As per the statement of Fundamental Theorem of Algebra, any polynomial which is non-constant accompanied by complex coefficients consist of not less than 1 complex root.
The theorem also means that any polynomial having complex coefficients belonging to n degree contains n complex roots, which are calculated with multiplicity.
The field F, having the property which every polynomial having coefficients within F contains a root within F, which is known as “Algebraically Closed”, therefore the “Fundamental Theorem of Algebra” gives you that statement that:
Complex Numbers, C’s field is Algebraically Closed.
x²+1 polynomial consists of no real roots, but it has two complex roots i and −i
x²+ i polynomial is having 2 complex roots namely, ±2 / 1−i
One may expect that the polynomials accompanied by complex coefficients are having problems with the non-existence of the roots which are identical to real polynomials; i.e. it’s not logical to imagine that a few of polynomials such as
x3 + ix² − (1+πi) x −e
will not be having the complex root, plus to find such root which needs to look out in a bit larger field that encompasses the field of complex numbers.
As per Fundamental Theorem of Algebra, this is not that very case: all roots of polynomial accompanied by complex coefficients are witnessed existing within the already established field of complex numbers.
This section of the article includes a much more precise statement of the fundamental theorem of algebra’s various equivalent forms. This needs a definition of root’s multiplicity of the polynomial.
Definition: Root’s multiplicity ‘r’ of the polynomial f(x) exists as the largest or the greatest positive integer “k” in a way that (x-r)^k divides the function, f(x).
Similarly, it’s the positive integer which is the smallest one – k in a way that f^(k) (r) ≠ 0 where,
f^(k) represents = kth derivative of f.
Assuming F as the field and following are the equivalents:
- Every polynomial with a -accompanied with coefficients within F contains root in ‘F”.
- Every polynomial with a non-constant having n degree with coefficients within F contains ‘n’ roots in F which are calculated with multiplicity.
- Every polynomial non-constant value with the coefficients within F entirely splits as linear factors’ product with the coefficients within F.
Evidently, (3)⇒(2)⇒(1), therefore, the exclusive non-trivial section is (1)
To view this, introduce into n’s degree of f(x). The fundamental case of n=1 is clarified.
Now, let us assume the result is for the polynomials of n-1 degree. Accordingly, letting the f(x) be the polynomial of ‘n’ degree. By (1), now, f(x) contains root a.
The argument of the standard division algorithm presents that “x-a” is f (x)’s factor:
On dividing, f(x) by x−a to achieve f(x)=(x−a) q(x) + r, where r is the constant polynomial.
Filling in ‘a’ to both of the sides & give a value, 0=(a−a) q (a)+r, so r = 0. So, f(x)=(x−a) q(x). But, q(x) is a polynomial of degree n−1, so it divides into linear factors’ product by an “Inductive Hypothesis” or “Inductive Assumption”. Thus, f(x) performs as well. Therefore, the result is verified by induction.
The fundamental theorem of algebra states that field C belonging to complex numbers contains property (1), therefore, going by theorem stated , it should have (1), (2), and (3) properties.