# Entire function is a polynomial or constant? [closed]

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

• The point is that the zeroes of a nonzero holomorphic function are isolated, so there are countably many of them. – Mindlack Nov 9 '19 at 18:25
• Zeros of a non constant function are always isolated. – Sachin Nov 9 '19 at 18:29
• @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 $0$) – rawbacon Nov 10 '19 at 10:12

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.
• It would be nice to tell how to choose $n$ as it is not fixed in the statement. – Alan Muniz Nov 10 '19 at 1:06
• @AlanMuniz $n$ is just given in the problem, no? – rawbacon Nov 10 '19 at 10:11
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.