# zeroes of analytic functions are isolated

The zeroes of a non-constant analytic function^{} on $\u2102$ are isolated.
Let $f$ be an analytic function
defined in some domain $D\subset \u2102$ and
let $f({z}_{0})=0$ for some ${z}_{0}\in D$. Because $f$ is analytic,
there is a Taylor series^{} expansion for $f$ around ${z}_{0}$ which
converges on an open disk $$. Write it as
$f(z)={\mathrm{\Sigma}}_{n=k}^{\mathrm{\infty}}{a}_{n}{(z-{z}_{0})}^{n}$, with ${a}_{k}\ne 0$ and $k>0$
(${a}_{k}$ is the first non-zero term).
One can factor the series so that
$f(z)={(z-{z}_{0})}^{k}{\mathrm{\Sigma}}_{n=0}^{\mathrm{\infty}}{a}_{n+k}{(z-{z}_{0})}^{n}$ and define
$g(z)={\mathrm{\Sigma}}_{n=0}^{\mathrm{\infty}}{a}_{n+k}{(z-{z}_{0})}^{n}$ so that $f(z)={(z-{z}_{0})}^{k}g(z)$.
Observe that $g(z)$ is analytic on $$.

To show that ${z}_{0}$ is an isolated zero of $f$, we must find $\u03f5>0$ so that $f$ is non-zero on $$. It is enough to find $\u03f5>0$ so that $g$ is non-zero on $$ by the relation $f(z)={(z-{z}_{0})}^{k}g(z)$. Because $g(z)$ is analytic, it is continuous at ${z}_{0}$. Notice that $g({z}_{0})={a}_{k}\ne 0$, so there exists an $\u03f5>0$ so that for all $z$ with $$ it follows that $$. This implies that $g(z)$ is non-zero in this set.

Title | zeroes of analytic functions are isolated |
---|---|

Canonical name | ZeroesOfAnalyticFunctionsAreIsolated |

Date of creation | 2013-03-22 13:38:10 |

Last modified on | 2013-03-22 13:38:10 |

Owner | brianbirgen (2180) |

Last modified by | brianbirgen (2180) |

Numerical id | 8 |

Author | brianbirgen (2180) |

Entry type | Result |

Classification | msc 30C15 |

Synonym | zeros of analytic functions are isolated |

Related topic | Complex |

Related topic | LeastAndGreatestZero |

Related topic | IdentityTheorem |

Related topic | WhenAllSingularitiesArePoles |