This seems to be a favorite question everywhere, including Princeton quals. How many ways are there?
Please give a new way in each answer, and if possible give reference. I start by giving two:
-
Ahlfors, Complex Analysis, using Liouville's theorem.
-
Courant and Robbins, What is Mathematics?, using elementary topological considerations.
I won't be choosing a best answer, because that is not the point.
Best Answer
Here is the proof of the equivalent statement "Every complex non-constant polynomial $p$ is surjective".
1) Let $C$ be the finite set of critical points , i.e. $p'(z)=0$ for all $z\in C$. $C$ is finite by elementary algebra.
2) Remove $p(C)$ from the codomain and call the resulting open set $B$ and remove from the domain its inverse image $p^{-1}\left( p (C) \right)$, and call the resulting open set $A$. Note that the inverse image is again finite.
3) Now you get an open map from $A$ to $B$, which is also closed, because any polynomial is proper (inverse images of compact sets are compact). But $B$ is connected and so $p$ is surjective.
I like this proof because you can try it for real polynomials and it breaks down at step 3) because if you remove a single point from the line you disconnect it, while you can remove a finite set from a plane leaving it connected.