Theorem (Fundamental Theorem of Algebra) Every polynomial of degree n \ge 1  with complex coefficients has a zero in C.

Proof

Let p(z) = z^n + a_{n-1}z^{n-1} + \ldots + a_1z + a_0  be a polynomial of degree n \ge 1  and assume that p(z) \ne 0  for all z \in  C.

Define g: [0, \infty) \times [0, 2\pi] \rightarrow C by g(r, \theta) = {1}/{{p}({r}e^{i\theta})}.   Then the function g  is continuous on (0, \infty) \times (0, 2\pi) satisfying \frac{\partial g}{\partial \theta} = ir\times \frac{\partial g}{\partial r}.

Define now F:\rightarrow [0,\infty)\rightarrow C by F(r) = \int_0^{2\pi} g(r, \theta)d\theta . Then by Leibniz's rule for differentiation under the integral sign we have for all r \gt 0

irF'(r) = ir\int_0^{2\pi}\frac{\partial g}{\partial r}d\theta = \int_0{2\pi}\frac{\partial g}{\partial \theta}d\theta = g(r, 2\pi ) - g(r, 0) = 0.

Hence F'(r) = 0 for all r \gt 0 . This implies that F constant on [0, \infty) with F(r) = F(0) = \frac{2\pi }{p(0)} \ne 0 . On the other hand, |p(z)|\rightarrow 0  as r\rightarrow \infty , which is a contradiction.

Reference

  • A. R. Schep, A Simple Complex Analysis and an Advanced Calculus Proof of the Fundamental Theorem of Algebra, Am Math Monthly 116, Jan 2009, 67-68.