Zeroes of Analytic Function are Isolated

Theorem
Let $$U \subset \C \ $$ be some open set and let $$f \ $$ be an analytic function defined on $$U \ $$.

Then either $$f \ $$ is a constant function, or the set $$\left\{{z\in U|f(z)=0}\right\} \ $$ is totally disconnected.

Proof
Suppose $$f \ $$ has no zeroes in $$U \ $$; then the set described in the theorem is the empty set, and we're done.

So we suppose $$\exists z_0 \in U \ $$ such that $$f(z_0) = 0 \ $$.

Since $$f \ $$ is analytic, there is a Taylor series for $$f \ $$ at $$z_0 \ $$ which converges for $$|z-z_0|<R \ $$. Now, since $$f(z_0)=0, \ $$ we know $$a_0 =0 \ $$. Other $$a_j \ $$ may be $$0 \ $$ as well. So let $$k \ $$ be the number such that $$a_j =0 \ $$ for $$0\leq j < k$$, and $$a_k\neq 0 \ $$. Then we can write the Taylor series for $$f \ $$ about $$z_0 \ $$ as:


 * $$\sum_{n=k}^\infty a_n (z-z_0)^n = (z-z_0)^k \sum_{n=0}^\infty a_{n+k}(z-z_0)^n \ $$

where $$a_k \neq 0 \ $$ (otherwise, we'd just start at $$k+1 \ $$) and define a new function, $$g(z) \ $$, as the sum on the right hand side, which is clearly analytic in $$|z-z_0|0 \ $$ so that $$\forall z \ $$ such that $$|z-z_0|<\epsilon, |g(z)-a_k|<\frac{|a_k|}{2} \ $$.

But then $$g(z) \ $$ can't possibly be $$0 \ $$ in that disk.

Hence the result.