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Find the cube roots of 27(cos 279° + i sin 279°).

User Uduse
by
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1 Answer

6 votes

The solution would be like this for this specific problem:

27(cos 279° + i sin 279°).
(cosx +i sinx) = cox(nx)) + i sin(nx)
(27×(cos 279+i sin 279))1/3=27 1/3×(cos 279+i sin 279)1/3
2713=27−−√3=?
3×(cos279+i sin279)1/3
3×(cos 279+i sin 279)13=3(cos 279/3+i sin 279/3)

279/3 = 93
cube root =
3(cos 93 + i sin 93)

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

User Ivan Ferrer Villa
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8.8k points

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