Removable singularities can, as the name suggests, be removed.
Suppose that has a removable singularity at . Then, can be holomorphically extended to all of , i.e. there exists a holomorphic such that for all .
Proof. Let be a circle centered at , oriented counterclockwise, and sufficiently small so that and its interior are contained in . For in the interior of , set
converges uniformly for . Thanks to the uniform convergence, the order of the derivative and the integral operations can be interchanged. Hence, we may deduce that exists for all in the interior of . Furthermore, by the Cauchy integral formula we have that for all , and therefore furnishes us with the desired extension.
|Date of creation||2013-03-22 12:56:01|
|Last modified on||2013-03-22 12:56:01|
|Last modified by||rmilson (146)|