step 3. Apply Step 1 to every term on the right. The exponent n−m−1n - m - 1 equals −1-1 exactly when n=mn = m, and every other term integrates to zero. So the entire infinite sum collapses to the single surviving term: ∮Cf(z)(z−a)−m−1 dz=cm⋅2πi\oint_C f(z)(z-a)^{-m-1}\,dz = c_m \cdot 2\pi i.

Step 4. Divide by 2πi2\pi i and rename mm as nn to get cn=12πi∮Cf(z)(z−a)n+1 dzc_n = \frac{1}{2\pi i}\oint_C \frac{f(z)}{(z-a)^{n+1}}\,dz — valid for every integer nn, positive, negative, or zero, and for both the Taylor case (ff analytic at aa) and the Laurent case (annulus, ff possibly singular at aa). For n≥0n \ge 0 with ff analytic at aa, this integral is exactly the derivative form of Cauchy's Integral Formula, so it equals f(n)(a)/n!f^{(n)}(a)/n!; for n<0n < 0 that reduction is impossible because f(n)(a)f^{(n)}(a) may not even exist, which is precisely why negative-index coefficients must come from this integral (or, in practice, from algebra) and never from differentiating ff.

Worked example

Problem. Find the Laurent series of f(z)=1z(z−1)f(z) = \dfrac{1}{z(z-1)} in the annulus 0<∣z∣<10 < |z| < 1, and then in the exterior region ∣z∣>1|z| > 1, and compare the two.

Step 1. Partial fractions give f(z)=−1z+1z−1f(z) = \dfrac{-1}{z} + \dfrac{1}{z-1}, valid everywhere except z=0,1z=0,1; only the second term needs expanding, and how it expands depends on which region we are in.

Step 2. In 0<∣z∣<10 < |z| < 1: write 1z−1=−11−z\dfrac{1}{z-1} = -\dfrac{1}{1-z} and, since ∣z∣<1|z|<1, use the geometric series 11−z=∑n=0∞zn\dfrac{1}{1-z} = \sum_{n=0}^{\infty} z^n. So f(z)=−1z−∑n=0∞zn=−1z−1−z−z2−⋯f(z) = -\dfrac1z - \sum_{n=0}^{\infty} z^n = -\dfrac1z - 1 - z - z^2 - \cdots. This matches the inner-annulus result: the principal part is the single term −1/z-1/z, giving residue c−1=−1c_{-1} = -1 at z=0z=0, and the analytic part is −∑n≥0zn-\sum_{n\ge0} z^n.

Step 3. In ∣z∣>1|z| > 1: now ∣z∣<1|z|<1 no longer holds, so the previous geometric series diverges. Instead write 1z−1=1z⋅11−1/z\dfrac{1}{z-1} = \dfrac{1}{z}\cdot\dfrac{1}{1-1/z}, and since ∣1/z∣<1|1/z| < 1 in this region, expand 11−1/z=∑n=0∞z−n\dfrac{1}{1-1/z} = \sum_{n=0}^{\infty} z^{-n}. So 1z−1=1z∑n=0∞z−n=∑n=0∞z−n−1=1z+1z2+1z3+⋯\dfrac{1}{z-1} = \dfrac1z\sum_{n=0}^{\infty} z^{-n} = \sum_{n=0}^{\infty} z^{-n-1} = \dfrac1z + \dfrac{1}{z^2} + \dfrac{1}{z^3} + \cdots.

Step 4. Combine with the −1/z-1/z term: f(z)=−1z+1z+1z2+1z3+⋯=1z2+1z3+⋯=∑n=2∞1znf(z) = -\dfrac1z + \dfrac1z + \dfrac{1}{z^2} + \dfrac{1}{z^3} + \cdots = \dfrac{1}{z^2} + \dfrac{1}{z^3} + \cdots = \sum_{n=2}^{\infty} \dfrac{1}{z^n}, valid for ∣z∣>1|z|>1. The exterior Laurent series is f(z)=∑n=2∞z−nf(z) = \sum_{n=2}^{\infty} z^{-n}, purely a principal part with no analytic part at all, and its residue is c−1=0c_{-1} = 0 — strikingly different from the inner series, where c−1=−1c_{-1} = -1. Both series expand the same f(z)f(z); the annulus is not a detail, it is part of the answer.

Competitive stretch

Notice the residue itself changed between annuli, −1-1 inside versus 00 outside. This is not a contradiction: the residue used in residue calculus is always the coefficient of (z−a)−1(z-a)^{-1} in the series valid on the specific annulus immediately surrounding the isolated singularity at aa (here, 0<∣z∣<10<|z|<1 around a=0a=0), never in a series valid "outside everything," which is why the next topic always specifies the region before quoting a residue.

Common traps

  • Using f(n)(a)/n!f^{(n)}(a)/n! for a negative-index Laurent coefficient. That shortcut only comes from Cauchy's derivative formula, which requires ff analytic at aa; for c−1,c−2,…c_{-1}, c_{-2}, \dots, or whenever ff is singular at aa, only the contour integral or algebraic manipulation works.
  • Quoting "the" Laurent series of a function without stating the annulus. As the worked example shows, 1/[z(z−1)]1/[z(z-1)] has one series for 0<∣z∣<10<|z|<1 and a completely different one for ∣z∣>1|z|>1 — always state and respect the region of validity.
  • Mixing up a Taylor disk with a Laurent annulus. A Taylor series is tied to one disk out to the nearest singularity; a Laurent series is tied to a specific ring between two singularities (or a singularity and infinity), and neither series extends across a singularity it was built around.
  • Sign or algebra slips when substituting into ∑zn\sum z^n. Rewriting 1/(z−1)1/(z-1) as −1/(1−z)-1/(1-z) versus 1/z⋅1/(1−1/z)1/z \cdot 1/(1-1/z) is easy to get backwards; always check the first two or three terms of a new series by direct division.

Ask the tutor: If the shift between the inner-annulus and exterior-annulus expansions of the same rational function still feels slippery, ask the AI tutor to walk through one more function with three poles and three annuli.

Why cant I just always use f superscript n of a over n factorial to find Laurent coefficients?

That formula is a shortcut derived from Cauchys Integral Formula, and it only works when f is actually analytic at the expansion point a, because it requires differentiating f there n times. In a genuine Laurent expansion, f is typically singular at a itself, so f of a and its derivatives may not even exist. For negative-index coefficients especially, you must fall back on the general contour-integral definition, or in practice on algebraic manipulation of known series, never on differentiating a function that is not even defined at that point.

How do I know which annulus to expand in if the problem does not say?

Look at where the function is analytic. A rational function like 1 over z times z minus 1 is analytic everywhere except at its poles, here z equals 0 and z equals 1, so the plane splits into three natural regions bounded by those poles and infinity: 0 less than absolute z less than 1, absolute z greater than 1, and any punctured disk around the other pole. A problem should specify which region it wants; if it does not, state the region you are choosing before writing the series, since as shown above the same function has genuinely different series in each region.

Why does the residue of the same function come out different in two different annuli?

Because the residue is not a property of the function alone at a point, it is a property of the specific Laurent series valid on a specific annulus around that point. The inner annulus 0 less than absolute z less than 1 is the one that actually surrounds the isolated singularity at z equals 0, so its coefficient c minus one, here minus 1, is the residue used in residue calculus. The exterior annulus absolute z greater than 1 does not immediately surround that singularity, so its c minus one coefficient answers a different question and should not be quoted as the residue at z equals 0.

Before this

Where this leads