# Answer on Complex Analysis Question for shasha

Question #42431

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

Help me please

Help me please

Expert's answer

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.

(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.

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