# Answer to Question #42429 in Complex Analysis 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

Need a fast expert's response?

Submit orderand get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

## Comments

## Leave a comment