Since this is true for every z0 c it must be the case

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Unformatted text preview: C C Combining them gives w - z0 1Û Ù f(w) - f(z0) = O f(z) dz ı 2πi (z - w)(z - z0) C Noting that the term w - z0 is constant, and dividing by it gives f(w) - f(z0) Û f(z) Ù O ı (z - w)(z - z ) dz w - z0 0 C Now the integrand is a continuous function of w, so letting wÆz0 gives = 1 2πi f(w) - f(z0) 1Û f(z) Ù f'(z0) = lim = O ı (z - z )2 dz, 2πi wÆz0 w - z0 0 C showing the case for n = 1. To show the proof for n = 2, use the same technique as for n = 1, except that we use the formula for n = 1 instead of the Cauchy integral formula. Then continue the proof inductively. Corollary 5.4 (An important Inequality) n!M (n) |f (z0)| ≤ n for all n ≥ 0 r where M is an upper bound of |f(z)| on a circle centered at z0 with radius r. Proof: Ô n! (n) |f (z0)| = Ô Ô2πi Ô But, for z on C, Ô Ô Û n! Ô Û f(z) f(z) Ù Ù O dzÔ = 2π ÔO dzÔ ı (z - z )n+1 Ô Ô ı (z - z0)n+1 Ô 0 Ô ÔC Ô C 18 Ô Ô f(z) M M Ô Ô n+1 ≤ n+1 = n+1 Ô(z - z0) Ô |z-z0| r where r is the radius of the circle C. Therefore Ô n! M n! Ô...
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This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.

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