removable singularityRemovableSingularity2002-08-13 07:27:112003-03-29 22:36:18Definition\usepackage{amsmath}
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\newcommand{\nl}[1]{{\PMlinkescapetext{{#1}}}}
\newcommand{\pln}[2]{{\PMlinkname{{#1}}{#2}}}Let $U\subset\cnums$ be an open neighbourhood of a
point $a\in \cnums$. We say that a function
$f:U\backslash\{a\}\rightarrow \cnums$ has a \emph{removable singularity} at
$a$, if the complex derivative $f'(z)$ exists for all $z\neq a$, and
if $f(z)$ is bounded near $a$.
Removable singularities can, as the name suggests, be removed.
\begin{theorem}
Suppose that $f:U\backslash\{a\}\rightarrow \cnums$ has a removable
singularity at $a$. Then, $f(z)$ can be holomorphically extended to
all of $U$, i.e.
there exists a holomorphic $g:U\rightarrow\cnums$ such that
$g(z)=f(z)$ for all $z\neq a$.
\end{theorem}
\emph{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 $g'(z)$ 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.