HW10.EGM4313 - Page 657, problem 9: ∫z C dz =∫ −1 C 2...

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Unformatted text preview: Page 657, problem 9: ∫z C dz =∫ −1 C 2 1 z − 1dz = 2πi 1 = −π i z +1 −1 −1 Page 657, problem 15: ∫ C Ln( z − 1) dz = 2π i ⋅ ln ( 4 ) z −5 Page 672, problem 21: Convergent, but not absolutely convergent. Real and imaginary parts satisfy the alternating sign test. Page 677, problem 9: ( n − i) n → ∞ as n → ∞ ⇒ Radius of convergence =0. Page 677, problem 11: ( − 1) n+1 lim ( − 1n) n →∞ n+2 = lim n →∞ n +1 = 1 ⇒ Radius of convergence = 1. Center is at z=0. n n +1 Page 682, problem 3: Center is at z=-i. lim n →∞ n 2n = 2 ⇒ R 2 = 2 ⇒ R = 2 n +1 2 n +1 I don’t see any advantage to integrating or differentiating to get an easier convergence problem so feel free to do either. Page 682, problem 4: ( − 1) n lim (2n )+ 1 −1 n →∞ n +1 =1⇒ R = π 2n + 3 ∞ z Differentiate series to get ∑ ( − 1) π n =1 2n n ⇒ R = π by inspection. Page 690, problem 8: f ( z ) = Ln(1 − z ) ⇒ f ( n ) ( z ) = − ( n − 1)! (1 − z ) n ∞ ⇒ f ( z ) = Ln(1 − i ) + ∑ n =1 Radius of convergence = distance from i to 1 = −1 (1 − i ) n n ( z − i) n 2. Page 707, problem 9: 1 1 1 1 ∞ ( − 1) f ( z) = 2 = = ∑ 2 n+1 i n+1 ( z − i ) n z + 1 z − i z + i z − i n =0 n Closest singularity at z=-i, implies R=2. Page 707, problem 15: 1 = 1 + z 3 + z 6 + z 9 + ⋅ ⋅ ⋅, z < 1 3 1− z Let w=1/z. 1 1 1 1 1 1 1 = = −w3 = − w 3 − w 6 − w 9 − ⋅ ⋅ ⋅ = − 3 − 6 − 9 − 12 − ⋅ ⋅ ⋅, z > 1. 3 3 1 1− z z z z z 1− w 1− 3 w ...
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HW10.EGM4313 - Page 657, problem 9: ∫z C dz =∫ −1 C 2...

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