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 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.

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.