Zeroes of Analytic Function are Isolated

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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 least 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:

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

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

Now we define a new function $g(z)$, as the sum on the right hand side, which is clearly analytic in $|z-z_0|<R$.

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

Since $g(z_0) = a_k \neq 0, \exists \epsilon >0$ so that $\forall z$ such that $|z-z_0|<\epsilon, |g(z)-a_k|<\dfrac{|a_k|}{2}$.

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

Hence the result.

$\blacksquare$