Fundamental Theorem Of Algebra

Fundamental Theorem of Algebra:

The Fundamental Theorem of Algebra: Let p(z) be a nonconstant (holomorphic) polynomial. Then p has a root. That is, there exists an α $\in \!\,$ C such that p(α) = 0.

Proof: Suppose not. Then g(z) = 1/p(z) is entire. Also when |z| → $\infty \!\,$ , then |p(z)| → $\infty \!\,$ . Thus 1/ |p(z)| → 0 as |z| → $\infty \!\,$ ; hence g is bounded. By Liouville’s Theorem, g is constant, hence p is constant. Contradiction: The polynomial p has degree k ≥ 1, then let α₁ denote the root provided by the Fundamental Theorem. By the Euclidean algorithm (see [HUN]), we may divide z − α₁ into p with no remainder to obtain;

Here p1 is a polynomial of degree k −1 . If k − 1 ≥ 1, then, by the theorem, p1 has a root α₂ . Thus p1 is divisible by (z − α₂) and we have

for some polynomial p₂ (z) of degree k − 2. This process can be continued until we arrive at a polynomial p_k of degree 0; that is, p_kis constant. We have derived the following fact: If p(z) is a holomorphic polynomial of degree k, then there are k complex numbers α₁, . . .α_k (not necessarily distinct) and a non-zero constant C such that,

If some of the roots of p coincide, then we say that p has multiple roots. To be specific, if m of the values α_j1 , . . . , α_jm are equal to some complex number , then we say that p has a root of order m at (or that p has a root of multiplicity m at ). It is an easily verified fact that the polynomial p has a root of order m at α if p(α) = 0, p'(α) = 0, . . . p^(m−1)(α) = 0 (where the parenthetical exponent denotes a derivative).

Figure of Complete Set:

An example will make the idea clear: Let,

Then we say that p has a root of order 3 at 5, a root of order 8 at −2, and it has roots of order 1 at 3i and at −6. We also say that p has simple roots at 1 and −6.

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