Entire function $f: \mathbb{C} \to \mathbb{C}$ maps every unbounded sequence to an unbounded sequence then $f$ is a polynomial? I came across this problem. Hopefully my last for the exam tomorrow.
(i) Entire function $f: \mathbb{C} \to \mathbb{C}$ maps every unbounded sequence to an unbounded sequence then $f$ is a polynomial?
The other question is
(ii) Is the same conclusion true if $f$ is holomorphic in $ \mathbb{C} \setminus\{ 0\}?  $
My thoughts go to the Liouville's Theorem first, since $f$ is unbounded it can't be constant. From there I would say that the function $f$ is analytic (bc holomorphic) with $\sum\nolimits_{n=0}^ \infty a_n(z - z_0)^n$ and some $a_n \neq 0$ for an $n > 0$ since $f$ is not constant. And therefore it must be a polynomial? But I dont have the gut feeling that I am here right or am I? :D On the other question (ii) I can't find a thing why it not should work like my thought befor actually.
Thank you guys

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*Consider $\infty$ as a singularity point. It's known that it's a pole of an entire function $f$ iff $f$ is a polynomial. Let's show it is a pole. Obviously it's not a removable singularity since then $f$ is bounded. If it's an essential singularity, by the Casserti Weirstrass thm we can find an unbounded sequence $a_n\to\infty$ with $f(a_n)\to 0$, contradiction. Therefore it is a pole and $f$ is a polynomial.


*This is false - consider $f(z)=\frac{z^2+1}{z}$. If $a_n$ is unbounded, take $a_{n_k}\to\infty$. Since $\infty$ is a pole of $f$, $f(a_{n_k})\to\infty$ and $f(a_n)$ is unbounded. So $f$ maps unbounded sequences to unbounded sequences, but is not a polynomial.
In fact, one can show that the only functions that are holomorphic in the punctured plane and satisfy this property must be rational with a pole at infinity (this is if and only if).
