Showing an entire function f(z) is monomial

Question

Given f is entire and satisfying $|f(z)| \leq 3|z|^{\alpha}$, show that $f(z) = cz^{\alpha}$ for some constant $c$ if $\alpha$ is a positive integer, and $f(z) = 0$ if $\alpha$ is not an integer.

I already showed that $f(z) = a_1z + a_2z^2 + ... + a_nz^n$, where $n$ is the largest positive integer less than $\alpha$. I tried to consider the function $g(z) = \frac{f(z)}{z^{\alpha}}$, which has modules less than or equal to $3$, but I'm not sure how to proceed from there.

Thank you.

EDIT: I should write the title "monomial", not "monic".

Answer 1

f has a power series expansion $f(z)=\sum_{n=0}^\infty a_n z^n$. Using Cauchy's inequalities, (which are simple consequences of Cauchy's integral formula) we have $$ \forall\,r>0,\quad |a_n|r^n\leq M(r) $$ where $M(r)=\sup\{|f(re^{i\theta})|:\theta\in\Bbb{R}\}$. By assumption we have $M(r)\leq 3r^\alpha$, thus $$ \forall\,r>0,\quad |a_n| \leq 3 r^{\alpha-n} $$

• If $n>\alpha$, by letting $r\to\infty$, we obtain $a_n=0$.
• If $n<\alpha$, by letting $r\to0$, we obtain $a_n=0$. So, if $\alpha\notin\Bbb{N}$, there are no $a_n$'s left, and $f\equiv0$. on the other hand if $\alpha$ in an integer we conclude from the above discussion that $f(z)=a_\alpha z^\alpha$, with $|a_\alpha|\leq 3$.