Is an analytic one-to-one function on the whole plane necessarily a polynomial? (Can it be disproved?) I had to show what a one-to-one analytic function from the plane to itself could possibly be.
So, I studied the behavior of such a function at infinity:
Case 1: Such a function cannot have no singularity (or a removable singularity) at infinity, since then the function would be entire and bounded, and by Liouville's Theorem, is constant, which contradicts the one-to-one assumption of f(z).
Case 2: Such a function cannot have an essential singularity at infinity, since then the function would attain every complex value (except for maybe one value) infinitely often in a (every) neighborhood of infinity, which again contradicts the one-to-one assumption of f(z).  This comes from Picard's theorem.
Cases 1 & 2 imply that the function must have a pole at infinity.
We conclude that f(z) must be a polynomial.
Is this ok? 
My concern is that I feel that I've made a jump from saying that a 1-1 entire function with a pole at infinity is a polynomial -- as if it were a definition.  Can I treat this property of a polynomial as a definition?  (Which would then make my proof complete.)
Thanks,
 A: A meromorphic function over the Riemann sphere is necessarily a rational function (subtract its Laurent pieces and apply Liouville to the resulting function), so indeed, if your entire function is meromorphic at infinity, it should have a pole at infinity, and is hence a polynomial.
You may want to see it as a meromorphic function that is holomorphic at infinity: then it has a finite number of poles, which lie on the complex plane. This you can do by simple inversion  ($z\mapsto 1/(z-\alpha)$).
Then you might want to see it as a differentiable automorphism of the Riemann sphere, and its degree is exactly the degree of the polynomial.  C'est fini!
I forgot to say that, given a rational function $f(z)=P(z)/Q(z)$ with irreducible fraction, one can see that the degree of $f$ as an analytic map of the Riemann sphere onto itself is precisely $\mbox{max}(\mbox{deg }P,\mbox{deg }Q).$  In fact, Lüroth's theorem (which is an exercise in Elementary Abstract Algebra) shows precisely that in terms of field extensions: if $T$ is a transcendental variable over a field $k,$ then every subextension
$$k(T)|L|k$$ is either trivial or $L=k(\frac{P(T)}{Q(T)})$ with $P$ or $Q$ nonconstant, where the degree 
$$[k(T):L]=\mbox{max}(\mbox{deg }P,\mbox{deg }Q).$$
