# Complex function with infinitely many zeros on bounded domain

I found a similar post that discussed complex functions that have infinitely many complex zeros here: Examples of complex functions with infinitely many complex zeros

I was wondering, if one of these functions is in a bounded domain (and its infinitely many zeros are also in the bounded domain) does f have to be identically $$0$$? Or can it be a non-constant function still?

• It is essential to say something about the regularity of the function. Commented Feb 9, 2021 at 16:02
• $\sin \frac{\pi}{1-z}$ on the unit disc has roots at $z=1-1/n, n \ge 1$; they converge to the boundary so no contradiction; there are even bounded functions on the unit dosc with infinitely many zeroes inside (read about infinite Blaschke products) Commented Feb 9, 2021 at 16:02
• @Conrad Awesome thank you! Commented Feb 9, 2021 at 16:04
• @PierreCarre what do you mean by that? Commented Feb 9, 2021 at 16:05
• happy to be of help - usually various conditions on $f$ impose conditions on the zeroes; for example for $f$ bounded in the unit disc, the zeroes $z_n$ ordered say in increasing modulus and with repeats for multiplicty must satisfy $\sum (1-|z_n|) < \infty$; this shows that the example above must be unbounded since $\sum 1/n =\infty$ Commented Feb 9, 2021 at 16:06

Let me put this as an answer to bring all the comments together and make things clear:

As a concrete example $$f(z)=\sin \frac{\pi}{1-z}$$ is the easiest example of a (nontrivial) holomorphic function on the unit disc that has infinitely many zeroes $$z_n=1-1/n$$. However $$\sum (1-(1-1/n))=\infty$$ and that by a general result forces $$f$$ to be unbounded which is clear since it has an essential singularity at $$1$$.

If the zeroes would be say $$1-1/n^2$$ or any $$z_n$$ with $$\sum (1-|z_n|) < \infty$$ one can actually construct a bounded function on the unit disc with zeroes precisely at $$z_n$$ (Blaschke products) and that is one of the simplest theorem of the type $$f$$ satisfies some growth condition (here bounded on the unit disc), then its zeroes satisfy some growth restrictions (here $$\sum (1-|z_n|) < \infty$$) and those are sufficient.

Going even more general, we know that given any sequence in the plane $$z_n$$ with no (finite) accumulation point, one can construct an entire function $$f$$ with zeroes precisely at $$z_n$$ and then there is a factorization theorem that tells us that all such functions are of a specific form (though one needs growth restrictions like finite order for the theorem to be useful as otherwise there are too many choices) so one can ask same about any domain the plane, bounded or not and indeed the analogous result is true though it requires some subtle topological arguments in full generality, while the factorization theorem is even less useful outside specific cases like the unit disc and bounded functions say.

So given any domain (open connected) $$G$$ and any sequence $$z_n$$ with no accumulation point in $$G$$, there is $$f$$ holomoprhic on $$G$$ with zeroes at precisely $$z_n$$

• This helped a lot thank you very much! Commented Feb 9, 2021 at 16:27
• happy to be of help Commented Feb 9, 2021 at 16:28

If you only consider a complex function, without holomorphicity, anything can happen. Simply consider the indicator function $$1_{U\setminus\mathbb{Q}}(z)$$ where $$U$$ is the domain.

If $$f:U\to\mathbb{C}$$ has infinitely many zeros in the bounded domain $$U$$ and can be extended as an entire function, then yes, $$f$$ must be zero, since zeros of non-trivial holomorphic functions are isolated.

If $$f:U\to\mathbb{C}$$ is holomorphic but not necessarily the restriction of an entire function, then Conrad's comment gives an example. In this case, see also this question: Can a non-constant analytic function have infinitely many zeros on a closed disk?

• no because the zeroes can converge to the boundary Commented Feb 9, 2021 at 16:03
• @Conrad: Sure. Edited. Thanks.
– user9464
Commented Feb 9, 2021 at 16:06
• @Conrad : Boundary? Entire function?
– MPW
Commented Feb 9, 2021 at 16:07
• @mrsamy thank you! Commented Feb 9, 2021 at 16:29