AEMC19 - 19 Series and Residues Exercises 19.1 1 5i 5 5i 5...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
19 Series and Residues Exercises 19.1 1. 5 i , 5, 5 i ,5 ,5 i 2. 2 i ,1 ,2+ i ,3 ,2 i 3. 0, 2, 0, 2, 0 4. 1+ i ,2 i , 2+2 i , 4, 4 4 i 5. Converges. To see this write the general term as 3 i +2 /n 1+ i . 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 . 7. Converges. To see this write the general term as ( i +2 /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+ 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 +1 2as n →∞ , and Im( z n )= 6 n 2 4 n 4 n 2 +1 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 0as n →∞ since 2 / 4 < 1. 13. S n = 1 1+2 i 1 2+2 i + 1 2+2 i 1 3+2 i + 1 3+2 i 1 4+2 i + ··· + 1 n +2 i 1 n +1+2 i = 1 1+2 i 1 n +1+2 i Thus, lim n →∞ S n = 1 1+2 i = 1 5 2 5 i . 14. By partial fractions, i k ( k +1) = i k i k +1 and so S n = i i 2 + i 2 i 3 + i 3 i 4 + ··· + i n i n +1 = i i n +1 . Thus lim n →∞ S n = i . 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 =1 / 3. Since | z | =1 / 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 | =1 / 2 < 1 the series converges to i/ 2 1 i/ 2 = 1 5 + 2
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Exercises 19.1 18. We identify a =1 / 2 and z = i . Since | z |− 1 the series is divergent. 19. We identify a = 3 and z =2 / (1+2 i ). Since | z | =2 / 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 | =1 / 2 < 1 the series converges to 1 1+ i 1 i 1+ 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 1+ i n +1 1 n µ i 1+ i n ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = lim n →∞ n n +1 ¯ ¯ ¯ ¯ i 1+ i ¯ ¯ ¯ ¯ = 1 2 we see that the radius of convergence is R = 2. The circle of convergence is | z | = 2. 23.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 23

AEMC19 - 19 Series and Residues Exercises 19.1 1 5i 5 5i 5...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online