# 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

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

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 ${\displaystyle {\frac {\sin(z)}{z}}}$ shows that

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

## Riemann's theorem

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

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

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

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

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

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

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

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

${\displaystyle h(z)=c_{0}+c_{1}(z-a)+c_{2}(z-a)^{2}+c_{3}(z-a)^{3}+\cdots \,.}$

We have c0 = h(a) = 0 and c1 = hTemplate:'(a) = 0; therefore

${\displaystyle h(z)=c_{2}(z-a)^{2}+c_{3}(z-a)^{3}+\cdots \,.}$

Hence, where z≠a, we have:

${\displaystyle f(z)=h(z)/(z-a)^{2}=c_{2}+c_{3}(z-a)+\cdots \,.}$

However,

${\displaystyle g(z)=c_{2}+c_{3}(z-a)+\cdots \,.}$

is holomorphic on D, thus an extension of f.

## 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 ${\displaystyle m}$ such that ${\displaystyle \lim _{z\rightarrow a}(z-a)^{m+1}f(z)=0}$. If so, ${\displaystyle a}$ is called a pole of ${\displaystyle f}$ and the smallest such ${\displaystyle m}$ is the order of ${\displaystyle a}$. So removable singularities are precisely the poles of order 0. A holomorphic function blows up uniformly near its poles.
2. If an isolated singularity ${\displaystyle a}$ of ${\displaystyle f}$ is neither removable nor a pole, it is called an essential singularity. It can be shown that such an ${\displaystyle f}$ maps every punctured open neighborhood ${\displaystyle U\setminus \{a\}}$ to the entire complex plane, with the possible exception of at most one point.