Complex Analysis - Rational Functions I'm studying for my final exam and came across this problem:
Let f and g be entire analytic functions and |f(z)|<|g(z)| when |z|>1. Show that f/g is a rational function. 
I'm not quite sure where to start at all, but it does seem like a rather interesting result that I wouldn't expect.
 A: First, $g$ can only have a finite number of zeros say $a_1,\ldots, a_n$. More importantly $g(z)=(z-a_1)\cdots(z-a_n)\cdot h$ where $h$ is some entire nonzero function. But then $|\frac{f(z)}{h(z)}|<|(z-a_1)\cdots(z-a_n)|$ for every $|z|>1$. That is $\frac{f}{h}$ is bounded by a polynomial. Thus, $\frac{f}{h}$ is a polynomial, which shows both that $h$ must be a constant and that $f$ is a polynomial. Thus both $f$ and $g$ are polynomials. 
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Theorem. Let $f$ be an entire function and $n\in\mathbb{N}$ such that $|f|\le |z|^n$. Then $f$ must be a polynomial. 
Proof. Let $f(z)=\sum_0^\infty a_k z^k $, we will show that the $a_k$'s must be zero eventually; they will in fact be zero when $k>n$. We perform the following calculation ($n$ is fixed, $m$ is arbitrary): 
$\displaystyle |a_{n+m}|=\left|\frac{f^{(n+m)}(0)}{(n+m)!}\right|=\lim_{R\to\infty} \left|\frac{1}{2\pi i} \int_{B(0,R)} \frac{f(z)}{z^{n+m+1}} dz\right|\le\lim_{R\to\infty} \frac{1}{2\pi} \underset{z\in B(0,R)}{max}\ \left|\frac{f(z)}{z^{n+m+1}} \right|\cdot 2\pi R  \le\lim_{R\to\infty} \underset{z\in B(0,R)}{max}\frac{|z|^n}{|z|^{n+m+1}} \cdot R  \le\lim_{R\to\infty} \frac{R^n}{R^{n+m+1}} \cdot R  = 0$
Thus, $a_k=0$ for $k>n$ and so $f$ must be a polynomial. 
For the general result notice that any polynomial $|p|\le M\cdot |z|^n$ for some $M\in\mathbb{R}$ and $n\in\mathbb{N}$. 
Interestingly, using the real or imaginary version of cauchy integral formula for taylor coefficients, we can extend this result to the real or imaginary part of a function. That is if $f=u+vi$ and $|u|\le |p|$ for some polynomial $p$ then $f$ must a polynomial. We can even (shockingly) drop the absolute values. 
Theorem (Markushevich - Volume 2 - Page 265) Let $f=u+vi$. Suppose that $u(z)\le|z|^n$ for some $n\in\mathbb{N}$, then $f$ must be a polynomial. 
