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Answer to Question #42431 in Complex Analysis 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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