I am studying complex analysis from Theodore Gamelin's text and Exercise 1 of chapter IX.2 says that if $f$ is analytic inside the open unit disk and continuous on its boundary that satisfies $|f(z)| = 1$ for $|z| = 1$, then $f$ is a finite Blaschke product. Clearly, this would imply that $f$ has only finitely many zeros in the open unit disk. But the proof of it already assumes this fact. So my question is that is it trivial that such an $f$ has finitely many zeros in the open unit disk?

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    $\begingroup$ every zero is isolated because $f(a+z) = \sum_n c_n z^n$. and since every bounded sequence has a convergent subsequence, if $f$ had infinitely many zeros on a finite region, one of the zero would be an accumulation point of some sequence of zeros (it wouldn't be isolated). you can also integrate $f'/f$ on a closed contour without zero, it is finite hence $f$ (if it is analytic inside) has a finite number of zeros (counted with multiplicity) $\endgroup$
    – reuns
    Mar 25, 2016 at 0:08
  • $\begingroup$ @user1952009 both arguments makes perfect sense. Thanks! $\endgroup$
    – dezdichado
    Mar 25, 2016 at 0:12
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    $\begingroup$ just to say that appart maybe the great Picard theorem, nearly every theorems on holomorphic/meromorphic/analytic functions are very easy to prove (once you know the trick / main argument), that's what is so magic with the complex analysis $\endgroup$
    – reuns
    Mar 25, 2016 at 0:14
  • $\begingroup$ @user1952009 The zeros could - theoretically - accumulate to the boundary... But $|f(z)| = 1$ there. So, no. ;-) $\endgroup$ Mar 25, 2016 at 0:47
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    $\begingroup$ I found such a function (by googling): $f(z) = \sin(1/(1+z))$ is holomorphic in $\mathbb D$ and has infinitely many zeros there. The sequence converges to $-1$, where the function has an essential singularity. $\endgroup$ Mar 25, 2016 at 1:27

1 Answer 1


Let $\mathbb{D}$ denote the open unit disc. In general, an analytic function $f:\mathbb{D}\to\mathbb{C}$ is allowed to have countably many zeros in $\mathbb{D}$. As Friedrich has pointed out, $$ \sin\left(\frac{1}{1+z}\right) $$ is an example of a function that is analytic on $\mathbb{D}$ and has infinitely zeros inside $\mathbb{D}$.

However, if we assume that $f$ is continuous on $\mathbb{D}$, and also that $|f(z)| = 1$ for $|z|=1$, then the story changes. Suppose $f$ has countably many zeros $z_n$ in $\mathbb{D}$. Then by compactness, the set $\{z_n\}$ has a limit point in $\overline{\mathbb{D}}$.

The zeros cannot have a limit point on boundary of the unit disc, since if $z_{n_k}\to z_\infty\in\partial\mathbb{D}$ then $f(z_{n_k})\to f(z_\infty)$ by continuity, but $|f(z_{n_k})| = 0$ and $|f(z_\infty)| = 1$, contradiction.

So the limit point in $\mathbb{\overline{\mathbb{D}}}$ must lie inside $\mathbb{D}$. But then $f$ has a sequence of zeros converging inside its domain of definition, and since $f$ is analytic it follows that $f \equiv 0$. This is a contradiction if $f$ is assumed nontrivial.

Therefore it follows that if $f$ is nontrivial, then $f$ can only have finitely many zeros inside $\mathbb{D}$. At this point one can express $f$ as a product of finitely many Blaschke factors using a consequence of the Schwarz lemma.

  • $\begingroup$ If $f$ is only assumed to be analytic in the open unit disk and $|f(z)|\rightarrow 1$ as $|z|\rightarrow 1$, then would $f$ still have finitely many zero? I am thinking we can extend $f$ to be analytic in a neighborhood of the boundary of the unit disk and then apply the same argument as above. $\endgroup$
    – dezdichado
    Apr 4, 2016 at 4:34

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