removable singularity

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

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

Theorem 1

Suppose that $f\mathrm{:}U\mathrm{\backslash \backslash }\mathrm{\{}a\mathrm{\}}\mathrm{\to }\mathrm{C}$ has a removablesingularity at $a$. Then, $f\mathit{}\mathrm{(}z\mathrm{)}$ can be holomorphically extended toall 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, andsufficiently 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 uniformconvergence, the order of the derivative and the integral operationscan be interchanged. Hence, we may deduce that $g^{\prime }(z)$ existsfor all $z$ in the interior of $C$. Furthermore, by the Cauchyintegral formula we have that $f(z)=g(z)$ for all $z\ne a$, and therefore$g(z)$ furnishes us with the desired extension.