Fundamental Theorem of Algebra

An n-th order algebraic equation, Pn(z) = 0, has at least one root where
Pn(z) ≡ an zn + an−1zn−1 + …a1 z + a0.

(Proof): Let
f(z) ≡ 1
P(z)
 .
If we assume that Pn(z) is never equal to 0 (i.e. Pn(z) has no zeros), then, f(z) is analytic everywhere (the denominator never goes to 0) and it is bounded (P(z)→ ∞ as z → ∞ along the real axis). Thus, according to the Liouville theorem, P(z) must be a constant, which is not the case, hence, the assumption (Pn(z) never goes to 0) is wrong !!! 1
If Pn(z) has a zero, say, a1, it can be factorized as
Pn(z) = (z − a1)(bn−1zn−1 + bn−2 zn−2 + …b1 z + b0).
(1)
Applying the fundamental theorem of calculus recursively, it is shown that an n-th order algebraic equation has n roots (including multiple roots).

Footnotes:

1 Unfortunately, the fundamental theorem of algebra does not tell you how to actually solve the algebraic equation.

File translated from TEX by TTH, version 4.03.
On 18 Jan 2021, 22:03.