Question #42429
Find the cube roots of 27(cos 279° + i sin 279°).
Help me please
Help me please
Expert's answer
Using thr De Moivre's formula we get
(27(cos 279° + i sin 279°))^(1/3)=27^(1/3)(cos 279°/3 + i sin 279°/3)=3(cos 93 + i sin 93)=(approximately) -0.157+3i
(27(cos 279° + i sin 279°))^(1/3)=27^(1/3)(cos 279°/3 + i sin 279°/3)=3(cos 93 + i sin 93)=(approximately) -0.157+3i
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#339914 on Dec 2023
#339914 on Dec 2023
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