proof of Riemann’s removable singularity theorem

Suppose that f is holomorphic on U{a} and limza(z-a)f(z)=0. Let


be the Laurent seriesMathworldPlanetmath of f centered at a. We will show that ck=0 for k<0, so that f can be holomorphically extended to all of U by defining f(a)=c0.

For any non-negative integer n, the residue of (z-a)nf(z) at a is


This is equal to zero, because

||z-a|=δ(z-a)nf(z)dz| 2πδmax|z-a|=δ|(z-a)nf(z)|
= 2πδnmax|z-a|=δ|(z-a)f(z)|

which, by our assumptionPlanetmathPlanetmath, goes to zero as δ0. Since the residue of (z-a)nf(z) at a is also equal to c-n-1, the coefficients of all negative powers of z in the Laurent series vanish.

Conversely, if a is a removable singularityMathworldPlanetmath of f, then f can be expanded in a power seriesMathworldPlanetmath centered at a, so that


because the constant term in the power series of (z-a)f(z) is zero.

A corollary of this theorem is the following: if f is bounded near a, then


for some M>0. This implies that (z-a)f(z)0 as za, so a is a removable singularity of f.

Title proof of Riemann’s removable singularity theorem
Canonical name ProofOfRiemannsRemovableSingularityTheorem
Date of creation 2013-03-22 13:33:03
Last modified on 2013-03-22 13:33:03
Owner pbruin (1001)
Last modified by pbruin (1001)
Numerical id 5
Author pbruin (1001)
Entry type Proof
Classification msc 30D30