814hw5sol - Math 814 HW 5 December 11, 2007 p. 87: 6, 7 p....

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Unformatted text preview: Math 814 HW 5 December 11, 2007 p. 87: 6, 7 p. 96: 8a, 10, 11 p. 110: 1bchi, 5,13. p.87, no. 6. Let f be analytic on D = B (0 , 1) and suppose | f ( z ) | 1 on D . Show that | f (0) | 1 . Proof: Let < r < 1 and let r ( t ) = re it for t 2 . By the Cauchy Integral Formula, we have f (0) = 1 2 i Z r f ( w ) w 2 dw, so | f (0 | 1 2 1 r 2 2 r = 1 r . Taking the limit as r 1 , we have | f (0) | 1 . p.87, no. 7. Let ( t ) = 1 + e it for t 2 and let n N . Find Z z z- 1 n dz. Apply C.I.F. to f ( z ) = z n , to get Z z z- 1 n dz = 2 i ( n- 1)! f n- 1 (1) = 2 ni. 1 You can also do this without C.I.F., by computing directly: Z z z- 1 n dz = Z 2 1 + e it e it n ie it dt = i Z 2 ( 1 + e- it ) n e it dt = i Z 2 1 + ne- it + n 2 e- 2 it + e it dt = 2 ni, since R 2 e kit dt = 0 for k a nonzero integer. p.96, no. 8a. We must integrate ( z- a )- 1 and ( z- b )- 1 over the path , which can be written as a sum of six paths, two of which are closed and have a,b in their -components, hence have zero integral and two pairs of non-closed paths. One pair starts at the leftmost crossing point, each goes around a in opposite directions, and they meet at the middle crossing point. The other one pair starts at the middleand they meet at the middle crossing point....
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This note was uploaded on 10/19/2009 for the course MATH 814 taught by Professor Cong during the Three '09 term at University of Adelaide.

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814hw5sol - Math 814 HW 5 December 11, 2007 p. 87: 6, 7 p....

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