Mapping of Analytic Functions I have another question that is really catching me off guard, but it looks very promising and wholesome as it combines complex analysis theory and algebra. The question is as below:
Let $f$ be entire and have the property that 
if $B \subset \mathbb{C}$ is any bounded set,$\hspace{1.7in}$
then $f^{-1}(B)$ is bounded (or perhaps empty).
Could it be shown that for any $\omega \in \mathbb{C}$, 
there $\hspace{0.7in}$ exists $z \in \mathbb{C}$ such that $f(z)=\omega$.
Also, can one show that $f(\mathbb{C})$ is both open and $\hspace{1.1in}$ 
closed and deduce that $f(\mathbb{C})=\mathbb{C}$. Apply this result to
polynomials to deduce $\hspace{1.4in}$ yet another proof of the Fundamental
Theorem of Algebra.
 A: I'm not sure what theorems you've seen, but if you know the open mapping theorem, then you know that $f(\mathbb C)$ is open.  Let $(z_n)$ be a sequence in $f(\mathbb C)$ converging to $z\in\mathbb C$.  Then there is a sequence $(w_n)$ with $f(w_n)=z_n$.  Since convergent sequences are bounded, $(z_n)$ is a bounded sequence.  Since the sequence $(w_n)$ lies in $f^{-1}(\{z_n\})$, $(w_n)$ is also a bounded sequence, and therefore there is a subsequence $(w_{n_k})_k$ converging to some $w\in\mathbb C$.  By continuity of $f$, $f(w)=\lim_k f(w_{n_k})=\lim_k z_{n_k}=z$.  Therefore $f(\mathbb C)$ is closed.  Since $\mathbb C$ is connected, it follows that $f$ is onto.
If $f$ is a nonconstant polynomial, then $\lim_{z\to\infty}f(z)=\infty$, which implies that $f$ satisfies the hypothesis.  Therefore $f$ is onto, and in particular $0$ is in its image, which is the fundamental theorem of algebra.
An alternative solution to the original problem that is probably overkill and actually uses the fundamental theorem of algebra goes as follows.  If $f$ is a nonconstant polynomial, then it is onto by the fundamental theorem of algebra.  If $f$ is an entire function that is not a polynomial, then it has an essential singularity at $\infty$, and hence $f(\{z:|z|>M\})$ is dense in $\mathbb C$ for all $M>0$ by the Casorati-Weierstrass theorem.  In particular, $f^{-1}(\{z:|z|<1\})$ intersects $\{z:|z|>M\}$ for all $M>0$, and therefore it is unbounded.  (I guess this is a good approach if for some reason you want to avoid using the open mapping theorem.)
Edit: Removed remark about assuming that $f$ is nonconstant, since that is implied by the hypothesis as Theo pointed out.
A: Such a map $f$ is called proper. It is easy to show (using local compactness of $\mathbb{C}$, continuity of $f$ and compactness of $f^{-1}(K)$ for each compact $K$) that $f$ maps closed sets to closed sets and since $f$ is not constant it maps open sets to open sets by the open mapping theorem. Hence it does what you want: the image of $f$ is both open and closed and of course non-empty in $\mathbb{C}$, hence it is all of $\mathbb{C}$ by connectedness.
Using the growth lemma for polynomials you can easily show that non-constant polynomials are proper, hence are onto. I leave the details for you as they are rather straightforward.
A: Building on my comment:
Picard's theorem states that an entire function which misses more than two values is constant. Our function is clearly not constant, as mentioned before. Suppose it is not onto. Then there exists only one element $z \notin f(\Bbb{C})$. Pick a disk $D=D(z,r)$ centred in $z$. Since $D$ is bounded, it follows that $f^{-1}(D)$ is bounded. There exists $z_n \in D,\ z_n \to z$ for which there exist $w_n$ with the property that $f(w_n)=z_n$. The sequence $w_n$ is bounded, therefore it has a convergent subsequence $(w_{n_k})\to w$ and finally $f(w)=z$. Contradiction.
This means that $f$ is onto.
This is a valid argument, but clearly, not as simple as the first one.
