Suppose that $f:\mathbb C \to \mathbb C$ is an analytic function. Then $f$ is a polynomial if...

Question

Suppose that $f:\mathbb C \to \mathbb C$ is an analytic function. Then $f$ is a polynomial if (select all that apply):

(A) For any point $a\in \mathbb C$, if $f(z)=\sum_{n=0}^\infty a_n(z-a)^n$ is a power series expansion at $a$, Then $a_n=0$ for atleast one $n$.

(B) $\lim _{\left|z\right|\to \infty }\left|f\left(z\right)\right|=M$

(C) $\lim _{\left|z\right|\to \infty }\left|f\left(z\right)\right|=\infty $

(D) $\left|f\left(z\right)\right|\le M\left|z\right|^{n\ }$ for $|z|$ sufficiently large and for some $n$.

I know that (B) is false and (C) is true. I can prove it. (D) is also true. How do I prove (A) is true? Given in the Answer key that (A) is true.

For any point $a\in \mathbb C$, if $f(z)=\sum_{n=0}^\infty a_n(z-z_o)^n$ is a power series expansion at $a$, Then $a_n=0$ for at least one $n$.

Consider the power series with $a_{2n+1}=0$ and $a_{2n}=1$ this is not a polynomial. How do (A) is true?

Answer 1

$B$ is true, by Liouville's theorem.

$A$ is sufficient (see the other answer).

$C$ is apparently true (see the link in the comments).

For $D$ it means $\mid f(z)\mid=O(\mid z^n\mid)$. It's sufficient. See this.

Answer 2

For A, Consider $$B:=\{z\in \Bbb C: f^{(n)}(z)=0\;\text{for some}\; n \in \Bbb N \}=\bigcup_n\{z\in \Bbb C: f^{(n)}(z)=0\}$$

Here $B$ is uncountable. That means, atleast one set in the union is uncountable. Thus, $\exists k$ so that $\{z: f^{(k)}(z)=0\}$ is uncountable, so it has limit point in $\Bbb C$ and hence result follows!