Apply Rouché’s Theorem to prove the Fundamental Theorem of Algebra:
If \[ P_n(z) = a_0 + a_1z + \cdots + a_{n-1}z^{n-1} + a_nz^n\quad (a_n \neq 0) \] is a polynomial of degree \(n\), then it has \(n\) zeros in \(\mathbb{C}\).
Solution
Goal: Apply Rouch'e’s theorem to prove the Fundamental Theorem of Algebra: a polynomial \(P_n(z) = a_0 + a_1 z + \cdots + a_{n-1}z^{n-1} + a_n z^n\) with \(a_n \ne 0\) has exactly \(n\) zeros in \(\CC\) counting multiplicity.
1. Setup: compare \(P_n\) with its leading term \(a_n z^n\) on a large circle.
write \(P_n(z) = a_n z^n + Q(z)\) with \(Q(z) = a_0 + \cdots + a_{n-1}z^{n-1}\) a polynomial of degree \(\le n-1\).
2. For sufficiently large \(R\), \(|Q(z)| < |a_n z^n|\) on \(|z| = R\).
on \(|z| = R\), \(|a_n z^n| = |a_n|R^n\) and \(|Q(z)| \le \sum_{k=0}^{n-1}|a_k|R^k \le C R^{n-1}\) for \(R \ge 1\); choose \(R\) with \(C R^{n-1} < |a_n| R^n\), i.e. \(R > C/|a_n|\).
3. \(P_n\) has exactly \(n\) zeros in \(|z| < R\) counting multiplicity.
Rouch'e’s theorem on \(|z| = R\) with \(f = a_n z^n\) and \(g = Q\): by 2, \(|Q| < |a_n z^n|\), so \(P_n = f + g\) has the same number of zeros in \(|z| < R\) as \(a_n z^n\), which has exactly \(n\) (all at \(0\), with multiplicity \(n\)).
4. \(P_n\) has exactly \(n\) zeros in \(\CC\) counting multiplicity.
by 3, \(P_n\) has \(n\) zeros (counting multiplicity) in \(|z| < R\); since \(R\) can be chosen arbitrarily large and a degree-\(n\) polynomial has at most \(n\) zeros counting multiplicity, these are all of them.
5. Q.E.D.
<1>1–<1>4 prove the Fundamental Theorem of Algebra via Rouch'e’s theorem.