Prove the fundamental theorem of Algebra using
Rouche’s Theorem.
The maximum modulus principle.
Solution
Let \(P(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0\) be a non-constant polynomial with complex coefficients, where \(n \geq 1\) and \(a_n \neq 0\).
(a) Rouché’s theorem. Let \(f(z) = a_n z^n\) and \(g(z) = a_{n-1} z^{n-1} + \cdots + a_1 z + a_0\), so that \(P(z) = f(z) + g(z)\).
On the circle \(C_R = \{z \in \CC : |z| = R\}\) for \(R > 0\): \[ |f(z)| = |a_n| R^n, \] \[ |g(z)| \leq |a_{n-1}| R^{n-1} + \cdots + |a_1| R + |a_0|. \] Dividing by \(|f(z)|\): \[ \frac{|g(z)|}{|f(z)|} \leq \frac{|a_{n-1}|}{|a_n| R} + \cdots + \frac{|a_0|}{|a_n| R^n}. \] As \(R \to \infty\), the right-hand side approaches \(0\). Therefore, there exists a sufficiently large \(R > 0\) such that for all \(z \in C_R\): \[ |g(z)| < |f(z)|. \] By Rouché’s theorem, \(P(z) = f(z) + g(z)\) and \(f(z) = a_n z^n\) have the same number of zeros (counted with multiplicity) in the open disk \(D(0, R) = \{z \in \CC : |z| < R\}\). Since \(f(z) = a_n z^n\) has exactly \(n \geq 1\) zeros in \(D(0, R)\) (a zero of multiplicity \(n\) at the origin), \(P(z)\) has exactly \(n \geq 1\) zeros in \(D(0, R) \subset \CC\). In particular, \(P(z)\) has at least one complex root.
(b) Maximum modulus principle. Suppose towards a contradiction that \(P(z)\) has no zeros in \(\CC\). Then the function: \[ h(z) = \frac{1}{P(z)} \] is entire (holomorphic on all of \(\CC\)).
For \(|z| = R\) with \(R\) large enough that \(\frac{|a_{n-1}|}{|a_n| R} + \cdots + \frac{|a_0|}{|a_n| R^n} \le \frac12\): \[ |P(z)| = |a_n| R^n \left| 1 + \frac{a_{n-1}}{a_n z} + \cdots + \frac{a_0}{a_n z^n} \right| \geq |a_n| R^n \left( 1 - \frac{1}{2} \right) = \frac{|a_n|}{2} R^n \to \infty \quad \text{as } R \to \infty. \] Therefore: \[ \lim_{|z| \to \infty} |h(z)| = \lim_{|z| \to \infty} \frac{1}{|P(z)|} = 0. \] This implies that \(|h(z)|\) attains a global maximum on \(\CC\): Since \(h(0) = 1/a_0 \neq 0\) (if \(a_0 = 0\), \(P(0) = 0\) is already a root), there exists \(R > 0\) such that \(|h(z)| < |h(0)|\) for all \(|z| \geq R\). On the compact closed disk \(\overline{D(0, R)}\), the continuous function \(|h(z)|\) attains its maximum at some point \(z_0 \in D(0, R)\), which is therefore a global maximum of \(|h|\) on the entire complex plane \(\CC\).
By the maximum modulus principle, since \(h\) is holomorphic on \(\CC\) and \(|h|\) attains an interior local maximum at \(z_0\), \(h\) must be constant. If \(h(z)\) is constant, then \(P(z) = 1/h(z)\) is constant, contradicting that \(\deg(P) = n \geq 1\). Thus \(P(z)\) must have at least one root in \(\CC\).