ISSN: 2250-3005 - ijcer

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International Journal Of Computational Engineering Research (ijceronline.com) Vol. 2 Issue. 8So that we have:f(x) = (x – α 1 )f'(x)But we know f'(x) has at least one solution α 2f'(x)/(x–α 2 ) = f''(x) where deg f''(x) = n-2. Andf(x) = (x – α 1 )(x – α 2 )f''(x)Eventually we get to the point where the degree offn(x)=1 .In this case, f n (x) = x – α n . This establishes that there are n roots for a given equation f(x) where thedegree is n. Putting this all together gives us:f(x) = (x – α 1 )(x – α 2 ). . . (x – α n )Now, since f(x)=0 only when one of the values α i =x, we see that the n roots α i are the only solutions.So, we have proven that each equation is equal to n roots.One important point to remember is that the n roots are notnecessarily distinct. That is, it is possible that α i = α j where i ≠ j. That proves the theoremFundamental theorem of Algebra due to CauchyWe will provef(z) = z n + a 1 z n-1 + a 2 z n-1 + . . . + a n = 0where a i are complex numbers n ≥ 1 has a complex root.Proof:- let a n ≠ 0 denote z = x + iy x, y real Then the functiong(x ,y)= │f(z) │= │f(x + iy) │Is defined and continuous in R 2Letcn a jj1We make the estimationfit is +ve using the triangle inequalitya a2 ... z zn 1( z)z 12a1a2af ( z) 1 ... z zzn c 1f ( z) z 1 zz 2z max 1, 2cBeing true for diskaznnnnn denote r max 1,2c,n 2 a n: conside thex2 y2 r Because it is compact the function g(x,y) attains at a point (x 0 y 0 ) of the disk its absoluteminimum value (infimum) in the disk if │z│> r we haven2 anan01 n 1 n 1g ( x,y) f ( z) z r 2 2 2Thusg(x,y) g(0,0) an f ( z)for z Hence g ( x 0, y0)is the absolute minimum of g(x,y) in the whole complex planewe show that g(x 0 y 0 ) = 0 therefore we make the antitheses that g(x 0 y 0 ) > 0rDenotez0 x0 iy0 z0UBf( z)f( z0 u) bn bn1U... b U1n1UnThen b n = f(z 0 ) ≠ 0 by the antithesis||Issn 2250-3005(online)|| ||December|| 2012 Page 313
International Journal Of Computational Engineering Research (ijceronline.com) Vol. 2 Issue. 8More over denotecjbbAnd assume that c n-1 = c n-2 = . . . = c n-k+1 = 0But c n-k ≠ 0 then we may writejn( i 12... n),c 01bnk1f ( z) bn(1 cnk cnku cnk1u... c0uk)Ifcn kukp(cos isin) & ue(cos sin)Thenkcnkupen(cos k) isin( k)By Demorvie‟s identity choosingcc1 we get &u pekk kn k& can make the estimatenk1uk 1... c0uncnk1ek 1...c0cncnk1...c0ek 1Rek 1Where R is constantLet now i p min k 1, .we obtain p 2Rf( z) b 1npekhuf( z)bn1pekh(u)f( z)bn1pekRek1bn1ekpRp2Rb2nbnf(z 0)Which results is impossible since │f(z) │ was absolute minimum. Consequently the antithesis is wrong & the proof is settled.13. Another Proof Of Fundamental Theorem Of AlgebraThe proof depends on the following four lemmasLemma1: Any odd-degree real polynomial must have a real root.Proof:We know from intermediate value theorem, suppose p( x) R x with degree p(x) = 2k +1 andsuppose the leading coefficient a n > 0( the proof is almost identical if a n > 0). Then||Issn 2250-3005(online)|| ||December|| 2012 Page 314
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ISSN: 2250-3005 - ijcer

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