Answer to Question #42429 in Complex Analysis for shasha

Question

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

Help me please

Answer 1

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