(i) Let $f$ be an entire function satisfying $$|f(z)|\leq K|z|^{5/2}$$ for $K>0$ and $|z|>1$. Prove that $f$ is a polynomial of degree at most $2$.
(ii) Now suppose the bound in (i) is valid for all $z\in\mathbb{C}$. Prove that $f$ must be the zero function.
For (i), I write $f(z)=\sum^{\infty}_{n=0}\frac{f^{(n)}(0)}{n!}z^n$. By Cauchy's Estimate, $$|f^{(n)}(0)|\leq \frac{n!KR^{5/2}}{R^n}$$ for $|z|=R>1$.
Letting $R\rightarrow \infty$, we get $f^{(n)}(0)=0$ for $n\geq 3$.
Hence $f$ is a polynomial of degree at most $2$.
For (ii), I write $f(z)=az^2+bz+c$. Since $|f(0)|\leq K0=0$. We have $c=0$. So we have $f(z)=az^2+bz$. I have no idea how to proceed so that $a=b=0$.
We have
$$K|z|^{5/2} \ge |az^2 + bz| = |z| \cdot |az + b| \implies |az + b| \le K |z|^{3/2}$$
for all $z \ne 0$. This implies $b = 0$, and a repeated reduction leads to $a = 0$.