# removable singularity

Let $U\subset \u2102$ be an open neighbourhood of a
point $a\in \u2102$. We say that a function
$f:U\backslash \{a\}\to \u2102$ has a *removable singularity ^{}* at
$a$, if the complex derivative

^{}${f}^{\prime}(z)$ exists for all $z\ne a$, and if $f(z)$ is bounded

^{}near $a$.

Removable singularities can, as the name suggests, be removed.

###### Theorem 1

Suppose that $f\mathrm{:}U\mathrm{\backslash}\mathrm{\{}a\mathrm{\}}\mathrm{\to}\mathrm{C}$ has a removable singularity at $a$. Then, $f\mathit{}\mathrm{(}z\mathrm{)}$ can be holomorphically extended to all of $U$, i.e. there exists a holomorphic $g\mathrm{:}U\mathrm{\to}\mathrm{C}$ such that $g\mathit{}\mathrm{(}z\mathrm{)}\mathrm{=}f\mathit{}\mathrm{(}z\mathrm{)}$ for all $z\mathrm{\ne}a$.

*Proof.*
Let $C$ be a circle centered at $a$, oriented counterclockwise, and
sufficiently small so that $C$ and its interior are contained in
$U$. For $z$ in the interior of $C$, set

$$g(z)=\frac{1}{2\pi i}{\oint}_{C}\frac{f(\zeta )}{\zeta -z}\mathit{d}\zeta .$$ |

Since $C$ is a compact set, the defining limit for the derivative

$$\frac{d}{dz}\frac{f(\zeta )}{\zeta -z}=\frac{f(\zeta )}{{(\zeta -z)}^{2}}$$ |

converges uniformly for $\zeta \in C$. Thanks to the uniform
convergence^{}, the order of the derivative and the integral operations
can be interchanged. Hence, we may deduce that ${g}^{\prime}(z)$ exists
for all $z$ in the interior of $C$. Furthermore, by the Cauchy
integral formula^{} we have that $f(z)=g(z)$ for all $z\ne a$, and therefore
$g(z)$ furnishes us with the desired extension.

Title | removable singularity |
---|---|

Canonical name | RemovableSingularity |

Date of creation | 2013-03-22 12:56:01 |

Last modified on | 2013-03-22 12:56:01 |

Owner | rmilson (146) |

Last modified by | rmilson (146) |

Numerical id | 5 |

Author | rmilson (146) |

Entry type | Definition |

Classification | msc 30E99 |

Related topic | EssentialSingularity |