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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 nonclosed 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.
 Three '09
 Cong
 Math

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