Singularities

What Laurent Series Reveal Near an Excluded Point

The Laurent Expansion on an Annulus from the previous tutorial gives a way to describe a holomorphic function near a point that is excluded from its domain. The terms with negative powers record how the function behaves as it approaches that point. Depending on those terms, the excluded point may be filled in holomorphically, may be a pole, or may be an essential singularity.

The distinction begins with the domain of the function. A function can be holomorphic at every point of a punctured disk while being undefined at its center. The Laurent series on that punctured disk then provides the information needed to classify the center. In this tutorial, $a$ is the center, and all neighborhoods under consideration are small enough to contain no other excluded points.

Isolated Singularities and Their Three Types

Definition: Let $f$ be a complex-valued function. A point $a$ is an isolated singularity of $f$ if there exists $R>0$ such that $f$ is defined and holomorphic on the punctured disk $0<|z-a|<R$. The definition does not require that $f$ fail to extend holomorphically to $a$. An isolated singularity is removable if $f$ has a holomorphic extension to a neighborhood of $a$; it is a pole if, for some positive integer $m$, $(z-a)^m f(z)$ extends holomorphically to $a$ with a nonzero value there; and it is essential if it is neither removable nor a pole.

The definition of an isolated singularity concerns the punctured disk, not the behavior at its center. The function might have been assigned an arbitrary value at $a$, or might not be defined there at all. Such a value, if present, has no bearing on the classification: what matters is the behavior on $0<|z-a|<R$.

For a pole, the integer $m$ in the definition is its order. The condition that the extended product be nonzero at $a$ ensures that $m$ is the exact order, rather than merely an exponent large enough to cancel the singularity. We will see that the Laurent series makes these distinctions precise.

Worked Example: A Removable Singularity at the Origin

Consider $f(z)=\sin z/z$ for $z\neq 0$. The Taylor expansion of $\sin z$ gives

\frac{\sin z}{z} =\frac{z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots}{z} =1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots.

There are no negative powers, and the resulting power series defines a holomorphic function at $0$ with value $1$. Thus $0$ is an isolated singularity of the originally defined function on the punctured disk, and it is removable. Defining $f(0)=1$ gives its holomorphic extension.

The Boundedness Criterion for Removability

A key test for a removable singularity uses only the size of the function near the excluded point. Boundedness here must hold throughout some punctured neighborhood, not merely along one sequence approaching $a$.

Theorem (Riemann’s Removable Singularity Theorem): Suppose $f$ is holomorphic on $0<|z-a|<R$. Then $f$ has a holomorphic extension to $a$ if and only if $f$ is bounded on some punctured disk $0<|z-a|<r$, where $0<r<R$.

Proof. First suppose $f$ has a holomorphic extension to a neighborhood of $a$. The extension is continuous, so it is bounded on a sufficiently small closed disk centered at $a$. In particular, $f$ is bounded on the punctured disk inside it.

For the converse, suppose $|f(z)|\leq M$ whenever $0<|z-a|<r$, for some finite $M$. By the Laurent Expansion on an Annulus, $f$ has a Laurent expansion there,

f(z)=\sum_{n=-\infty}^{\infty}c_n(z-a)^n.

Fix a positive integer $k$. The Laurent coefficient formula, applied on a circle of radius $t$ with $0<t<r$, gives

c_{-k} =\frac{1}{2\pi i}\int_{|\zeta-a|=t} f(\zeta)(\zeta-a)^{k-1}\,d\zeta.

The circle has length $2\pi t$; on it, $|f(\zeta)|\leq M$ and $|\zeta-a|^{k-1}=t^{k-1}$. Therefore

|c_{-k}| \leq \frac{1}{2\pi}(2\pi t)Mt^{k-1} =Mt^k.

The coefficient $c_{-k}$ is independent of the permitted circle radius, by the Laurent coefficient formula. Letting $t$ decrease to $0$ in the bound $ |c_{-k}|\leq Mt^k$ shows that $c_{-k}=0$. This holds for every positive integer $k$, so all negative-power coefficients vanish.

The Laurent series is therefore a power series with only nonnegative powers. It defines a holomorphic function at $a$ as well as on the punctured disk, and agrees with $f$ there. Hence it gives the required holomorphic extension. $\square$

Worked Example: Boundedness Identifies a Removable Point

Let $g(z)=(1-\cos z)/z^2$ for $z\neq 0$. Since

1-\cos z =\frac{z^2}{2!}-\frac{z^4}{4!}+\frac{z^6}{6!}-\cdots,

division by $z^2$ gives

g(z)=\frac{1}{2!}-\frac{z^2}{4!}+\frac{z^4}{6!}-\cdots.

This series converges near $0$ and defines the extension there, with value $1/2$. In particular, $g$ is bounded near $0$, so the removable-singularity theorem also guarantees that the point is removable. Direct substitution verifies the extension’s value: the constant term is $1/2!=1/2$.

Classification by the Negative Laurent Coefficients

The Laurent series gives a complete classification. If there are no negative powers, the singularity is removable. If there are finitely many negative powers and at least one, the point is a pole. If there are infinitely many negative powers, the point is essential.

Theorem (Classification of Isolated Singularities): Suppose $f$ is holomorphic on $0<|z-a|<R$, with Laurent expansion $$ f(z)=\sum_{n=-\infty}^{\infty}c_n(z-a)^n. $$ Exactly one of the following holds:

Removable: $c_n=0$ for every $n<0$.
Pole of order $m$: $c_{-m}\neq 0$ for some positive integer $m$, and $c_n=0$ for every $n<-m$.
Essential: infinitely many coefficients with negative indices are nonzero.

Proof. If all negative coefficients vanish, the Laurent series is a power series and extends holomorphically to $a$. Thus the singularity is removable. Conversely, if the singularity is removable, the Riemann Removable Singularity Theorem shows that its Laurent coefficients with negative indices all vanish.

Now suppose there are finitely many nonzero negative coefficients and at least one. Let $m\geq 1$ be the largest positive integer for which $c_{-m}\neq 0$. Then the terms with negative powers are $c_{-m}(z-a)^{-m}+\cdots+c_{-1}(z-a)^{-1}$. Multiplying the Laurent expansion by $(z-a)^m$ gives a power series near $a$, with value $c_{-m}\neq 0$ at $a$. Thus $a$ is a pole of order $m$, according to the definition.

Conversely, if $a$ is a pole of order $m$, then $(z-a)^m f(z)$ extends holomorphically to $a$ with nonzero value there. Its Taylor series near $a$, divided by $(z-a)^m$ on the punctured disk, shows that $f$ has no terms with exponents below $-m$, and that its coefficient $c_{-m}$ is nonzero. Finally, if infinitely many negative coefficients are nonzero, the singularity is neither removable nor a pole, by the first two cases. It is therefore essential. These alternatives exhaust the possibilities for the negative coefficients and cannot overlap. $\square$

Worked Example: A Pole of Order Two

Consider $f(z)=e^z/(z-1)^2$ near $a=1$. The function is holomorphic on every sufficiently small punctured disk about $1$. Set $g(z)=e^z$. Then $g$ is holomorphic at $1$, and $g(1)=e\neq 0$, while

f(z)=\frac{g(z)}{(z-1)^2}.

This already verifies from the definition that $1$ is a pole of order two. The Laurent expansion confirms the leading coefficient: writing $w=z-1$, we have $e^z=e^{1+w}=e(1+w+w^2/2!+\cdots)$, so

\frac{e^z}{(z-1)^2} =e\left(\frac{1}{w^2}+\frac{1}{w}+\frac{1}{2!}+\frac{w}{3!}+\cdots\right).

The coefficient of $(z-1)^{-2}$ is $e\neq0$, and there are no powers below $-2$. This is precisely the Laurent-series criterion for a pole of order two.

Essential Singularities and the Importance of the Full Series

An essential singularity is characterized by an infinite collection of nonzero negative-power terms. It is not enough to notice that a function is undefined at $a$, or that it has some negative powers: a finite principal part indicates a pole, while no principal part indicates a removable singularity. The entire pattern of negative powers matters.

Worked Example: An Essential Singularity

Define $h(z)=\sin(1/z)$ for $z\neq 0$. Substituting $1/z$ into the Taylor series for sine gives

\sin(1/z) =\frac{1}{z}-\frac{1}{3!z^3} +\frac{1}{5!z^5}-\frac{1}{7!z^7}+\cdots.

This Laurent expansion has infinitely many nonzero negative-power coefficients. By the classification theorem, $0$ is an essential singularity. The conclusion follows from the series itself; no claim about a value assigned at $0$ is needed.

A useful local factorization follows directly from the pole case. If $a$ is a pole of order $m$, then $f(z)=(z-a)^{-m}g(z)$, where $g$ is holomorphic near $a$ and $g(a)\neq0$. By continuity, $|g(z)|$ is bounded below by a positive number sufficiently close to $a$. Thus $|f(z)|$ grows like a positive constant times $|z-a|^{-m}$, and in particular the function is unbounded near the pole.

This also distinguishes the two main tests: boundedness on a punctured neighborhood proves removability, while a pole necessarily gives unboundedness of a specific power-law form. An essential singularity is the remaining case: its Laurent series has infinitely many negative powers. Do not infer boundedness or unboundedness merely from the function’s being undefined at the center; use the Laurent coefficients or a valid local estimate.

The classification is useful because it turns local behavior into an algebraic question about a series. It also prepares for the next tutorial: when a function has an isolated singularity, its Laurent coefficient of $(z-a)^{-1}$ will play a particularly important role in computing contour integrals.

Check Your Understanding

Use the definitions, the boundedness criterion, and the negative Laurent coefficients to answer the following questions.

Why can a removable point still be an isolated singularity under the definition used here?
What must be true of the negative-power coefficients for a singularity to be removable?
Suppose the Laurent series has a nonzero $c_{-4}$ and no nonzero coefficients with indices below $-4$. What is the singularity, and what is its order?
Why does boundedness along just one sequence approaching $a$ not suffice for the Riemann Removable Singularity Theorem?
Classify the singularity at $0$ of $1/(z^2(1+z))$ using its local form, and state its order.

What You'll Learn