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Fundamental Theorem of Algebra

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Submitted By utkarshg
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Author : Utkarsh Garg

FUNDAMENTAL THEOREM OF ALGEBRA
The name suggests that it is some starting theorem of algebra or the basis of algebra. But it is not so, the theorem just say something interesting about the polynomials. Definition: The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This includes polynomials with real coefficients, since every real number is a complex number with zero imaginary part. PROOF: This is an algebraic proof. I am doing this for a 2 degree polynomial . It can be extended for any degree polynomial. We know that the roots of a quadratic equation az 2 + bz + c = 0 are given by the formula

irrespective of the fact whether a, b, c are real or complex numbers. Also it is clear that in this case there are two roots, say α + β = −b/a, αβ = c/a and az 2 + bz + c = a(z − α)(z − β) . Also any α, β satisfying α + β = a, αβ = b are given as roots of the quadratic equation z 2 − az + b = 0 . Now we will show that in order to prove the fundamental theorem of algebra it is sufficient to prove that any non-constant polynomial with real coefficients has a complex root. Let us then assume that every non-constant polynomial with real coefficients has a complex root. Let f (z) = a0 z n + a1 z n−1 + ⋯ + an−1 z + an be a polynomial with complex coefficients. Let g(z) be the polynomial obtained from f (z) by replacing the coefficients with their conjugates. Clearly the polynomial h(z) = f (z)g(z) has real coefficients and hence has a complex root α. Now h(α) = 0 implies that f (α)g(α) = 0 so that either f (α) = 0 or g(α) = 0. If f (α) = 0 then f (z) has a complex root α. On the other hand if g(α) = 0 then a0 αn + a1 αn−1 + ⋯ + an−1 α + an = 0 ⇒ a0 αn + a1 αn−1 + ⋯ + an−1 α + an = 0 ⇒ a0 αn + a1 αn−1 + ⋯ + an−1 α + an = 0 ⇒ ¯α is a root of f(z) We now prove the result that every non-constant polynomial with real coefficients has a complex root. Let f (z) denote such a polynomial of degree n. Let us write n = 2k m where k ≥ 0 is an integer and m is odd positive integer. We will use induction on k. Clearly for k = 0 the result holds as n = m

is odd and the coefficients of f (z) are real. Now let us assume that the result holds whenever n = 2k−1 m′ where m′ is any odd positive integer. It can be easily shown that for any polynomial f (z) over any field F there is a field L ⊇ F which contains all roots of f (z). Thus for our polynomial f (z) with real coefficients of degree n = 2k m we have a field L ⊇ R containing all the roots z1 , z2 , … , zn. The challenge is to prove that at least one of these roots lies in the field C of complex numbers. For any real number t let us define a polynomial (with coefficients in L) gt (z) = ∏ (z − zi − zj − tzi zj)

1≤i

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