# 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}|. 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}|.

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 ZeroesOfAnalyticFunctionsAreIsolated 2013-03-22 13:38:10 2013-03-22 13:38:10 brianbirgen (2180) brianbirgen (2180) 8 brianbirgen (2180) Result msc 30C15 zeros of analytic functions are isolated Complex LeastAndGreatestZero IdentityTheorem WhenAllSingularitiesArePoles