Show that an entire function $f$ s.t. $|f(z)|>1$ for $|z|>1$ is a polynomial I have been struggling on the following problem.
Suppose $f$ is an entire analytic function such that $|f(z)|>1$ if $|z|>1$. Show that $f$ is a polynomial.
My idea is as followed: all zeros of $|f(z)|$ lie inside $|z|\leq 1$. Applying Argument Principle, we can show that number of zeros of $f$ is bounded. So we can assume
$f(z)=(z-z_1)...(z-z_M)g(z)$
where g is entire analytic without any zeros.
Then I would like to apply Liouville's Theorem: the point is that it isn't too clear to me why $|\dfrac{1}{g(z)}|$ is a bounded function.
 A: If the Taylor series about 0 does not terminate, $f(1/z)$ has an essential singularity at $0$ (why?)
Then from the Casorati–Weierstrass theorem (have you learnt this?) you know $f(1/z)$ cannot be bounded from below near 0. Contradiction!
A: The function $1/f$ is bounded at infinity, hence has a fake singularity there and can be holomorphically extended through infinity.
Hence $1/f$ is actually rational ($\iff$ meromorphic on the whole Riemann sphere)  and so also $f$ is rational.
But since $f$ does not  have any pole at finite distance (since it is an entire function) it is a polynomial (and has a pole at infinity if it isn't a constant).
A: The number of zeros of $f$ can be finite as the set of zeros of $f$ will have a accumulation point and $f$ will be equivalent to $0$ (not possible) then
$$g=\frac{f}{(z-a)\cdots(z-x)}$$
where $a$ to $x$ are the finite number of zeros of $f$ then $g$ is analytic consider $h=1/g$ see that $h(z)$ is never zero as we have deleted the zeros of $f$ and also $f$ tends to infinity as $z$ tends to infinity (by the given condition) which implies
$$|h(z)| \le a+|z|^n,$$
where $n$ is the number of zeros of $f$, but by fundamental theorem of algebra we get $h(z)=c$ (constant) thus
$$f=\frac{(z-a)\cdots(z-x)}{k}.$$
