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 $a_k \ne 0$ and $k > 0$ ($a_k$ is the first non-zero term). One can factor the series so that $f(z) = (z-z_0)^k \Sigma_{n=0}^{\infty} a_{n+k} (z-z_0)^n$ 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 $g(z_0)=a_k \ne 0$ , 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.