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Therefore fz fz0 o z z dz length of c 2 0 c the

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Unformatted text preview: is to O ı C 1 2 dz where C is the circle |z| = 3. 1+z Exercise Set 4 p. 832 #1–7 odd, 17, 23, 29 p. 837 # 1, 5, 9, 11, 13, 15 p. 842 # 1, 3, 5, 7, 11, 21 5. Cauchy's Integral Formula This theorem gives the value of an analytic function at a point in terms of its values in a contour surrounding that point. 16 Theorem 5.1 (Cauchy's Integral Formula) Let f be analytic on simply connected D , let z0 é D and let C be any simple closed path in D around z0. Then. 1 Û f(z) Ù f(z0) = O dz 2πi ı z - z 0 C Proof The trick is replace f(z) by the constant f(z0). So: f(z) = f(z0) + f(z) - f(z0). The integrand becomes f(z ) + f(z) - f(z0) f(z) =0 z - z0 z - z0 f(z0) f(z) - f(z0) = + z - z0 z - z0 The integral of the first term is 2πif(z0) by Example (D) of the previous section. This will give us the result if we can show that the integral of the second term is zero. By Consequence 3, we can use a circle about z0 as small as we like. Choose œ > 0 as small as you like. Since f is analytic, we have f(z) - f(z0) = f(z) - f(z0) - f'(z0) + f'(z0) z - z0 z - z0 Since the integral of the constant term f'(z0) is zero, we are left with the integral of f(z) - f(z0) - f'(z0) z - z0 whose magnitude is less than œ for z sufficiently close to z0 (which we can assume by choosing a small enough circle). Therefore ÔÛ f(z) - f(z0) Ô Ù ÔO z - z dzÔ ≤ œ...
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This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.

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