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I think not, however my proof is quite sketchy so far.. My attempt: Suppose an analytic function f has infinitely many zeros on some closed disk D. Then there exists a sequence of zeros in D with a limit point in D. Thus by the identity theorem (Let D be a domain and f analytic in D. If the set of zeros Z(f) has a limit point in D, then f ≡ 0 in D.), f is identically zero and thus constant.

My main reasons for confusion (other than having a weak understanding of the identity theorem):

-Couldn't such a function f have a finite number of distinct zeros, each with infinite multiplicity? in this case there wouldn't be a convergent sequence of zeros...

-What is the relevance of the fact that D is closed?

Any help in understanding this problem would be greatly appreciated! Thanks

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Your proof is correct, you just need to realize that when you say "has infinitely many zeros" you mean "has infinitely many points where it evaluates to $0$", so one is not talking about multiplicities here. The importance of $D$ being a closed disk is that it is then compact, and that implies the existence of a convergent sequences of zeros of the functions, which allows you to invoke the identity theorem (which you certainly do want to understand fully).

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  • $\begingroup$ What does multiplicity have to do it? What can go wrong? If there are infinitely many zeros "counting multiplicity" then there are infinitely many distinct zeros, or else there is a zero of infinite multiplicity, right? And in either case, everything is fine? What am I missing? $\endgroup$ – bof Apr 13 '15 at 11:11
  • $\begingroup$ Multiplicity is irrelevant since the distinctness of the zeros is implied. The other interpretation of "has infinitely many zeros" that I was considering was "has an infinite number of zeros counting multiplicity", in which case there COULD be say exactly 2 zeros in the disk both with infinite multiplicity, in which case the rest of the proof falls apart. $\endgroup$ – Will Apr 13 '15 at 13:16
  • $\begingroup$ but as quid pointed out, this is an impossible case anyways $\endgroup$ – Will Apr 13 '15 at 13:17
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The relevance of the fact that the set is closed is that the limit point must be in the set. For example $\sin (1/(z+1))$ has infinitely many zeros in the open unit disc, but is not zero. (The sole point of accumulation is $-1$, outside the domain, and the function is not analytic there.)

On the multiplicity: first I think distinct zeros are meant. Second, a non-zero analytic function (on a connected domain) cannot vanish to infinite order anywhere; this would give the series there is $0$ and so the function is zero.

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Since the disk $\mathcal{D}$ is bounded and closed it is compact, so, by Heine-Borel it fulfils the Bolzano-Weierstrass property: every infinite set has at least one limit point. So the set of $\mathcal{Z}$ of zeros has a limit point, and you're right to go with the identity theorem. So you can see the reason for closure: it lets you call on the Heine-Borel theorem.

Incidentally, before about 1930 and Pavel Alexandrof's school, "compact" meant by definition having the Bolzano-Weierstrass property.

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