ISSN: 2250-3005 - ijcer

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International Journal Of Computational Engineering Research (ijceronline.com) Vol. 2 Issue. 8We draw a circle γ with center at b and radius δ. Then (3) shows that│f (z) │< M - Z inside γ …………(4)2Again we draw another circle Ѓ with center at a and radiusBy Cauchy‟s Integral formula .1 f zdzi ( )f(a) 2z aon Ѓ, z – a = re iө21f ( a) f a rei ( 20iirie d)irieIf we measure in anti-clock wise direction &if│b – a │= r< QPR = thenf ( a)12 02 f ( a rei) d1i1if ( a) f ( a) re d f ( a re ) d220221 1. f a) M d2 2 20 2 f ( a ) M M M 2 224( MdThen M = │f (a) │ < M -“or” M < M - 4For M cannot be less than M -Hence the Required results follow. 4. A contradiction 4||Issn 2250-3005(online)|| ||December|| 2012 Page 305
International Journal Of Computational Engineering Research (ijceronline.com) Vol. 2 Issue. 84. Minimum Modulus PrincipleSTATEMENT:- Suppose f(z) is analytic within and on a closed contour C and Let f(z) ≠ 0 inside C suppose further that f(z)is not constant . Then │f(z) │ attains its minimum value at a point on the boundary of “C” that is to say, if M is theminimum value of │f(z) │ inside and on C. Then │f(z) │> m z inside CPROOF:f(z) is analytic within and on C and f(z) ≠ 0 inside C. It follows that1f ( z)is analytic within C. By MaximumModulus principal 1 attains its Maximum value on the boundary of C. So that │ f (z)│ attains its Minimumf ( z)value on the boundary of C. Hence the theorem5. Proof Of The Fundamental Theorem Via MaximumModulus Principle Proof: Assume p(z) is non-constant and never zero. M such that |p(z)| |a 0 | ≠ 0 if |z| > M. Since |p(z)| is continuous, it achieves its minimum on a closed interval. Let z 0 be the value in the circle of radius M where p (z) takesits minimum value.So |p(z 0 )| |p(z)| for all z C, and in particular|p(z 0 )| |p(0)| = |a 0 |.Translate the polynomial.Let p(z) = p((z - z 0 ) + z 0 );Let p(z) = Q(z - z 0 ).Note the minimum of Q occurs atz = 0: |Q(0)| |Q(z)| for all z C.Q(z) = c 0 + c j z j + · · · + c n z n ,Where j is such that cj is the first coefficient (after c 0 ) that isNon-zero. I must show Q(0) = 0 Note if c 0 = 0, we are done.We may rewrite such thatQ(z) = c 0 + c j z j + z j+1 R(z)We will extract roots.Letre iө = -c 0c jFurther, Letz1r1jc ezij cjSo, j0Let 0 be a small real number. ThenQ(z1)c0 cjjzj1 j1zj11R(z1)Q(z1)c0 cjjzjj j1z1j1R(z1)cjj1j10 c0 z1N,Where N chosen such that N R z ) , and is chosen so that(1||Issn 2250-3005(online)|| ||December|| 2012 Page 306
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ISSN: 2250-3005 - ijcer

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