# proof of Cauchy residue theorem

Being $f$ holomorphic by Cauchy-Riemann equations^{} the differential form^{} $f(z)dz$ is closed. So by the lemma about closed differential forms on a simple connected domain we know that the integral ${\int}_{C}f(z)\mathit{d}z$ is equal to ${\int}_{{C}^{\prime}}f(z)\mathit{d}z$ if ${C}^{\prime}$ is any curve which is homotopic to $C$.
In particular we can consider a curve ${C}^{\prime}$ which turns around the points ${a}_{j}$ along small circles and join these small circles with segments. Since the curve ${C}^{\prime}$ follows each segment two times with opposite orientation it is enough to sum
the integrals of $f$ around the small circles.

So letting $z={a}_{j}+\rho {e}^{i\theta}$ be a parameterization of the curve around the point ${a}_{j}$, we have $dz=\rho i{e}^{i\theta}d\theta $ and hence

$${\int}_{C}f(z)\mathit{d}z={\int}_{{C}^{\prime}}f(z)\mathit{d}z=\sum _{j}\eta (C,{a}_{j}){\int}_{\partial {B}_{\rho}({a}_{j})}f(z)\mathit{d}z$$ |

$$=\sum _{j}\eta (C,{a}_{j}){\int}_{0}^{2\pi}f({a}_{j}+\rho {e}^{i\theta})\rho i{e}^{i\theta}\mathit{d}\theta $$ |

where $\rho >0$ is chosen so small that the balls ${B}_{\rho}({a}_{j})$ are all disjoint and all contained in the domain $U$. So by linearity, it is enough to prove that for all $j$

$$i{\int}_{0}^{2\pi}f({a}_{j}+{e}^{i\theta})\rho {e}^{i\theta}\mathit{d}\theta =2\pi i\mathrm{Res}(f,{a}_{j}).$$ |

Let now $j$ be fixed and
consider now the Laurent series^{} for $f$ in ${a}_{j}$:

$$f(z)=\sum _{k\in \mathbb{Z}}{c}_{k}{(z-{a}_{j})}^{k}$$ |

so that $\mathrm{Res}(f,{a}_{j})={c}_{-1}$. We have

$${\int}_{0}^{2\pi}f({a}_{j}+{e}^{i\theta})\rho {e}^{i\theta}\mathit{d}\theta =\sum _{k}{\int}_{0}^{2\pi}{c}_{k}{(\rho {e}^{i\theta})}^{k}\rho {e}^{i\theta}\mathit{d}\theta ={\rho}^{k+1}\sum _{k}{c}_{k}{\int}_{0}^{2\pi}{e}^{i(k+1)\theta}\mathit{d}\theta .$$ |

Notice now that if $k=-1$ we have

$${\rho}^{k+1}{c}_{k}{\int}_{0}^{2\pi}{e}^{i(k+1)\theta}\mathit{d}\theta ={c}_{-1}{\int}_{0}^{2\pi}\mathit{d}\theta =2\pi {c}_{-1}=2\pi \mathrm{Res}(f,{a}_{j})$$ |

while for $k\ne -1$ we have

$${\int}_{0}^{2\pi}{e}^{i(k+1)\theta}\mathit{d}\theta ={\left[\frac{{e}^{i(k+1)\theta}}{i(k+1)}\right]}_{0}^{2\pi}=0.$$ |

Hence the result follows.

Title | proof of Cauchy residue theorem |
---|---|

Canonical name | ProofOfCauchyResidueTheorem |

Date of creation | 2013-03-22 13:42:04 |

Last modified on | 2013-03-22 13:42:04 |

Owner | paolini (1187) |

Last modified by | paolini (1187) |

Numerical id | 7 |

Author | paolini (1187) |

Entry type | Proof |

Classification | msc 30E20 |