Find the solution of the following equation whose argument is strictly between
120° to 180°
Round your answer to the nearest thousandth.
z^10 = i
So I'm having trouble solving its Angle
This is the formula I use (θ+2kπ)/n so when I use this formula
[90 + 2(180)]/10 = 45°
so 45° is not between 120 to 180. I need to know how can I get the angle that will suitable between 120 to 180..
Powers of complex numbers problem..?
2020-05-01 10:53 am
回答 (2)
2020-05-01 11:20 am
✔ 最佳答案
z^10 = iz^10 = cos(90°) + i sin(90°)
z = [cos(90°) + i sin(90°)]^(1/10)
By De Moivre's theorem:
z = cos[(90° + 360k°)/10] + i sin[(90° + 360k°)/10]
where k = 0, 1, 2, …… 9
120° < (90° + 360k°)/10 < 180°
1200 < 90 + 360k < 1800
1110 < 360k < 1710
3.08 < k < 4.75
Hence, k = 4
When k = 4:
z = cos[(90° + 360°*4)/10] + i sin[(90° + 360*°4)/10]
z = cos(153°) + i sin(153°)
z = -0.891 + 0.454i
2020-05-01 10:58 am
z^10 = i
z^10 = 0 + i
z^10 = cos(pi/2 + 2pi * k) + i * sin(pi/2 + 2pi * k)
k is an integer
z = cos((pi/2 + 2pi * k) / 10) + i * sin((pi/2 + 2pi * k) / 10)
z = cos((1/20) * (pi + 4pi * k)) + i * sin((1/20) * (pi + 4pi * k))
Or in degrees
z = cos((1/20) * (180 + 720 * k)) + i * sin((1/20) + (180 + 720 * k))
z = cos(9 + 36 * k) + i * sin(9 + 36 * k)
120 < 9 + 36 * k < 180
111 < 36k < 171
37 < 12k < 57
37/12 < k < 57/12
k is an integer, so k = 4 is the only answer that works
9 + 36 * k =>
9 + 36 * 4 =>
9 + 144 =>
153
z^10 = 0 + i
z^10 = cos(pi/2 + 2pi * k) + i * sin(pi/2 + 2pi * k)
k is an integer
z = cos((pi/2 + 2pi * k) / 10) + i * sin((pi/2 + 2pi * k) / 10)
z = cos((1/20) * (pi + 4pi * k)) + i * sin((1/20) * (pi + 4pi * k))
Or in degrees
z = cos((1/20) * (180 + 720 * k)) + i * sin((1/20) + (180 + 720 * k))
z = cos(9 + 36 * k) + i * sin(9 + 36 * k)
120 < 9 + 36 * k < 180
111 < 36k < 171
37 < 12k < 57
37/12 < k < 57/12
k is an integer, so k = 4 is the only answer that works
9 + 36 * k =>
9 + 36 * 4 =>
9 + 144 =>
153
收錄日期: 2021-04-18 18:30:00
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200501025321AA5fNAG