I am working on a textbook exercise. A similar question: An analytic function in a compact region has finitely many zeros, but it's not quite clear to me and I also have possibly another approach? I want to prove basically the same question, that if $f$ is analytic inside and on a simple closed contour $C$ (except possibly for poles inside $C$), and if all zeros of $f$ are inside $C$ and of finite order, then the zeros must be finitely many.

Hopefully my attempt below can be verified or corrected.

My attempt:

Suppose otherwise. Then by Bolzano-Weierstrass, the set $S$ of all zeros of $f$ (which is infinite) contains an accumulation point inside $C$. Let's say it is $z_0$. This $z_0$ is also a zero of $f$ since it is the limit a subsequence of zeros in $S$ and $f$ is analytic (hence continuous too). By assumption, it is a zero of finite order, say $m$.

I claim that in any neighborhood $N$ of $z_0$, $f$ cannot be identically zero. To see this, write $f(z)=(z-z_0)^mg(z)$ where $g$ is nonzero and analytic at $z_0$. Hence by these properties of $g$, there is a neighborhood around $z_0$ (intersected with $N$) where $g$ is nonzero. However, this neighborhood contains another (different) zero, say $z'$, of $f$ by definition of accumulation point. Hence, $0=f(z')=(z'-z_0)^mg(z')$, implying that $g$ can be zero in this neighborhood, a contradiction.

Now by a theorem in the textbook, since $f$ is analytic and zero at $z_0$, but not identically zero in any neighborhood of $z_0$, there must be a deleted neighborhood of $z_0$ where $f$ is identically nonzero. But again, in this deleted neighborhood contains a zero of $f$, say $z''$, by definition of accumulation point, contradicting $f$ being identically nonzero there. QED.

So my questions would be:

  1. Is the above valid? If not, which part should be improved?

  2. Are there any other approaches?

Usually Q2 is more interesting, but I highly appreciate if Q1 is answered too. Thanks a lot!

EDIT: Now that I think about it after some comment inputs:

My first paragraph should be fine.

  1. As for my second paragraph until conclusion, I should do it like this:

As $z_0$ is of order $m$, we can write $f(z) = (z-z_0)^m g(z)$ where $g$ is analytic and nonzero at $z_0$. By continuity of $g$ and being nonzero at $z_0$, there is a neighborhood at $z_0$ where $g$ is identically nonzero. Deleting $z_0$ there, $f$ is then nonzero in that deleted neighborhood. However, this contradicts the fact that $z_0$ is an accumulation point of zeros. Done?


  1. Another method, I can also say: Either $f$ is not identically zero in any neighborhood $N$ of $z_0$ , or $f$ is identically zero in some neighborhood $N$ of $z_0$ . For the former, my original third paragraph follows to conclude. For the latter, by identity theorem $f$ must be identically zero inside $C$. By analyticity, their derivatives of all order are zero, showing infinite order. Done?
  • $\begingroup$ You wrote "In any neighborhood $N$ of $z_0$, $f$ cannot be identically zero" . Why? This is almost like assuming what you want to prove, which is: an analytic non-zero function in a non-empty compact region can have only a finite number of zeros in it". $\endgroup$
    – DonAntonio
    Commented Dec 19, 2020 at 10:14
  • $\begingroup$ The reason is in the next sentence. I think I wrote it not in a natural way. I just edited it, adding "I claim that". $\endgroup$ Commented Dec 19, 2020 at 10:37
  • $\begingroup$ Your understanding is correct, except that in the second paragraph you proved that if the zeros have an accumulation point where $f$ is analytic then $f$ must be identically $0$ around it, thus by analytic continuation / identity theorem it must be identically $0$ in the whole connected component. $\endgroup$
    – reuns
    Commented Dec 19, 2020 at 15:26
  • $\begingroup$ @DonAntonio I think I realize my mistake. I have put two edits, two methods but similar conclusion. $\endgroup$ Commented Dec 20, 2020 at 5:08
  • 1
    $\begingroup$ The "except " part is something I was missing. $\endgroup$
    – DonAntonio
    Commented Dec 20, 2020 at 13:16

1 Answer 1


I propose the following: let us prove that if a function $f$ is analytic in the region $R$ consisting of all points inside and on a simple closed contour $C$, except possibly for poles inside $C$, and if all the zeros of $f$ in $R$ are interior to $C$ and are of finite order, then those zeros must be finite in number. I think we must add the condition that $\;f\;$ isn't identically equal to zero in any non-trivial open, connected subset of $\;R\;$ . This is from a book (I already found a paper about this from 1981...) which I still cannot locate and it seems to be something very close to what you actually want. Observe that the conditions above for the function $\;f\;$ actually say the function's meromorphic on the domain enclosed by $\;C\;$ .

Proof: Suppose there are infinite zeros $\;\{z_1,z_2,...\}\;$ of $\;f\;$ inside $\;C\;$ . Then by Bolzano-Weierstrass, there exists $\;z_0\;$ on $\;R\;$ s.t. $\;\lim\limits_{n\to\infty}z_n=z_0\;$ . By continuity of $\;f\;$ , we get that $\;f(z_0)=0\;$ , too.

Since we're assuming all the zeros of $\;f\;$ on $\;R\;$ are of finite order and isolated, there exists $\;m\in\Bbb N\;$ s.t. $\;f(z)=(z-z_0)^mg(z)\;$ , in some open neighborhood $\;U\;$ of $\;z_0\;$ and for some meromorphic function $\;g\;$ s.t. $\;g(z)\neq0\;\;\forall\,z\in U\;$ . Since the possible poles of $\;f\;$ inside $\;C\;$ are isolated, we can take a neighborhood $\;V\;$ of $\;z_0\;$ where there are no poles of $\;f\;$ inside $\;V\;$ , and take the above relation $\;f(z)=(z-z_0)^mg(z)\;$ in $\;U':=U\cap V\;$, and this time $\;g\;$ is non-zero and analytic in $\;U'\;$ .

Thus we are almost thru, since then by the identity theorem of analytic functions we'd get that $\;f\;$ would be identically zero in some connected neighborhood of $\;z_0\;$ , since this point is an accumulation point of a set where $\;f\;$ and the zero function coincide, and this contradicts the further condition added above.

  • $\begingroup$ Thanks. It kinda looks similar but more organized. I will accept it. $\endgroup$ Commented Dec 20, 2020 at 13:22

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