s5_95-100aSOLUTION

# s5_95-100aSOLUTION - ACM95a/100a Last updated November 9...

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Unformatted text preview: ACM95a/100a Last updated November 9, 2009 Problem Set V Solutions Prepared by Faisal Amlani 1. (20 pts.) Let f ( z ) = ∑ ∞ k =1 k 3 ( z 3 ) k . Compute each of the following integrals giving justification in each case (10 pts. each): ( i ) integraldisplay | z | =1 e iz f ( z ) dz ( ii ) integraldisplay | z | =1 f ( z ) z 4 dz SOLUTION : The radius of convergence of the series for f ( z ) is R = lim n →∞ vextendsingle vextendsingle vextendsingle vextendsingle k 3 / 3 k ( k + 1) 3 / 3 k +1 vextendsingle vextendsingle vextendsingle vextendsingle = 3 lim n →∞ vextendsingle vextendsingle vextendsingle vextendsingle k 3 ( k + 1) 3 vextendsingle vextendsingle vextendsingle vextendsingle = 3 . Thus f ( z ) is a function which is analytic inside the circle of radius 3. (i) The integrand is clearly analytic so by Cauchy’s theorem, contintegraldisplay | z | =1 e iz f ( z ) dz = 0 . (ii) By Cauchy’s integral formula, contintegraldisplay | z | =1 f ( z ) z 4 dz = 2 πi 3! f (3) (0) = 2 πi 3! 3! 3 3 3 3 = 2 πi. 2. (15 pts.) Find the circle of convergence of the following series (5 pts. each). ( i ) ∞ summationdisplay 1 n 2 n ( z − i ) n , ( ii ) ∞ summationdisplay n n z n , ( iii ) ∞ summationdisplay 1 [3 + ( − 1) n ] n z n . SOLUTION : (i) From the ratio test, R = lim n →∞ vextendsingle vextendsingle vextendsingle vextendsingle n/ 2 n ( n + 1) / 2 n +1 vextendsingle vextendsingle vextendsingle vextendsingle = 2 lim n →∞ vextendsingle vextendsingle vextendsingle vextendsingle n n + 1 vextendsingle vextendsingle vextendsingle vextendsingle = 2 or, from the root test, R = 1 lim n →∞ n radicalbig | n/ 2 n | = 2 lim n →∞ n √ n = 2 ....
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## This note was uploaded on 01/14/2010 for the course ACM 1 taught by Professor Prof during the Fall '09 term at Caltech.

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s5_95-100aSOLUTION - ACM95a/100a Last updated November 9...

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