When proper functions are rational functions. 
Let $f:\mathbb{C}-A\rightarrow \mathbb{C}$ be analytic and proper, where $A$ is a finite point set in $\mathbb{C}$.
I.e. $f^{-1}(C)$ is compact for all compact sets $C\subset \mathbb{C}$.
Is it true that $f$ is a rational function, i.e. $f(z)=\frac {p(z)}{q(z)}$ for complex polynomials $p(z)$ and $q(z)$, such that $f$ has poles at $A$ and $\infty$  ? Does the converse hold as well? I.e. if $f$ is a rational function with poles at $A$ and $\infty$, then $f$ is proper.

I am pretty sure this is true, but am new with the concept of poles and am not sure where to start. I would appreciate any hints/help if possible. I apologize for the lack fo work. Thanks in advance!
 A: The first part is not true without $f$ also being holomorphic. To see this, observe that if any such proper map existed at all, holomorphic or not, we could turn it into a non-holomorphic map by adding a radial bump function around some far away point. This is still proper since we're changing nothing in most places, and if we do this radially, the pre-image of any non-zero point is changed only by adding some small circle, which is compact. Bump functions are not real analytic, so this means that we must assume your map is holomorphic.
So we'll assume $f$ is holomorphic.
View $f$ as a map from the Riemann sphere $\mathbb{CP}^1 \to \mathbb{CP}^1$. This is possible since the extension of $f$ to the sphere cannot have an essential singularity at $\infty$, as Picard's theorem would imply $f$ is not proper (preimage of any point is now an infinite set). Now that the map is holomorphic, it is a rational function, as all holomorphic maps from the sphere to itself are rational functions, or is identically $\infty$. This is elementary enough, so I'll explain why below, if you'd like to see a proof.
There are at most finitely many zeros and poles of this map, if it is not identically $0$ or $\infty$. This is because infinitely many such points plus the compactness of the sphere would give a subsequence converging to something in the sphere along which the function would be $0$ or $\infty$, and the identity theorem forbids this. Now let $g(z) = \prod(z-a_i)$, where the $a_i$ are the poles. We are not assuming all of these are distinct, as poles can have order bigger than $1$. We can consider the product $h(z) = f(z)g(z)$. This is also a holomorphic function from the sphere to itself as $f$ and $g$ are. However, $h$ has a finite value at every pole of $f$ by construction of $g$.
Now this implies that $h$ is a polynomial in $z$. To see this, work in the coordinates on the sphere containing $0$ and not $\infty$. At every finite value, $h$ is finite. At $\infty$, we either have $h(\infty) \neq \infty$, in which case, compactness of the sphere plus the maximum modulus principle implies $h$ is constant, and it is a polynomial, or $h(\infty) = \infty$. 
Now let's switch to looking near $\infty$, and consider $\tilde{h}(w) = 1/(h(1/w)$ near $0$. Clearly $\tilde{h}(0) = 0$. Now let $n$ be the order of $0$ at $0$, so that $\tilde{h}(w)$ can be written as $\tilde{h}(w) = \sum_{k=n}^\infty a_k w^k$. We get a $\delta>0$ so that when $|w| < \delta$, $|\tilde{h}(w)| \geq |a_nw^n|/2$. 
For $h$, this means that $|z| \geq 1/\delta$ implies $|g(z)| \leq 2 |z^n|/|a_n|$. Now we use the integral formula for $j > n$ around a contour $\gamma$ containing a ball of radius $R > 0$, and we get:
$$g^{(j)}(0) = \frac{j!}{2 \pi i} \int_{\gamma} \frac{g(z)}{z^{j+1}} dz $$
Now taking $R > 1/\delta$ and taking norms, we get the estimate using $LM$ (length times maximum value):
$$|g^{(j)}(0)| \leq \frac{ j}{2 \pi |a_n|} \int_\gamma \frac{|z|^n}{|z^{j+1}|} dz \leq \frac{2j!}{ |a_n|R^{j-n}} $$
Taking $R \to \infty$ now, we see that for $j \geq n+1$, the derivatives all identically vanish (note that we're just imitating the proof of Liouville's theorem here), and this establishes $h$ is a polynomial.
Now that we've established $h$ is a polynomial, $h = fg$ implies $f = h/g$, and $g$ is clearly a polynomial. This completes the proof.
A: For the easy direction, suppose $f$ has poles at $A$ and $\infty$, and let $N>0$. Then $\frac{1}{f}$ is continuous there with value $0$, so there is an open set around each point in this finite collection in which $\frac{1}{f}$ takes values of magnitude at most $\frac{1}{N}$. Then $f^{-1}(B_N(0))$ is a subset of the complement of these open sets, which is compact, so $f$ is proper.
