math185f08-hw7sol - MATH 185: COMPLEX ANALYSIS FALL 2008/09...

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Unformatted text preview: MATH 185: COMPLEX ANALYSIS FALL 2008/09 PROBLEM SET 7 SOLUTIONS For any S C , recall that S denotes its closure and S denotes its boundary. For any real or complex-valued function f , we will write f k ( z ) := [ f ( z )] k for k N . We will use the shorthand | z- a | = r to denote the circle of radius r centered at a , parameterized by z : [0 , 2 ] C , z ( t ) = a + re it . 1. Let be a region and D (0 , 1) . Let f : C be analytic. (a) Show that 2 Z 2 f ( e i )cos 2 2 d = 2 f (0) + f (0) . Solution. By the generalized Cauchy integral formula, f (0) = 1 2 i Z | z | =1 f ( z ) z dz = 1 2 Z 2 f ( e i ) d, (1.1) f (0) = 1 2 i Z | z | =1 f ( z ) z 2 dz = 1 2 Z 2 f ( e i ) e- i d. (1.2) By Cauchys theorem, 0 = 1 2 i Z | z | =1 f ( z ) dz = 1 2 Z 2 f ( e i ) e i d. (1.3) If we take 2 (1.1) + (1.2) + (1.3), we get 2 f (0) + f (0) = 1 2 Z 2 f ( e i )(2 + e i + e- i ) d = 1 Z 2 f ( e i )(1 + cos ) d = 2 Z 2 f ( e i )cos 2 2 d as required. (b) Show that for n N , n ! Z 2 [Re f ( e i )] e- in d = f ( n ) (0) . Solution. By the generalized Cauchy integral formula, f ( n ) (0) = n ! 2 i Z | z | =1 f ( z ) z n +1 dz = n ! 2 Z 2 f ( e i ) e- in d. Now observe that the given integral may be written as n ! Z 2 [Re f ( e i )] e- in d = n ! 2 Z 2 [ f ( e i ) + f ( e i )] e- in d = f ( n ) (0) + n !...
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This note was uploaded on 06/15/2009 for the course MATH 185 taught by Professor Lim during the Fall '07 term at University of California, Berkeley.

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math185f08-hw7sol - MATH 185: COMPLEX ANALYSIS FALL 2008/09...

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