# proof of Riemann’s removable singularity theorem

Suppose that $f$ is holomorphic on $U\setminus \{a\}$ and ${lim}_{z\to a}(z-a)f(z)=0$. Let

$$f(z)=\sum _{k=-\mathrm{\infty}}^{\mathrm{\infty}}{c}_{k}{(z-a)}^{k}$$ |

be the Laurent series^{} of $f$ centered at $a$. We will show that ${c}_{k}=0$ for $$, so that $f$ can be holomorphically extended to all of $U$ by defining $f(a)={c}_{0}$.

For any non-negative integer $n$, the residue of ${(z-a)}^{n}f(z)$ at $a$ is

$$\mathrm{Res}({(z-a)}^{n}f(z),a)=\frac{1}{2\pi i}\underset{\delta \to {0}^{+}}{lim}{\oint}_{|z-a|=\delta}{(z-a)}^{n}f(z)dz.$$ |

This is equal to zero, because

$\left|{\displaystyle {\oint}_{|z-a|=\delta}}{(z-a)}^{n}f(z)dz\right|$ | $\le $ | $2\pi \delta \underset{|z-a|=\delta}{\mathrm{max}}|{(z-a)}^{n}f(z)|$ | ||

$=$ | $2\pi {\delta}^{n}\underset{|z-a|=\delta}{\mathrm{max}}|(z-a)f(z)|$ |

which, by our assumption^{}, goes to zero as $\delta \to 0$. Since the residue of ${(z-a)}^{n}f(z)$ 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 singularity^{} of $f$, then $f$ can be expanded in a power series^{} centered at $a$, so that

$$\underset{z\to a}{lim}(z-a)f(z)=0$$ |

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

$$|(z-a)f(z)|\le |z-a|M$$ |

for some $M>0$. This implies that $(z-a)f(z)\to 0$ as $z\to a$, 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 |