Removable singularity

A graph of a parabola with a removable singularity at x = 2

In complex analysis, a removable singularity (sometimes called a cosmetic singularity) of a holomorphic function is a point at which the function is undefined, but it is possible to define the function at that point in such a way that the function is regular in a neighbourhood of that point.

For instance, the function

$f(z) = \frac{\sin z}{z}$

has a singularity at z = 0. This singularity can be removed by defining f(0) := 1, which is the limit of f as z tends to 0. The resulting function is holomorphic. In this case the problem was caused by f being given an indeterminate form. Taking a power series expansion for $\frac{\sin(z)}{z}$ shows that

$f(z) = \frac{1}{z}\left(\sum_{k=0}^{\infty} \frac{(-1)^kz^{2k+1}}{(2k+1)!} \right) = \sum_{k=0}^{\infty} \frac{(-1)^kz^{2k}}{(2k+1)!} = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \frac{z^6}{7!} + \cdots.$

Formally, if U is an open subset of the complex plane C, a is a point of U, and f: U − {a} → C is a holomorphic function, then a is called a removable singularity for f if there exists a holomorphic function g: U → C which coincides with f on U − {a}. We say f is holomorphically extendable over U if such a g exists.

Riemann's theorem

Riemann's theorem on removable singularities states when a singularity is removable:

Theorem. Let D be an open subset of the complex plane, a a point of D and f a holomorphic function defined on the set D \ {a}. The following are equivalent:

1. f is holomorphically extendable over a.
2. f is continuously extendable over a.
3. There exists a neighborhood of a on which f is bounded.
4. $\lim_{z\to a}(z - a) f(z) = 0$.

The implications 1 ⇒ 2 ⇒ 3 ⇒ 4 are trivial. To prove 4 ⇒ 1, we first recall that the holomorphy of a function at a is equivalent to it being analytic at a (proof), i.e. having a power series representation. Define

$h(z) = \begin{cases} (z - a)^2 f(z) & z \ne a ,\\ 0 & z = a . \end{cases}$

Clearly, h is holomorphic on D \ {a}, and there exists

$h'(a)=\lim_{z\to a}\frac{(z - a)^2f(z)-0}{z-a}=\lim_{z\to a}(z - a) f(z)=0$

by 4, hence h is holomorphic on D and has a Taylor series about a:

$h(z) = a_0 + a_1(z-a) + a_2 (z - a)^2 + a_3 (z - a)^3 + \cdots .$

We have a₀ = h(a) = 0 and a₁ = h'(a) = 0; therefore

$g(z) = a_2 + a_3 (z - a) + a_4 (z - a)^2 + \cdots$

is a holomorphic extension of f over a, which proves the claim.

Other kinds of singularities

Unlike functions of a real variable, holomorphic functions are sufficiently rigid that their isolated singularities can be completely classified. A holomorphic function's singularity is either not really a singularity at all, i.e. a removable singularity, or one of the following two types:

1. In light of Riemann's theorem, given a non-removable singularity, one might ask whether there exists a natural number m such that lim_{z → a}(z − a)^m+1f(z) = 0. If so, a is called a pole of f and the smallest such m is the order of a. So removable singularities are precisely the poles of order 0. A holomorphic function blows up uniformly near its poles.
2. If an isolated singularity a of f is neither removable nor a pole, it is called an essential singularity. It can be shown that such an f maps every punctured open neighborhood U − {a} to the entire complex plane, with the possible exception of at most one point.

See also

• Analytic capacity
• Removable discontinuity