This post assumes familiarity with some basic concepts in complex analysis, including continuity and entire (everywhere complex-differentiable) functions. This is likely the simplest proof of the theorem (at least, among those that this author has seen), although it stands on the shoulders of a highly nontrivial theorem.

The fundamental theorem of algebra has quite a few number of proofs (enough to fill a book!). In fact, it seems a new tool in mathematics can prove its worth by being able to prove the fundamental theorem in a different way. This series of proofs of the fundamental theorem also highlights how in mathematics there are many many ways to prove a single theorem, and in re-proving an established theorem we introduce new concepts and strategies. And perhaps most of all, we accentuate the unifying beauty of mathematics.

Problem: Let p(z):CC be a non-constant polynomial. Prove p(z) has a root in C.

Solution: Picard’s Little Theorem states that any entire function CC whose range omits two distinct points must be constant. (See here for a proof sketch, and other notes)

Suppose to the contrary that p has no zero. We claim p also does not achieve some number in the set {1k:kN}. Indeed, if it achieved all such values, we claim continuity would provide a zero. First, observe that as p(z) is unbounded for large enough |z|, we may pick a cR such that |p(z)|>1 whenever |z|>c. Then the points zk such that p(z)=1/k all lie within the closed disk of radius c. Because the set is closed and the complex plane is a complete metric space, we see that the sequence of zk converges to some z, and by continuity p(z)=0.

In other words, no such sequence of zk can exist, and hence there is some 1/k omitted in the range of p(z). As polynomials are entire, Picard’s Little theorem applies, and we conclude that p(z) is constant, a contradiction.


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DOI: https://doi.org/10.59350/fyg3n-z0q04