Pole — Definition, Formula & Examples

A pole is a type of isolated singularity of a complex function where the function's value grows without bound as you approach that point. Unlike essential singularities, poles have a predictable, structured blowup described by a finite number of negative-power terms in the Laurent series.

A function $f(z)$ has a pole of order $n$ at $z = z_0$ if $z_0$ is an isolated singularity and the Laurent series of $f$ about $z_0$ has the form $f(z) = \sum_{k=-n}^{\infty} a_k (z - z_0)^k$ where $a_{-n} \neq 0$ and $n$ is a positive integer. Equivalently, $\lim_{z \to z_0} (z - z_0)^n f(z)$ exists and is nonzero.

Key Formula

f(z) = \frac{a_{-n}}{(z - z_0)^n} + \frac{a_{-n+1}}{(z - z_0)^{n-1}} + \cdots + a_0 + a_1(z - z_0) + \cdots

Where:

$z_0$ = Location of the pole in the complex plane
$n$ = Order of the pole (a positive integer)
$a_{-n}$ = Leading Laurent coefficient, which must be nonzero

How It Works

To identify a pole and its order, look at where the denominator of a rational function equals zero (provided the numerator does not also vanish there). The order of the pole equals the multiplicity of that zero in the denominator. For more general functions, expand the Laurent series around the suspected singularity and count how many negative-power terms appear. A pole of order 1 is called a simple pole, and its residue is especially easy to compute:

\text{Res}_{z=z_0} f(z) = \lim_{z \to z_0} (z - z_0) f(z)

Worked Example

Problem: Find the poles and their orders for

f(z) = \dfrac{1}{z^2(z - 3)}

Step 1: Set the denominator equal to zero to find the singularities.

z^2(z - 3) = 0 \implies z = 0 \text{ or } z = 3

Step 2: Determine the order of each pole by looking at the multiplicity of each factor. The factor

z^2

gives a zero of multiplicity 2 at

z = 0

, and

(z-3)

gives a zero of multiplicity 1 at

z = 3

\text{Pole at } z = 0: \text{order } 2 \quad | \quad \text{Pole at } z = 3: \text{order } 1

Step 3: Verify the simple pole at

z = 3

by computing its residue.

\text{Res}_{z=3} f(z) = \lim_{z \to 3} (z-3) \cdot \frac{1}{z^2(z-3)} = \frac{1}{9}

Answer:

f(z)

has a pole of order 2 at

z = 0

and a simple pole (order 1) at

z = 3

, with residue

\frac{1}{9}

z = 3

Why It Matters

Poles are central to evaluating contour integrals via the residue theorem, one of the most powerful tools in complex analysis. In physics and engineering, identifying poles of transfer functions determines the stability and frequency response of electrical circuits and control systems.

Common Mistakes

Mistake: Calling a removable singularity a pole because the denominator is zero there.

Correction: If the numerator also vanishes at the same point and cancels the denominator factor entirely, the singularity is removable, not a pole. Always simplify or check the Laurent expansion first.