I'm learning about complex analysis and need help with the following problem:

Let $f: \mathbb{C} \to \mathbb{C}$ be analytic and non-constant. Show that for every $R > 0$, the complex function $f$ has finitely many zeros inside the disk $D(0, R)$.

I though I could prove the above statement by arguing by contradiction, that is supposing there is an infinite sequence of distinct points $z_k \in D$ with $f(z_k) = 0$ but I didn't get far.


1 Answer 1


The idea of the proof is this one :

As $D(0,R)$ is compact, your sequence $(z_k)$ has a subsequence that converge to some $z \in D(0,R)$. By continuity of $f$, $f(z) = 0$. But then, this zero is not isolated, so $f \equiv 0$

  • $\begingroup$ So this is a contradiction if $f$ is assumed nontrivial. Therefore it follows that if $f$ is nontrivial, then it can only have finitely many zeros inside $D$. Is that correct? $\endgroup$
    – glpsx
    Apr 16, 2016 at 10:47
  • $\begingroup$ @VonKar : yes, that's the idea $\endgroup$
    – Tryss
    Apr 16, 2016 at 12:14

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