Prove that the entire function $f$ is a polynomial if it maps every unbounded sequence to an unbounded sequence. While studying for an exam in complex analysis, I came across this problem. Unfortunately I was not able to solve it. Any help would be greatly appreciated.
Let $f$ be an entire function mapping every unbounded sequence in $\mathbb{C}$ to an unbounded sequence. Prove that $f$ is a polynomial.
 A: The hypothesis is equivalent to $f$ being entire and $\lim_{|z|\to \infty} |f(z)| = \infty$. 
Let $\phi(z) = f(\frac{1}{z})$, then $\phi$ is analytic on $\mathbb{C}\setminus \{0\}$ and, by hypothesis, has a pole at $z=0$. Hence for some $m$ we can write $\phi(z) = \frac{g(z)}{z^m}$, where $g$ is entire. Consequently, we have $f(z) = z^m g(\frac{1}{z})$ for $z\neq 0$. The set $K=\{\frac{1}{z} \,|\, |z| > 1 \}$, is bounded, and since $g$ is entire, there exists $M>0$ such that $|g(w)| \leq M$ for $w \in K$. Hence if $|z|>1$, we have $|f(z)| \leq M |z|^m$. Since $f$ is bounded on $|z|\leq 1$, we can (by adjusting $M$, if necessary) assume that $|f(z)| \leq M (|z|^m+1)$ for all $z$.
Now use Cauchy's estimate, let $R>0$, then we have $|f^{(k)}(0)| \leq n!\frac{M (R^m+1)}{R^k}$. Since $R>0$ is arbitrary, we see that if $k>m$, then $f^{(k)}(0)=0$. It follows that $f$ is a polynomial.
A: Let $ f $ be a non-constant entire function.

Case 1: $ f $ is a polynomial function.
If you factor out the largest power of $ z $ in $ f $, you will find that
$$
\lim_{n \to \infty} |z_{n}| = \infty \quad \Longrightarrow \quad \lim_{n \to \infty} |f(z_{n})| = \infty.
$$

Case 2: $ f $ is a transcendental function.
The Maclaurin series of $ f $ cannot terminate, otherwise $ f $ would be a polynomial function.
Consider the holomorphic function $ g: \mathbb{C} \setminus \{ 0 \} \to \mathbb{C} $ defined by
$$
\forall z \in \mathbb{C} \setminus \{ 0 \}: \quad g(z) \stackrel{\text{def}}{=} f \left( \frac{1}{z} \right).
$$
Note that $ g $ has an essential singularity at $ z = 0 $. This is because the left-tail of the Laurent series of $ g $ centered at $ z = 0 $ (obtained by making the substitution $ z \to \dfrac{1}{z} $ in the Maclaurin series of $ f $) does not terminate.
Applying the Weierstrass-Casorati Theorem, we see that for any neighborhood $ U $ of $ 0 $, the set $ g[U \setminus \{ 0 \}] $ is dense in $ \mathbb{C} $. Hence, for any $ w \in \mathbb{C} $, there exists a sequence $ (y_{n})_{n \in \mathbb{N}} $ in $ \mathbb{C} \setminus \{ 0 \} $ such that $ \displaystyle \lim_{n \to \infty} y_{n} = 0 $ and $ \displaystyle \lim_{n \to \infty} g(y_{n}) = w $. By setting
$$
(z_{n})_{n \in \mathbb{N}} := \left( \frac{1}{y_{n}} \right)_{n \in \mathbb{N}},
$$
we obtain $ \displaystyle \lim_{n \to \infty} |z_{n}| = \infty $ and $ \displaystyle \lim_{n \to \infty} f(z_{n}) = w $. Therefore, there exist unbounded sequences in $ \mathbb{C} $ that are mapped by $ f $ to a convergent sequence.

Conclusion: If $ f $ is a non-constant entire function, then $ f $ maps every unbounded sequence in $ \mathbb{C} $ to an unbounded sequence if and only if $ f $ is a polynomial function.
A: Hint: Suppose the contrary, that the entire function is not a polynomial and thus has an essential singularity at $z=\infty$.  How can Picard's big theorem help you?
