# proof of argument principle

Since $f$ is meromorphic, ${f}^{\prime}$ is meromorphic, and hence ${f}^{\prime}/f$ is meromorphic. The singularities of ${f}^{\prime}/f$ can only occur at the zeros and the poles of $f$.

I claim that all singularities of ${f}^{\prime}/f$ are simple poles^{}. Furthermore, if $f$ has a zero at some point $p$, then the residue of the pole at $p$ is positive and equals the multiplicity of the zero of $f$ at $p$. If $f$ has a pole at some point $p$, then the residue of the pole at $p$ is negative and equals minus the multiplicity of the pole of $f$ at $p$.

To prove these assertions, write $f(x)={(x-p)}^{n}g(x)$ with $g(p)\ne 0$. Then

$$\frac{{f}^{\prime}(x)}{f(x)}=\frac{n}{x-p}+\frac{{g}^{\prime}(x)}{g(x)}$$ |

Since $g(p)\ne 0$, the only singularity of ${f}^{\prime}/f$ at $p$ comes from the first summand. Since $n$ is either the order of the zero of $f$ at $p$ if $f$ has a zero at $p$ or minus the order of the pole of $f$ at $p$ if $f$ has a pole at $p$, the assertion is proven.

By the Cauchy residue theorem, the integral

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

equals the sum of the residues of ${f}^{\prime}/f$. Combining this fact with the characterization^{} of the poles of ${f}^{\prime}/f$ and their residues given above, one deduces that this integral equals the number of zeros of $f$ minus the number of poles of $f$, counted with multiplicity.

Title | proof of argument principle |
---|---|

Canonical name | ProofOfArgumentPrinciple |

Date of creation | 2013-03-22 14:34:32 |

Last modified on | 2013-03-22 14:34:32 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 11 |

Author | rspuzio (6075) |

Entry type | Proof |

Classification | msc 30E20 |

Synonym | Cauchy’s argument principle |