# Bolzano-Weierstrass and zeros of complex analytic function

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?

OR

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?
• 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". Dec 19, 2020 at 10:14
• 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". Dec 19, 2020 at 10:37
• 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. Dec 19, 2020 at 15:26
• @DonAntonio I think I realize my mistake. I have put two edits, two methods but similar conclusion. Dec 20, 2020 at 5:08
• The "except " part is something I was missing. Dec 20, 2020 at 13:16

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.