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
Best Answer
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.