# Answer on Complex Analysis Question for shasha

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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