solutionshw11-201141

solutionshw11-201141 - Solutions homework 11 (1) Evaluate...

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Unformatted text preview: Solutions homework 11 (1) Evaluate (without parametrizing, but using Cauchys Integral Theorem) R 1 1+ z 2 dz for a. ( t ) = 1 + e it (0 t 2 ). Solution: f ( z ) = 1 1+ z 2 = 1 2 i ( 1 z- i- 1 z + i ). Hence both z = i and z =- i are outside the curve and thus R f ( z ) dz = 0. b. ( t ) =- i + e it (0 t 2 ). Solution: In this case z = i is outside the curve and Ind (- i ) = 1, so R f ( z ) dz = 2 i (- 1 2 i ) =- . c. ( t ) = 2 e it (0 t 2 ). Solution: In this case both z = i and z =- i are inside the curve and Ind (- i ) = Ind ( i ) = 1. Thus R f ( z ) dz = 2 i ( 1 2 i- 1 2 i ) = 0. d. ( t ) = 3 i + 3 e it (0 t 2 ). Solution: In this case z =- i is outside the curve and Ind ( i ) = 1, so R f ( z ) dz = 2 i ( 1 2 i ) = . (2) Let C with | | 6 = 1. Compute Z 2 dt 1- 2 cos t + 2 by computing i Z 1 ( z- )( z- 1 ) dz, where ( t ) = e it with 0 t 2 ....
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solutionshw11-201141 - Solutions homework 11 (1) Evaluate...

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