# zero as contour integral

Suppose that $f$ is a complex function which is defined in some open set $D\subseteq \u2102$ which has a simple zero at some point $p\in D$. Then we have

$$p=\frac{1}{2\pi i}{\oint}_{C}\frac{z{f}^{\prime}(z)}{f(z)}\mathit{d}z$$ |

where $C$ is a closed path in $D$ which encloses $p$ but does not enclose or pass through any other zeros of $f$.

This follows from the Cauchy residue theorem. We have that the poles of
${f}^{\prime}/f$ occur at the zeros of $f$ and that the residue^{} of a pole of
${f}^{\prime}/f$ is $1$ at a simple zero of $f$. Hence, the residue of
$z{f}^{\prime}(z)/f(z)$ at $p$ is $p$, so the above follows from the residue theorem.

Title | zero as contour integral |
---|---|

Canonical name | ZeroAsContourIntegral |

Date of creation | 2013-03-22 16:46:42 |

Last modified on | 2013-03-22 16:46:42 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 5 |

Author | rspuzio (6075) |

Entry type | Corollary |

Classification | msc 30E20 |