zeroes of analytic functions are isolated

The zeroes of a non-constant analytic function on $ℂ$ are isolated.Let $f$ be an analytic functiondefined in some domain $D\subset ℂ$ andlet $f(z_0)=0$ for some $z_0\in D$. Because $f$ is analytic,there is a Taylor series expansion for $f$ around $z_0$ whichconverges 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 $ϵ>0$ so that $f$ is non-zero on .It is enough to find $ϵ>0$ so that $g$ is non-zeroon 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 $ϵ>0$ so that for all $z$ with it follows that .This implies that $g(z)$ is non-zero in this set.