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HW10.EGM4313 - Page 657 problem 9 ∫z C dz =βˆ βˆ’1 C 2 1...

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Unformatted text preview: Page 657, problem 9: ∫z C dz =∫ βˆ’1 C 2 1 z βˆ’ 1dz = 2Ο€i 1 = βˆ’Ο€ i z +1 βˆ’1 βˆ’1 Page 657, problem 15: ∫ C Ln( z βˆ’ 1) dz = 2Ο€ i β‹… ln ( 4 ) z βˆ’5 Page 672, problem 21: Convergent, but not absolutely convergent. Real and imaginary parts satisfy the alternating sign test. Page 677, problem 9: ( n βˆ’ i) n β†’ ∞ as n β†’ ∞ β‡’ Radius of convergence =0. Page 677, problem 11: ( βˆ’ 1) n+1 lim ( βˆ’ 1n) n β†’βˆž n+2 = lim n β†’βˆž n +1 = 1 β‡’ Radius of convergence = 1. Center is at z=0. n n +1 Page 682, problem 3: Center is at z=-i. lim n β†’βˆž n 2n = 2 β‡’ R 2 = 2 β‡’ 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: ( βˆ’ 1) n lim (2n )+ 1 βˆ’1 n β†’βˆž n +1 =1β‡’ R = Ο€ 2n + 3 ∞ z Differentiate series to get βˆ‘ ( βˆ’ 1) Ο€ n =1 2n n β‡’ R = Ο€ by inspection. Page 690, problem 8: f ( z ) = Ln(1 βˆ’ z ) β‡’ f ( n ) ( z ) = βˆ’ ( n βˆ’ 1)! (1 βˆ’ z ) n ∞ β‡’ f ( z ) = Ln(1 βˆ’ i ) + βˆ‘ n =1 Radius of convergence = distance from i to 1 = βˆ’1 (1 βˆ’ i ) n n ( z βˆ’ i) n 2. Page 707, problem 9: 1 1 1 1 ∞ ( βˆ’ 1) f ( z) = 2 = = βˆ‘ 2 n+1 i n+1 ( z βˆ’ i ) n z + 1 z βˆ’ i z + i z βˆ’ i n =0 n Closest singularity at z=-i, implies R=2. Page 707, problem 15: 1 = 1 + z 3 + z 6 + z 9 + β‹… β‹… β‹…, z < 1 3 1βˆ’ z Let w=1/z. 1 1 1 1 1 1 1 = = βˆ’w3 = βˆ’ w 3 βˆ’ w 6 βˆ’ w 9 βˆ’ β‹… β‹… β‹… = βˆ’ 3 βˆ’ 6 βˆ’ 9 βˆ’ 12 βˆ’ β‹… β‹… β‹…, z > 1. 3 3 1 1βˆ’ z z z z z 1βˆ’ w 1βˆ’ 3 w ...
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HW10.EGM4313 - Page 657 problem 9 ∫z C dz =βˆ βˆ’1 C 2 1...

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