AEM_3e_Chapter19

AEM_3e_Chapter19 - 19 Series and Residues EXERCISES 19.1...

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19 19 Series and Residues EXERCISES 19.1 Sequences and Series 1. 5 i , 5, 5 i ,5 i 2. 2 i ,1 ,2+ i ,3 ,2 i 3. 0, 2, 0, 2, 0 4. 1+ i i , 2+2 i , 4, 4 4 i 5. Converges. To see this write the general term as 3 i +2 /n i . 6. Converges. To see this write the general term as µ 2 5 n n 2 n i 1+3 n 5 n i . 7. Converges. To see this write the general term as ( i /n ) 2 i . 8. Diverges. To see this consider the term n n +1 i n and take n to be an odd positive integer. 9. Diverges. To see this write the general term as n µ 1 n i n . 10. Converges. The real part of the general term converges to 0 and the imaginary part of the general term converges to π . 11. Re( z n )= 8 n 2 + n 4 n 2 2as n →∞ , and Im( z n 6 n 2 4 n 4 n 2 3 2 as n . 12. Write z n = µ 1 4 + 1 4 i n in polar form as z n = à 2 4 ! n cos + i à 2 4 ! n sin .Now Re( z n à 2 4 ! n cos 0as n and Im( z n à 2 4 ! n sin n since 2 / 4 < 1. 13. S n = 1 1+2 i 1 i + 1 i 1 3+2 i + 1 i 1 4+2 i + ··· + 1 n i 1 n +1+2 i = 1 i 1 n i Thus, lim n →∞ S n = 1 i = 1 5 2 5 i . 14. By partial fractions, i k ( k +1) = i k i k and so S n = i i 2 + i 2 i 3 + i 3 i 4 + + i n i n = i i n . Thus lim n →∞ S n = i . 896
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19.1 Sequences and Series 15. We identify a = 1 and z =1 i . Since | z | = 2 > 1 the series is divergent. 16. We identify a =4 i and z / 3. Since | z | / 3 < 1 the series converges to 4 i 1 1 / 3 =6 i. 17. We identify a = i/ 2 and z = i/ 2. Since | z | / 2 < 1 the series converges to i/ 2 1 i/ 2 = 1 5 + 2 5 i. 18. We identify a / 2 and z = i . Since | z |− 1 the series is divergent. 19. We identify a = 3 and z =2 / (1+2 i ). Since | z | / 5 < 1 the series converges to 3 1 2 1+2 i = 9 5 12 5 i. 20. We identify a = 1 / (1 + i ) and z = i/ (1 + i ). Since | z | / 2 < 1 the series converges to 1 1+ i 1 i i = 1 . 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. 22. From lim n →∞ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 n +1 µ i i n +1 1 n µ i i n ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = lim n →∞ n n ¯ ¯ ¯ ¯ i i ¯ ¯ ¯ ¯ = 1 2 we see that the radius of convergence is R = 2. The circle of convergence is | z | = 2. 23. From lim n →∞ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ( 1) n +1 ( n + 1)2 n +1 ( 1) n n 2 n ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = lim n →∞ n 2( n +1) = 1 2 we see that the radius of convergence is R = 2. The circle of convergence is | z 1 i | =2. 24. From lim n →∞ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 ( n 2 (3 + 4 i ) n +1 1 n 2 (3+4 i ) n ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = lim n →∞ µ n n 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. 25. From lim n →∞ n p | 1+3 i | n = | i | = 10 we see that the radius of convergence is R / 10. The circle of convergence is | z i | / 10.
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AEM_3e_Chapter19 - 19 Series and Residues EXERCISES 19.1...

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