# 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.

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The "$f$ is locally constant ..." case does not need to be there, because they've already assumed $U$ is connected and $f$ is not constant. –  Robert Israel Aug 16 '12 at 0:04
"ensure you properly understand the precise meaning of all terminology before trying to consider a proof". +1: this should hang in gold letters in every mathematician's home! –  Georges Elencwajg Aug 16 '12 at 8:04

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.
... and the point is that $g(z) \ne 0$ in a neighbourhood of $z_0$, so $z_0$ is the only zero of $f(z)$ in that neighbourhood. –  Robert Israel Aug 16 '12 at 0:06
$z_0=0$ is the only zero of the identity function, and it is isolated.
Am I wrong to take the definition of isolated as the existence of some $\epsilon$ such that in a disk $D(z_0, \epsilon)$ centred at $z_0$ there exist no other members of the set to which $z_0$ belongs? Because would that then not contradict the definition of a continuous function? Forgive me if this sounds silly. –  providence Aug 16 '12 at 0:19
Yes, because $\epsilon$ needs to be positive. $\;\;$ Yes, because "the set to which $z_0$ belongs" is $\:\{w\in \mathbb{C} : f(w) = 0\}$. –  Ricky Demer Aug 16 '12 at 0:34
@RickyDemer Ahhh, they speak of isolated in the sense of other zeros! I was thinking the isolated in the sense of all values of $f$. I understand now! Thanks a bunch to you and to everyone. –  providence Aug 16 '12 at 0:48