# Complex analysis/Isolated singularity
## Definition

Let $G\subseteq \mathbb {C}$ be a domain and $z_{0}\in G$. If $f\colon G\setminus {z_{0}}\to \mathbb {C}$ is a holomorphic function, then $z_{0}$ is called an isolated singularity of $f$.

## Classification

Depending on the behavior of $f$ in the neighborhood of $z_{0}$, one distinguishes three different types of isolated singularities of $f$.

### Removable Singularities

If $f$ can be holomorphically extended to the entire domain $G$, then we say that $z_{0}$ is a removable singularity. According to the Riemann Removability Theorem, this is the case if $f$ is bounded in a neighborhood of $z_{0}$.

### Poles

If $z_{0}$ is not a removable singularity, but there exists an $n\geq 1$ such that $(\cdot -z_{0})^{n}\cdot f$ has a removable singularity at $z_{0}$, then we say that $f$ has a pole at $z_{0}$. The smallest such $n$ is called the order of the pole.

### Essential Singularities

If $z_{0}$ is neither removable nor a pole, then $z_{0}$ is called an essential singularity of $f$.

## Examples

* Since $\lim _{z\to 0}{\frac {\sin z}{z}}=1$, the function $f_{1}(z)={\frac {\sin z}{z}}$ has a removable singularity at $z_{0}=0$.
* The function $f_{2}(z)={\frac {1}{\sin z}}$ does not have $z_{0}=0$ a removable singularity at, since$f_{2}$ is unbounded at $0$, but $f_{2}$ has a first-order pole at $0$, because $f_{2}(z)\cdot (z-0)^{1}=f_{2}(z)z={\frac {z}{\sin z}}$ and $\lim _{z\to 0}{\frac {z}{\sin z}}=1$, which has a removable singularity at 0 .
* The function $f_{3}(z)=\sin {\frac {1}{z}}$ has an essential singularity at $z_{0}=0$, since for every $n\geq 1$, the function $f_{3}(z)z^{n}=z^{n}\sin {\frac {1}{z}}$ is unbounded in any neighborhood of $0$. To see this, consider$\sin z^{-1}={\frac {e^{iz^{-1}}-e^{-iz^{-1}}}{2i}}$.For $z=it$ with $t\in \mathbb {R}$ is also $f_{3}(it)(it)^{n}=(it)^{n}{\frac {e^{t^{-1}}-e^{-t^{-1}}}{2i}}$,which diverges as $t\to 0^{+}$ .

## Laurent Expansions

The type of isolated singularity can also be inferred from the Laurent Expansion of $f$ around $z_{0}$. Let
$f(z)=\sum _{n=-\infty }^{\infty }a_{n}(z-z_{0})^{n}$
be the Laurent Series of $f$ around $z_{0}$. We define
$o_{z}(f)=\sup\{n\in \mathbb {Z} |\forall k<n:a_{k}=0\}$
Then, $f$ has the following singularities:If $o_{z}(f)\geq 0$, i.e., all negative coefficients vanish, the main part of the series is zero, and the singularity is removable.
If $-\infty <o_{z}(f)<0$, i.e., only finitely many negative coefficients are nonzero, there is a pole of order $-o_{z}(f)$. If $o_{z}(f)=-\infty$, i.e., infinitely many negative coefficients are nonzero, the singularity is essential.

### Examples

Let us consider our three examples again:
It is $f_{1}(z)={\frac {\sin z}{z}}=\sum _{k=0}^{\infty }(-1)^{n}{\frac {z^{2n}}{(2n+1)!}}$, so $o_{0}(f_{1})=0$, a removable singularity.
It is
$f_{2}(z)={\frac {1}{\sin z}}={\frac {1}{z}}+{\frac {z}{6}}+{\frac {7}{360}}z^{3}+\ldots$
so $o_{0}(f_{2})=-1$, a pole of first order.
It is $f_{3}(z)=\sin z^{-1}=\sum _{n=-\infty }^{0}{\frac {(-1)^{n}}{(-2n+1)!}}z^{2n-1}$, so $o_{0}(f_{3})=-\infty$, an essential singularity.

## Page information
### Translation and Version Control

This page was translated based on the following Singularität Wikiversity source page and uses the concept of Translation and Version Control for a transparent language fork in a Wikiversity:
* Source: Kurs:Funktionentheorie/isolierte Singularität - URL:https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/isolierte Singularität
* Date: 11/20/2024