Section 56: Problem 1 Solution »

Section 56: The Fundamental Theorem of Algebra

The proof is, in fact, rather hard; the most difficult part is to prove that every polynomial equation of positive degree has at least one root... One can use only techniques of algebra; this proof is long and arduous. Or one can develop the theory of analytic functions of a complex variable to the point where it becomes a trivial corollary of Liouville’s theorem. Or one can prove it as a relatively easy corollary of our computation of the fundamental group of the circle...
(The Fundamental Theorem of Algebra) A polynomial of degree has at least one complex root.
If is given by , then is injective (if is a single loop, then ), and, hence is not nulhomotopic ( is injective because is a retract of , hence is injective and non-trivial).
If , then so that without loss of generality .
But then, if there is no root, then is nulhomotopic (because extendable to ) and homotopic to (via which is never zero because ). Contradiction.
  • If is such that , then all roots are in the interior of .