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# 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.

## Mathematics Subject Classification

30E99*no label found*

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## Recent Activity

new correction: Error in proof of Proposition 2 by alex2907

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