To show the proof for n 2 use the same technique as

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Unformatted text preview: ¥Length of C < 2πœ ı 0 ÔC Ô (the circle can be assumed to have a radius smaller than 1..) Since œ is arbitrarily small, the given integral must be zero, and we are done. Examples 5.2 Û Ù (A) Evaluate O ı C Û Ù (C) Evaluate O ı C ±π/2, ±3π/2, ... Û Ù (D) Evaluate O ı C z e z-2 dz, where z is any circle enclosing 2. tan z 2 z -1 dz where C is any simple contour enclosing 1 but non of the points z dz where C is the circle |z - 2i| = 4. z +9 2 17 z/(z+3i) [To evaluate this, rewrite the integrand as z-3i .] Corollary 5.3 (Analytic Functions have Derivatives of All Orders) Let f be analytic on simply connected D, let z0 é D and let C be any simple closed (n) path in D around z0. Then f (z0) exists, and n! Û f(z) (n) Ù f (z0) = O dz 2πi ı (z - z )n+1 0 C Proof: Let us start with n = 1: Write f(w) - f(z0) f'(z0) = lim wÆz0 w - z0 Applying the Integral Formula theorem to each term gives: 1 Û f(z) 1 Û f(z) Ù Ù f(w) = O and f(z0) = O dz ı z - w dz 2πi 2πi ı z - z 0...
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This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.

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