Answer to Question #42431 in Complex Analysis for shasha

Question

Find the cube roots of 125(cos 288° + i sin 288°).


Help me please

Answer 1

Using thr De Moivre's formula we get
(125(cos 288° + i sin 288°))^(1/3)=125^(1/3)(cos 288°/3 + i sin 288°/3)=5(cos 96° + i sin 96°)=(approximately)-0.5+4.97i.