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Unformatted text preview: , . . . . Proof. Similar to the previous proof, let g n ( z ) = 1 2 i ( zz ) n +1 , n = 0 , 1 , 2 , . . . . Then Z C g n ( z ) f ( z ) dz = 1 2 i Z C f ( z ) ( zz ) n +1 dz. However, we also have Z C g n ( z ) f ( z ) dz = X m = a m 1 2 i Z C 1 ( zz ) nm +1 dz. Since Z C 1 ( zz ) nm +1 dz = , if m 6 = n, 2 i, if m = n, 2 it follows that c n = 1 2 i Z C f ( z ) ( zz ) n +1 dz. 3...
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This note was uploaded on 11/23/2009 for the course MATHEMATIC Mathematic taught by Professor Dansloughter during the Spring '04 term at Furman.
 Spring '04
 DanSloughter
 Math, Taylor Series

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