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Let $f$ be an entire function. Consider $$ A = \{ z\in \mathbb{C} \mid f^{(n)}(z)=0 \text{ for some positive integer } n\}.$$ Then

  1. If $A=\mathbb{C}$ then $f$ is a polynomial or constant function.
  2. If $A$ is uncountable then $f$ is a polynomial or constant.

I'm not getting this problem. Please give some ideas

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    $\begingroup$ The point is that the zeroes of a nonzero holomorphic function are isolated, so there are countably many of them. $\endgroup$ – Mindlack Nov 9 '19 at 18:25
  • $\begingroup$ Zeros of a non constant function are always isolated. $\endgroup$ – Sachin Nov 9 '19 at 18:29
  • $\begingroup$ @Sachin No this is false, the function $x \mapsto x\sin(1/x)$ has zeros arbitrarily close to zero. (The problem is that this function is not holomorphic at $0$) $\endgroup$ – rawbacon Nov 10 '19 at 10:12
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It suffices to show that $f^{(n)}$ is constant zero. In this case, all higher derivatives are zero, so $f$ is a polynomial.

in 1., $f^{(n)}$ is constant zero by assumption. In 2., the zeros of $f^{(n)}$ must have an accumulation point, so $f^{(n)}$ must be constant by the identity theorem.

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  • $\begingroup$ It would be nice to tell how to choose $n$ as it is not fixed in the statement. $\endgroup$ – Alan Muniz Nov 10 '19 at 1:06
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    $\begingroup$ @AlanMuniz $n$ is just given in the problem, no? $\endgroup$ – rawbacon Nov 10 '19 at 10:11
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I'll only assume that $A$ is uncountable. First note that $$ A = \{ z\in \mathbb{C} \mid f^{(n)}(z)=0 \text{ for some positive integer } n\} = \cup_{n=1}^\infty B_n $$ where $$B_n = \{ z\in \mathbb{C} \mid f^{(k)}(z)=0 \text{ for some positive integer } k\leq n\}.$$ Then we make a natural decomposition $ A = \cup_{n=1}^\infty A_n $, where $A_n = B_n \backslash B_{n-1}$. As $A = \cup_{n=1}^\infty A_n$ is a countable disjoint union and $A$ is uncountable there exists $n$ such that $A_n$ is an uncountable set.

On the other hand $A_n \subset \{f^{(n)} = 0\}$. As $f^{(n)}$ is an entire function, the identity principle implies that $f^{(n)} \equiv 0$ which itself implies that $f$ is a polynomial or a constant.

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