Home
Articles
Sequences & Series
Laurent vs Taylor

// Article · Sequences & series

What Is a Laurent Series and How Is It Different from a Taylor Series? (Explained with Pictures)

Q: When should I use a Laurent series instead of a Taylor series?

When you need to expand a function around a point where it is not analytic (a singularity), or on a region that has a singularity inside it, such as the annulus 1 < |z| < 2 of 1/((z-1)(z-2)). If the function is analytic on the whole disc, the Laurent series simply becomes the Taylor series.

Q: What is the principal part?

It is the group of negative-power terms sum_{nge1} a_{-n}(z-c)^{-n} of a Laurent series. It describes how the function blows up near c: no principal part means a removable singularity, finitely many terms mean a pole, infinitely many mean an essential singularity.

Q: Why does the residue matter?

The residue is the coefficient a_{-1}, the only term that survives integration around a closed curve. That gives the residue theorem oint f dz = 2pi isumoperatorname{Res}, used for hard real integrals, the inverse Laplace transform and the inverse z-transform.

Q: How is the z-transform related to Laurent series?

The z-transform is defined as X(z) = sum x[n]z^{-n}, which is a Laurent series about 0. The signal x[n] is the list of coefficients and the region of convergence (ROC) is the annulus of the series. The same X(z) with a different ROC gives a different signal.

A Taylor series writes a function as an endless polynomial around a point where the function is well behaved. In engineering, though, we usually care about the points where a function blows up, such as the poles of a transfer function. A Laurent series fixes this with one idea: allow negative powers. This article compares the two series visually and ends with the reason electrical engineers use Laurent series every day, through the z-transform.

By Ajarn ModUpdated 24 Oct 2026

In 30 seconds

Taylor: $\sum_{n\ge0} a_n(z-c)^n$ converges on a disc · Laurent: $\sum_{n=-\infty}^{\infty} a_n(z-c)^n$ converges on an annulus
One function can have several Laurent series, one per annulus
The coefficient $a_{-1}$ is the residue · the z-transform is a Laurent series about 0

Side by side

	Taylor series	Laurent series
Form	$\sum_{n=0}^{\infty} a_n(z-c)^n$	$\sum_{n=-\infty}^{\infty} a_n(z-c)^n$
Region of convergence	Disc $\|z - c\| < R$	Annulus $r_1 < \|z - c\| < r_2$
Requires	f analytic at c	f analytic on the annulus (anything may happen at c)
Coefficients	$a_n = f^{(n)}(c)/n!$	$a_n = \frac{1}{2\pi i}\oint\frac{f(z)}{(z-c)^{n+1}}dz$
How many per function	One per centre c	One per annulus
Used for	Approximation, limits, linearisation	Classifying singularities, residues, integrals, the z-transform

Why negative powers are needed

Try to expand $\dfrac{e^z}{z^2}$ about 0 as a Taylor series: you cannot, because $f(0)$ is undefined. But divide the series $e^z = 1 + z + \tfrac{z^2}{2} + \cdots$ by $z^2$ and the expansion appears at once:

$\frac{e^z}{z^2} = \underbrace{\frac{1}{z^2} + \frac{1}{z}}_{\text{principal part}} + \underbrace{\frac12 + \frac{z}{6} + \frac{z^2}{24} + \cdots}_{\text{Taylor-like part}},\qquad 0 < |z| < \infty$

The first two negative-power terms describe everything that happens near 0: the function blows up like $1/z^2$ (a pole of order 2), and the $1/z$ term has coefficient 1, which is the residue. The rest is an ordinary, harmless power series.

One function, several series: it is all about the annulus

This is what confuses students most. $f(z) = \dfrac{1}{(z-1)(z-2)}$ has poles at 1 and 2. The circles through the poles split the plane around 0 into three regions, and each region has its own series.

Three regions about 0: (a) the disc |z| < 1 gives the Taylor series, (b) the annulus 1 < |z| < 2, (c) the outside |z| > 2.

There is one trick: split into partial fractions, $f = \dfrac{1}{z-2} - \dfrac{1}{z-1}$ , and expand each with the geometric series, where a pole inside the region gives negative powers and a pole outside gives positive powers.

(a) $|z| < 1$ : $f = \tfrac12 + \tfrac34 z + \tfrac78 z^2 + \cdots$ (no negative powers = Taylor)
(b) $1 < |z| < 2$ : $f = \cdots - \tfrac{1}{z^2} - \tfrac{1}{z} - \tfrac12 - \tfrac{z}{4} - \tfrac{z^2}{8} - \cdots$ (both directions)
(c) $|z| > 2$ : $f = \tfrac{1}{z^2} + \tfrac{3}{z^3} + \tfrac{7}{z^4} + \cdots$ (negative powers only)

You can check by substitution: at $z = 1.5$ series (b) converges to $f(1.5) = -4$ , but series (a) diverges there, even though f(1.5) is perfectly ordinary. Drag the test point yourself in the Laurent annulus lab (in Thai).

Reading the type of singularity from the principal part

Removable: bounded · pole: blows up from every direction · essential: the behaviour depends on the direction of approach.

No negative powers → removable, e.g. $\tfrac{\sin z}{z} = 1 - \tfrac{z^2}{6} + \cdots$
Finitely many negative powers, highest m → pole of order m, e.g. $\tfrac{\sin z}{z^4} = \tfrac{1}{z^3} - \tfrac{1}{6z} + \cdots$ (order 3)
Infinitely many negative powers → essential singularity, e.g. $e^{1/z} = 1 + \tfrac1z + \tfrac{1}{2z^2} + \cdots$

The residue: the one number left after integration

When you integrate a Laurent series around a circle inside its annulus, every term $(z-c)^n$ gives zero except $n = -1$ , which gives $2\pi i$ . So $\oint f\,dz = 2\pi i\,a_{-1}$ ; the number $a_{-1}$ is called the residue (what remains). It is also where the “cover-up” trick for partial fractions in the inverse Laplace transform (in Thai) comes from — it is really computing the residue at each pole.

Why electrical engineers need it: the z-transform

The z-transform of a digital signal, $X(z) = \sum x[n]z^{-n}$ , is literally a Laurent series about 0. The signal x[n] is the coefficient of $z^{-n}$ , and what signals-and-systems textbooks call the “ROC” is the annulus of convergence. Take $X(z) = \dfrac{z}{z - 0.5}$ :

ROC $|z| > 0.5$ (outside the pole): $X = 1 + 0.5z^{-1} + 0.25z^{-2} + \cdots$ ⇒ $x[n] = 0.5^n u[n]$ , a right-sided signal
ROC $|z| < 0.5$ (inside the pole): expand in positive powers ⇒ $x[n] = -0.5^n u[-n-1]$ , a left-sided signal

Same formula, two signals, different annuli. And the system is stable when the annulus contains the unit circle, which is exactly absolute convergence, $\sum|h[n]| < \infty$ . A full three-region example is in the z-transform section (in Thai).

Summary: which one to use

Approximating, linearising, finding limits → Taylor
Behaviour near a pole, residues, contour integrals, inverse transforms → Laurent
If the region of interest contains no singularity, the two are the same series

Continue in the library (in Thai)

Frequently asked questions

When should I use a Laurent series instead of a Taylor series?

When you need to expand a function around a point where it is not analytic (a singularity), or on a region that has a singularity inside it, such as the annulus $1 < |z| < 2$ of $1/((z-1)(z-2))$ . If the function is analytic on the whole disc, the Laurent series simply becomes the Taylor series.

What is the principal part?

It is the group of negative-power terms $\sum_{n\ge1} a_{-n}(z-c)^{-n}$ of a Laurent series. It describes how the function blows up near c: no principal part means a removable singularity, finitely many terms mean a pole, infinitely many mean an essential singularity.

Why does the residue matter?

The residue is the coefficient $a_{-1}$ , the only term that survives integration around a closed curve. That gives the residue theorem $\oint f\,dz = 2\pi i\sum\operatorname{Res}$ , used for hard real integrals, the inverse Laplace transform and the inverse z-transform.

How is the z-transform related to Laurent series?

The z-transform is defined as $X(z) = \sum x[n]z^{-n}$ , which is a Laurent series about 0. The signal x[n] is the list of coefficients and the region of convergence (ROC) is the annulus of the series. The same X(z) with a different ROC gives a different signal.