zeroes of analytic functions are isolated

The zeroes of a non-constant analytic function on ${\mathbb{C}}$ are isolated. Let $f$ be an analytic function defined in some domain $D\subset{\mathbb{C}}$ 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 $|z-z_{0}|<R$ . Write it as $f(z)=\Sigma_{n=k}^{\infty}a_{n}(z-z_{0})^{n}$ , with $a_{k}\neq 0$ and $k>0$ ( $a_{k}$ is the first non-zero term). One can factor the series so that $f(z)=(z-z_{0})^{k}\Sigma_{n=0}^{\infty}a_{n+k}(z-z_{0})^{n}$ and define $g(z)=\Sigma_{n=0}^{\infty}a_{n+k}(z-z_{0})^{n}$ so that $f(z)=(z-z_{0})^{k}g(z)$ . Observe that $g(z)$ is analytic on $|z-z_{0}|<R$ .

To show that $z_{0}$ is an isolated zero of $f$ , we must find $\epsilon>0$ so that $f$ is non-zero on $0<|z-z_{0}|<\epsilon$ . It is enough to find $\epsilon>0$ so that $g$ is non-zero on $|z-z_{0}|<\epsilon$ by the relation $f(z)=(z-z_{0})^{k}g(z)$ . Because $g(z)$ is analytic, it is continuous at $z_{0}$ . Notice that $g(z_{0})=a_{k}\neq 0$ , so there exists an $\epsilon>0$ so that for all $z$ with $|z-z_{0}|<\epsilon$ it follows that $|g(z)-a_{k}|<\frac{|a_{k}|}{2}$ . 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