ISSN: 2250-3005 - ijcer

International Journal Of Computational Engineering Research (ijceronline.com) Vol. 2 Issue. 8absolute value is still minimum at z = 0. It makes life simplest if we choose c to be a k th root of (-1/a k ). The new f we getwhen we substitute cz for z then turns out to be 1 – z k + a' k+1 z k+1 + ... + a' n z n . Thus, there is no loss of generality inassuming that a k = -1. Therefore, we may assume thatf(z) = 1 – z k + a k+1 z k+1 + ... +a n z n .If n = k then f(z) = 1 – z n and we are done, since f(1) = 0. Assume from here on that k is less than n. The main point. Weare now ready to finish the proof. All we need to do is show with f(z) = 1 – z k + a k+1 z k+1 + ...+ a n z n that the minimumabsolute value of f is less than 1, contradicting the situation we have managed to set up by assuming that f(r) is not 0. We shallshow that the absolute value is indeed less than one when z is a s mall positive real number (as we move in the complex pla neaway from the origin along the positive real axis or x-axis, the absolute value of f(z) drops from 1.) To see this assume that zis a positive real number with 0 < z < 1. Note that 1-z is then positive. We can then write|f(z)| =|1 – z k + a k+1 z k+1 + ... + a n z n |≤|1-z k |+|a k+1 z k +...+a n z n |= 1- z k + | a k+1 z k+1 + ... + a n z n |≤1- z k + | a k+1 |z k+1 + . . . + |a n |z n(keep in mind that z is a positive real number)= 1 - z k (1 – w z ), wherew z = | a k+1 |z + ... +|a n |z n-k .When z approaches 0 through positive values so does w z . Hence, when z is a s mall positive real number, so is z (1-w z ), and sofor z a s mall positive real number we have that0 < 1 – z k (1-w z ) < 1.Since |f(z)| < 1 – z k (1-w z )it follows that |f(z)| takes on values smaller than 1, and so |f(z)| is not minimum at z = 0 after all. This is a cont radiction. Itfollows that the minimum value of |f(z)| must be 0, and so f has a root .That proves the theorem.12. Fundemental Theorem of Algebra via Fermat’s Last TheoremFor the proof of the theorem we must know the following lemmasLemma 1:If an algebraic equation f(x) has a root α, then f(x) can be divided by x-α without a remainder and the degree of theresult f'(x) is less than the degree of f(x).Proof:Let f(x) = x n + a 1 x n-1 + . . . + a n-1 x + a nLet α be a root such that f(α) = 0Now, if we divide the polynomial by (x-α), we get the followingf(x)/(x - α) = f 1 (x) + R/(x-α)where R is a constant and f 1 (x) is a polynomial with order n -1.Multiplying both sides with x-α gives us:f(x) = (x - α)f 1 (x) + RNow, if we substitute α for x we get:f(α) = 0 which means that the constant in the equation is 0 so R = 0. That proves the lemma.Theorem: Fundamental Theorem of AlgebraFor any polynomial equation of order n, there exist n roots r i such that:xn + a 1 x n-1 + ... + a n-1 x + a n = (x – r 1 )(x – r 2 ) . . . (x – r n )Proof: Let f(x) = x n + a 1 x n-1 + ... + a n-1 x + anWe know that f(x) has at least one solution α 1 .Using Lemma 1 above, we know that:f(x)/(x – α 1 ) = f'(x) where deg f'(x) = n-1.||Issn 2250-3005(online)|| ||December|| 2012 Page 312