# Removable singularity

In complex analysis, a removable singularity of a holomorphic function is a point at which the function is undefined, but it is possible to redefine the function at that point in such a way that the resulting function is regular in a neighbourhood of that point.
For instance, the (unnormalized) sinc function, as defined by
    ${\text{sinc}}(z)={\frac {\sin z}{z}}$
has a singularity at z = 0. This singularity can be removed by defining ${\text{sinc}}(0):=1,$ which is the limit of sinc as z tends to 0. The resulting function is holomorphic. In this case the problem was caused by sinc being given an indeterminate form. Taking a power series expansion for sin⁡(z)/z {\textstyle {\frac {\sin(z)}{z}}} around the singular point shows that
    ${\text{sinc}}(z)={\frac {1}{z}}\left(\sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{(2k+1)!}}\right)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k}}{(2k+1)!}}=1-{\frac {z^{2}}{3!}}+{\frac {z^{4}}{5!}}-{\frac {z^{6}}{7!}}+\cdots .$
Formally, if $U\subset \mathbb {C}$ is an open subset of the complex plane $\mathbb {C}$, $a\in U$ a point of $U$, and $f:U\setminus \{a\}\rightarrow \mathbb {C}$ is a holomorphic function, then $a$ is called a removable singularity for $f$ if there exists a holomorphic function $g:U\rightarrow \mathbb {C}$ which coincides with $f$ on $U\setminus \{a\}$. We say $f$ is holomorphically extendable over $U$ if such a $g$ exists.

## Riemann's theorem

Riemann's theorem on removable singularities is as follows:
Theorem — Let $D\subset \mathbb {C}$ be an open subset of the complex plane, $a\in D$ a point of $D$ and $f$ a holomorphic function defined on the set $D\setminus \{a\}$. The following are equivalent:
* $f$ is holomorphically extendable over $a$.
* $f$ is continuously extendable over $a$.
* There exists a neighborhood of $a$ on which $f$ is bounded.
* $\lim _{z\to a}(z-a)f(z)=0$.
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
    $h(z)={\begin{cases}(z-a)^{2}f(z)&z\neq a,\\0&z=a.\end{cases}}$
Clearly, h is holomorphic on $D\setminus \{a\}$, and there exists
    $h'(a)=\lim _{z\to a}{\frac {(z-a)^{2}f(z)-0}{z-a}}=\lim _{z\to a}(z-a)f(z)=0$
by 4, hence h is holomorphic on D and has a Taylor series about a:
    $h(z)=c_{0}+c_{1}(z-a)+c_{2}(z-a)^{2}+c_{3}(z-a)^{3}+\cdots \,.$
We have c$_0$ = h(a) = 0 and c$_1$ = h'(a) = 0; therefore
    $h(z)=c_{2}(z-a)^{2}+c_{3}(z-a)^{3}+\cdots \,.$
Hence, where $z\neq a$, we have:
    $f(z)={\frac {h(z)}{(z-a)^{2}}}=c_{2}+c_{3}(z-a)+\cdots \,.$
However,
    $g(z)=c_{2}+c_{3}(z-a)+\cdots \,.$
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:
* In light of Riemann's theorem, given a non-removable singularity, one might ask whether there exists a natural number $m$ such that $\lim _{z\rightarrow a}(z-a)^{m+1}f(z)=0$. 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 other poles.
* If an isolated singularity $a$ of $f$ is neither removable nor a pole, it is called an essential singularity. The Great Picard Theorem shows that such an $f$ maps every punctured open neighborhood $U\setminus \{a\}$ to the entire complex plane, with the possible exception of at most one point.