Prove the fundamental theorem of Algebra using the maximum modulus principle.
Solution
Let \(p\) be a nonconstant complex polynomial. Suppose, for contradiction, that \(p\) has no zero. Then \[ g(z)=\frac1{p(z)} \] is entire. Since \(|p(z)|\to\infty\) as \(|z|\to\infty\), we have \(g(z)\to0\) as \(|z|\to\infty\).
Choose \(R>0\) so large that \[ |g(z)|<|g(0)| \qquad (|z|\ge R). \] On the compact disk \(\overline{D(0,R)}\), the continuous function \(|g|\) attains a maximum, say at \(z_0\). Because outside the disk \(|g|<|g(0)|\), this maximum is also the global maximum of \(|g|\) on \(\mathbb C\). Hence the entire function \(g\) attains its maximum modulus at an interior point. By the maximum modulus principle, \(g\) is constant, so \(p\) is constant, a contradiction.
Therefore every nonconstant complex polynomial has a zero. This is the fundamental theorem of algebra.