solutionshw11-201141

# solutionshw11-201141 - Solutions homework 11(1...

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Unformatted text preview: Solutions homework 11 (1) Evaluate (without parametrizing, but using Cauchy’s Integral Theorem) R γ 1 1+ z 2 dz for a. γ ( t ) = 1 + e it (0 ≤ t ≤ 2 π ). 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 γ f ( z ) dz = 0. b. γ ( t ) =- i + e it (0 ≤ t ≤ 2 π ). Solution: In this case z = i is outside the curve and Ind γ (- i ) = 1, so R γ f ( z ) dz = 2 πi (- 1 2 i ) =- π . c. γ ( t ) = 2 e it (0 ≤ t ≤ 2 π ). Solution: In this case both z = i and z =- i are inside the curve and Ind γ (- i ) = Ind γ ( i ) = 1. Thus R γ f ( z ) dz = 2 πi ( 1 2 i- 1 2 i ) = 0. d. γ ( t ) = 3 i + 3 e it (0 ≤ t ≤ 2 π ). Solution: In this case z =- i is outside the curve and Ind γ ( i ) = 1, so R γ f ( z ) dz = 2 πi ( 1 2 i ) = π . (2) Let α ∈ C with | α | 6 = 1. Compute Z 2 π dt 1- 2 α cos t + α 2 by computing i α Z γ 1 ( z- α )( z- 1 α ) dz, where γ ( t ) = e it with 0 ≤ t ≤ 2 π ....
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solutionshw11-201141 - Solutions homework 11(1...

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