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 C such that p(α) = 0.
Proof: Suppose not. Then g(z) = 1/p(z) is entire. Also when |z| → , then |p(z)| → . Thus 1/ |p(z)| → 0 as |z| → ; 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 pk of degree 0; that is, pk is constant. We have derived the following fact: If p(z) is a holomorphic polynomial of degree k, then there are k complex numbers α1, . . .αk (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 αj1 , . . . , αjm 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.