Sufficient condition for $f(z)$ to be polynomial I think it suffices to exhibit a sequence $\{R_n\}$ of positive real numbers such that $R_n \to \infty$ with $f(z) \neq 0,$ whenever $|z|=R_n$ and $\begin{align} \left|\int_{|z|=R_n}\dfrac{f'(z)}{f(z)}dz \right|\end{align} \leq M, \forall n$.
But I am loss at how to continue from here. Appreciate if someone could advise me.  Thank you. 
 A: I wouldn't use that approach personally (I doubt it would work as you have no information on $f'$). The theme of my approach is; first observe if $f$ had no zeroes, $1/f$ would be entire and bounded, hence constant by Liouville's theorem. What we can do in this case is multiply $1/f$ by a polynomial to make it entire, then use an extended version of Liouville's theorem that says polynomially bounded entire functions are polynomials, to get that it's a polynomial. 
First observe that $f$ has at most finitely many zeroes; if it had infinitely many, you could make a sequence of them, and as this sequence would be contained in closed ball of radius 1, which is compact, it'd have a convergent subsequence. This implies the set of zeroes of $f$ has an accumulation point; a contradiction, as $f$ cannot be identically zero. 
Now, let $p(z)$ be the polynomial with the same zeroes as $f$ (with the same multiplicities). I claim $$F(z) = p(z)/f(z)$$ is entire. It's holomorphic at all points $z$ with $f(z) \neq 0$ by the quotient rule. For each zero of $f$, say $z_0$ with multiplicity $m$, we can find an open neighbourhood of $z_0$ where we can write $f(z) = (z-z_0)^m h(z)$ with $h$ analytic and $h(z) \neq 0$ for all $z$ in the neighbourhood, and $p(z) = (z-z_0)^m q(z)$ with $q$ a polynomial. Then on this neighbourhood, $F(z) = q(z)/ h(z)$, showing it's analytic on the neighbourhood, hence at $z_0$.
We now apply this extended version of Liouville's theorem:

Suppose $f: \mathbb{C} \rightarrow \mathbb{C}$ is entire, and there exists $c,C>0$ and $R >0$ and $n\in \mathbb{N}$ such that for $|z| > c$, $\vert f(z) \vert < C \vert z \vert ^n$. Then, $f$ is a polynomial of degree at most $n$.

This applies to $F$ because when $|z|>1$, $|F(z)| < \frac{|p(z)|}{1}$. So $F$ is a polynomial, hence $f$ is rational. Then because $f$ is entire, it has to be a polynomial.
A: You could consider the holomorphic function $$g:D\setminus\lbrace0\rbrace\to D,\quad z\mapsto g(z)=f\left(\frac1z\right)^{-1}$$ where $D$ is the open unit disk of $\Bbb C$. By hypothesis, it is bounded by $1$, hence it extends holomorphically to $0$. Define $m\in\Bbb N$ by
$$g(z)=z^mh(z)$$
with $h$ holomorphic and $h(0)\neq 0$. Then $\frac{f(z)}{z^m}$ is converges to $\frac1{h(0)}$ as $|z|$ tends to infinity, and a standard consequence of Liouville's theorem mentioned in Matt Rigby's answer implies that $f$ is a polynomial of degree $m$.
