Polynomial growth, using the Cauchy Integral Formula, Is this a true statement in Complex Analysis?  
If a function grows like a polynomial, then it is a polynomial.
Or, is it really:  if a function grows like a polynomial at infinity, then it is a polynomial?
I am not sure which of these I saw in lecture, so I just wanted to post this question up to get some discussion going.  
Also, I am trying to prove this, using the Cauchy Integral Formula for Derivatives.  Can I just take the derivative at z=0, so that I am showing that $f^n$(0) = 0, when I let the radius R go to infinity?  Or does it have to be an arbitrary point, $z_0$?
Can someone provide a link to a proof?  I recall reading one from Gamelin...online, but I haven't been able to find it again.
One more question, actually: When I bound the integrand, should I put the modulus sign around the dz factor?  Or dz is always real-valued and hence it's not necessary? 
Thanks,
Edit: Or, feel free to share your own proof with me instead of giving me a link to another one.  My proof was pretty short, but I'm not entirely sure that it's correct.  If I let R go to infinity, and am using the Cauchy Integral Formula, which requires my function to be analytic on and inside this arbitrarily large disk, then I must be implying that f is entire, right?  So, I am really proving (or my professor, during lecture) that an entire function that grows like a polynomial at infinity is, in fact, a polynomial.  Does this sound ok?  Thanks,
 A: An entire function with polynomial growth is indeed a polynomial.
This is easy to see: If $|f(z)|\le M|z|^n$ for large $z$, recall that
$$f^{(n+1)}(z)=\frac1{2\pi i n!}\int_\Gamma \frac{f(\zeta)}{(\zeta-z)^{n+2}}\,d\zeta,$$
where $\Gamma$ is a circle surrounding $z$. Let $\Gamma$ be centered at the origin, and let its radius go to infinity to conclude $f^{(n+1)}(z)=0$.
Edit: I wasn't reading the question carefully enough. I see that this is precisely what OP suggested. I think this way is easiest. But if you do it only for $z=0$, then you conclude instead that the Taylor series is a polynomial, and then you have to rely on the convergence of the Taylor series to $f$. Admittedly, this is also one of the elementary results of the theory, but it still uses more machinery.
I am leaving this here even so, since someone might find it useful.
A: You're right:

Theorem 1 Let $f$ be an entire function. If there exists $n\in\mathbb{N}$, and $M,R>0$ such that $|f(z)| \leq M |z|^n$ for every $z\in\mathbb{C}\backslash D_R(0)$, then $f$ is a polynomial of degree at most $n$.

Proof
Consider the Taylor series of $f$:
$$
f(z) = \sum_{k=0}^\infty a_kz^k
$$
From the Cauchy Theorem  you can prove that (*), for $R\leq r$,
$$
|a_k| \leq r^{-k} \max_{|z|=r}\left|f(z)\right| \leq r^{-k} M r^n.
$$
Taking $k>n$ and  $r\to\infty$ we get that $a_k = 0$, hence $f$ is a polynomial.
See that Lioville's Theorem is just a particular case of this theorem for $n=0$.
(*) To show this remember that $a_k = \frac{f^{(k)}(z_0)}{k!}$ and use the Cauchy Inequality.
On the other side, we have the following theorem:

Theorem 2  Let $f$ be an entire function. If there exists $n\in\mathbb{N}$, and $M,R>0$ such that $|f(z)| \geq M |z|^n$ for every $z\in\mathbb{C}\backslash D_R(0)$, then $f$ is a polynomial of degree at least $n$.

Proof
Consider $z\in\mathbb{C}\backslash D_R(0)$, then $f(z) \geq MR^n$. Then, $0\not\in \overline{f(\mathbb{C}\backslash D_R(0))}$. If $f$ was transcendental this set would have to be dense, but it isn't. So $f$ is a polynomial.
I won't show the assertion about the degree of $f$ since it seems irrelevant for your question.
