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AEM_3e_Chapter_19

AEM_3e_Chapter_19 - 19 19 3 0 2 0 2 0 Series and Residues...

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19 19 Series and Residues EXERCISES 19.1 Sequences and Series 3. 0, 2, 0, 2, 0 6. Converges. To see this write the general term as µ 2 5 n 1+ n 2 n i 1+3 n 5 n i . 9. Diverges. To see this write the general term as n µ 1+ 1 n i n . 12. Write z n = µ 1 4 + 1 4 i n in polar form as z n = Ã 2 4 ! n cos + i Ã 2 4 ! n sin .Now Re( z n )= Ã 2 4 ! n cos 0as n →∞ and Im( z n )= Ã 2 4 ! n sin 0as n →∞ since 2 / 4 < 1. 15. We identify a = 1 and z =1 i . Since | z | = 2 > 1 the series is divergent. 18. We identify a =1 / 2 and z = i . Since | z |− 1 the series is divergent. 21. From lim n →∞ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 (1 2 i ) n +2 1 (1 2 i ) n +1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 1 | 1 2 i | = 1 5 we see that the radius of convergence is R = 5. The circle of convergence is | z 2 i | = 5. 24. From lim n →∞ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 ( n +1) 2 (3 + 4 i ) n +1 1 n 2 (3+4 i ) n ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = lim n →∞ µ n n +1 2 1 | 3+4 i | = 1 5 we see that the radius of convergence is R = 5. The circle of convergence is | z +3 i | =5. 27. From lim n →∞ n s ¯ ¯ ¯ ¯ 1 5 2 n ¯ ¯ ¯ ¯ = lim n →∞ 1 25 = 1 25 we see that the radius of convergence is R = 25. The circle of convergence is | z 4 3 i | = 25. 30. (a) The circle of convergence is | z | = 1. Since the series of absolute values X k =1 ¯ ¯ ¯ ¯ z k k 2 ¯ ¯ ¯ ¯ = X k =1 | z | k k 2 = X k =1 1 k 2 292

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19.2 Taylor Series converges, the given series is absolutely convergent for every z on | z | = 1. Since absolute convergence implies convergence, the given series converges for all z on | z | =1. (b) The circle of convergence is | z | = 1. On the circle, n | z | n →∞ as n →∞ . This implies nz n 6→ 0as n →∞ . Thus by Theorem 19 . 3 the series is divergent for every z on the circle | z | =1. EXERCISES 19.2 Taylor Series 3. DiFerentiating 1 1+2 z =1 2 z +2 2 z 2 2 3 z 3 + ··· gives 2 (1 + 2 z ) 2 = 2+2 · 2 2 z 3 · 2 3 z 2 + ··· .Thu s 1 (1+2 z ) =1 2 · (2 z )+3 · (2 z ) 2 −··· = X k =1 ( 1) k 1 k (2 z ) k 1 where R = 1 2 . 6. Replacing z in e z = X k =0 z k k ! by z 2 and multiplying the result by z gives ze z 2 = X k =0 ( 1) k k ! z 2 k +1 where R = . 9. Replacing z in cos z = X k =0 ( 1) k z 2 k (2 k )! by z/ 2 gives cos z 2 = X k =0 ( 1) k (2 k )! ³ z 2 ´ 2 k where R = . 12. Using the identity cos z = 1 2 (1 + cos2 z ) and the series cos z = X k =0 ( 1) k z 2 k (2 k )! gives cos 2 z = 1 2 + 1 2 X k =0 ( 1) k (2 z ) 2 k (2 k )!
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AEM_3e_Chapter_19 - 19 19 3 0 2 0 2 0 Series and Residues...

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