# 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 $\set {z \in U: \map f z = 0}$ 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 $\map f {z_0} = 0$.

Since $f$ is analytic, there is a Taylor series for $f$ at $z_0$ which converges for $\cmod {z - z_0} < R$.

Now, since $\map f {z_0} = 0,$ we know $a_0 =0$.

Other $a_j$ may be $0$ as well.

So let $k$ be the least number such that $a_j = 0$ for $0 \le j < k$, and $a_k \ne 0$.

Then we can write the Taylor series for $f$ about $z_0$ as:

$\displaystyle \sum_{n \mathop = k}^\infty a_n \paren {z - z_0}^n = \paren {z - z_0}^k \sum_{n \mathop = 0}^\infty a_{n + k} \paren {z - z_0}^n$

where $a_k \ne 0$ (otherwise, we'd just start at $k + 1$).

Now we define a new function $\map g z$, as the sum on the right hand side, which is clearly analytic in $\cmod {z - z_0} < R$.

Since it is analytic here, it is also continuous here.

Since $\map g {z_0} = a_k \ne 0, \exists \epsilon >0$ so that $\forall z$ such that $\cmod {z - z_0} < \epsilon, \cmod {\map g z - a_k} < \dfrac {\cmod {a_k} } 2$.

But then $\map g z$ cannot possibly be $0$ in that disk.

Hence the result.

$\blacksquare$