ISSN: 2250-3005 - ijcer

International Journal Of Computational Engineering Research (ijceronline.com) Vol. 2 Issue. 8This fact has led some to remark that the Fundamental Theorem of Algebra is neither fundamental, nor a the oremof algebra.Some proofs of the theorem only prove that any non-constant polynomial with real coefficients has some complexroot. This is enough to establish the theorem in the general case because, given a non -constant polynomial p(z)with complex coefficients, the polynomialq( z) p(z)p(z)has only real coefficients and, if z is a zero of q(z), then either z or its conjugate is a root of p(z).Different Proofs of the theorem: Statement of Fundamental theorem of algebra Every polynomial equation of degree n hasexactly n zeroes An expression of the formP(x) = a n x n + a n-1 x n-1 + . . . + a 1 x + a 0Where a 0, a 1 . . . an -1, a n ≠ 0are real or complex numbers and p(x) is called a polynomial equation of degree n and the equation p(x) = 0 is calleda polynomial equation of degree n .By a zero of the polynomial x or a root of the equation x = 0 , we mean a value of p(x) such that p(x) = 0.First proof: For the proof of the Theorem we must know the following Theorem known as Liouville‟s Theorem.STATEMENT: If a function f(z) is analytic for all finite values of z and is bounded, then f(z) is constant. “or”If f is regular in whole z-plane andif │f(z)│< K for all z then f(z) is constant.PROOF: Let a & b be arbitrary distinct points in z-plane and let C be a large circle with center z = 0 and radius Rsuch that C encloses a & b. The equation of C is│z│= R so that z = R e iөdz = ί R e iө│dz│ = R dөf(z) is bounded for all z │f(z) │ m for all z where m > 0By Cauchy integral formulaf(a) =f(b) =1 f ( z)2iz acd(z)1 f ( z) d(z)2iz bcf(a) –f(b) = 1 1 1 f ( Z)d(Z)2 i c z a z b f(a) –f(b) =│f(a) –f(b) │a b2cf ( z)d(z)( z a)(z b)a b f ( z)d(z)2 z a z b c│f(a) –f(b) │M a b.2R2 ( R a )( R b )MR a b│f(a) –f(b) │ ( R a )( R b 0asR ||Issn 2250-3005(online)|| ||December|| 2012 Page 300