Chapter 17

# Chapter 17 - 17 1 3 3i Functions of a Complex Variable...

• Notes
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17 Functions of a Complex Variable Exercises 17.1 1. 3 + 3 i 2. 4 i 3. i 8 = ( i 2 ) 4 = ( 1) 4 = 1 4. i 11 = i ( i 2 ) 5 = i ( 1) 5 = i 5. 7 13 i 6. 3 9 i 7. 7 + 5 i 8. 7 + 8 i 9. 11 10 i 10. 3 4 + 2 3 i 11. 5 + 12 i 12. 2 2 i 13. 2 i 14. i 1 + i · 1 i 1 i = i + 1 2 = 1 2 + 1 2 i 15. 2 4 i 3 + 5 i · 3 5 i 3 5 i = 14 22 i 34 = 7 17 11 17 i 16. 10 5 i 6 + 2 i · 6 2 i 6 2 i = 50 50 i 40 = 5 4 5 4 i 17. 9 + 7 i 1 + i · 1 i 1 i = 16 2 i 2 = 8 i 18. 3 i 11 2 i · 11 + 2 i 11 + 2 i = 35 5 i 125 = 7 25 1 25 i 19. 2 11 i 6 i · 6 + i 6 + i = 23 64 i 37 = 23 37 64 37 i 20. 4 + 3 i 3 + 4 i · 3 4 i 3 4 i = 24 7 i 25 = 24 25 7 25 i 21. (1 + i )(10 + 10 i ) = 10(1 + i ) 2 = 20 i 22. [(1 + i )(1 i )] 2 (1 i ) = 4 4 i 23. 20 + 23 i + 1 2 i · 2 + i 2 + i = 20 + 23 i + 2 5 + 1 5 i = 102 5 + 116 5 i 24. (2 + 3 i )( i ) 2 = 2 3 i 25. i 9 + 7 i · 9 7 i 9 7 i = 7 + 9 i 130 = 7 130 + 9 130 i 26. 1 6 + 8 i · 6 8 i 6 8 i = 6 8 i 84 = 1 14 2 21 i 27. x x 2 + y 2 28. x 2 y 2 29. 2 y 4 30. 0 31. ( x 1) 2 + ( y 3) 2 32. 36 x 2 + 16 y 2 33. 2 x + 2 yi = 9 + 2 i implies 2 x = 9 and 2 y = 2. Hence z = 9 2 + i . 34. x + 3 yi = 7 + 6 i implies x 7 and 3 y = 6. Hence z = 7 + 2 i . 35. x 2 y 2 + 2 xyi = 0 + i implies x 2 y 2 = 0 and 2 xy = 1. Now y = x implies 2 x 2 = 1 and so x = ± 1 / 2. The choice y = x gives 2 x 2 = 1 which has no real solution. Hence z = 1 2 + 1 2 i and z = 1 2 1 2 i . 757

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Exercises 17.1 36. x 2 y 2 4 x + ( 2 xy 4 y ) i = 0 + 0 i implies x 2 y 2 4 x = 0 and y ( 2 x 4) = 0. If y = 0 then x ( x 4) = 0 and so z = 0 and z = 4. If 2 x 4 = 0 or x = 2 then 12 y 2 = 0 or y = ± 2 3. This gives z = 2 + 2 3 i and z = 2 2 3 i . 37. | 10 + 8 i | = 164 and | 11 6 i | = 157. Hence 11 6 i is closer to the origin. 38. | 1 2 1 4 i | = 5 4 and | 2 3 + 1 6 i | = 17 6 . Since 5 4 < 17 6 , 1 2 1 4 i is closer to the origin. 39. | z 1 z 2 | = | ( x 1 x 2 ) + i ( y 1 y 2 ) | = ( x 1 x 2 ) 2 + ( y 1 y 2 ) 2 which is the distance formula in the plane. 40. By the triangle inequality, | z +6+8 i | ≤ | z | + | 6+8 i | . On the circle, | z | = 2 and so | z +6+8 i | ≤ 2+ 100 = 12. Exercises 17.2 1. 2(cos2 π + i sin2 π ) 2. 10(cos π + i sin π ) 3. 3 cos 3 π 2 + i sin 3 π 2 4. 6 cos π 2 + i sin π 2 5. 2 cos π 4 + i sin π 4 6. 5 2 cos 7 π 4 + i sin 7 π 4 7. 2 cos 5 π 6 + i sin 5 π 6 8. 4 cos 4 π 3 + i sin 4 π 3 9. 3 2 2 cos 5 π 4 + i sin 5 π 4 10. 6 cos π 6 + i sin π 6 11. z = 5 3 2 5 2 i 12. z = 8 + 8 i 13. z = 5 . 5433 + 2 . 2961 i 14. z = 8 . 0902 + 5 . 8779 i 15. z 1 z 2 = 8 cos π 8 + 3 π 8 + i sin π 8 + 3 π 8 = 8 i ; z 1 z 2 = 1 2 cos π 8 3 π 8 + i sin π 8 3 π 8 = 2 4 2 4 i 16. z 1 z 2 = 6 cos π 4 + π 12 + i sin π 4 + π 12 = 6 2 + 3 2 2 i z 1 z 2 = 6 3 cos π 4 π 12 + i sin π 4 π 12 = 2 2 + 6 6 i 17. 3 2 cos 7 π 4 + i sin 7 π 4 10 cos π 3 + i sin π 3 = 30 2 cos 7 π 4 + π 3 + i sin 7 π 4 + π 3 = 40 . 9808 + 10 . 9808 i 18. 4 2 cos π 4 + i sin π 4 2 cos 3 π 4 + i sin 3 π 4 = 8 cos π 4 + 3 π 4 + i sin π 4 + 3 π 4 = 8 19.
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