Complex analytic functions and their zeros. I am having trouble with the following statement found in a textbook:
"Let $U$ be a connected open set. Let $f$ be a complex analytic function on $U$ and not constant. 
Either $f$ is locally constant and equal to $0$ in a neighbourhood of a zero $z_0$, or $z_0$ is an isolated zero."
I just cannot see how this is the case. For instance, isn't the identity function analytic, not isolated, and not locally constant at $z_0 = 0$?
Edit: While the accepted answer to this question is a good proof of why the above statement is true, the actual trouble I ran into was not properly understanding the terminology. I mistook the meaning "isolated" in this case. If this may serve to help anyone in future, I say this: ensure you properly understand the precise meaning of all terminology before trying to consider a proof.
 A: 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.
A: $z_0=0$ is the only zero of the identity function, and it is isolated. 
