Question 115817: change -sqrt3 + i in trigonometric form and find
1. (-sqrt3+i)^5, answer in rectangular form such as a+bi?
2. (-sqrt3 + i)^1/3, write all answers separately in trigonometric form using degree angles?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! change -sqrt3 + i in trigonometric form and find
1. (-sqrt3+i)^5, answer in rectangular form such as a+bi?
r = sqrt[(-sqrt3)^2 + 1^2] = sqrt[4] = 2
theta = tan^-1[1/(-srt3)] = 150 degrees
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Therefore: (-sqrt3+i) = [2cis(150)]^5
= 32cis(5*150)
= 32 cis (750)
= 32cis(30)
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2. (-sqrt3 + i)^1/3, write all answers separately in trigonometric form using degree angles?
(-sqrt3 + i)^(1/3) = [2cis(150)]^(1/3)
= 2^(1/3)cis((1/3)150)
=2^(1/3)cis(50)
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I'm assuming you want the principal root, not all three cube roots
If you want all three roots you need to use (1/3)[150+360k) where
k = 0, then k=1, then k=2
You would have angles of 50, 50+120, and 50+240
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Cheers,
Stan
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