# removable singularity

Let $U\subset\mathbb{C}$ be an open neighbourhood of a point $a\in\mathbb{C}$. We say that a function $f:U\backslash\{a\}\rightarrow\mathbb{C}$ has a removable singularity  at $a$, if the complex derivative  $f^{\prime}(z)$ exists for all $z\neq a$, and if $f(z)$ is bounded  near $a$.

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

###### Theorem 1

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

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

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

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 $z\neq a$, and therefore $g(z)$ furnishes us with the desired extension.

Title removable singularity RemovableSingularity 2013-03-22 12:56:01 2013-03-22 12:56:01 rmilson (146) rmilson (146) 5 rmilson (146) Definition msc 30E99 EssentialSingularity