Question

PLEASE HELP!!!!!!! SO CONFUSED!
Find the cube roots of 27(cos 279° + i sin 279°).

Answer 1

Do you know De Moivre's Theorem?

Answer 2

Yes (cosx +isinx) = cox(nx)) + isin(nx)

Answer 3

Good! Then we can express the question as:
\[(27\times (\cos 279+i \sin 279))^{\frac{1}{3}}=27^{\frac{1}{3}}\times (\cos 279+i \sin 279)^{\frac{1}{3}}\]
Can you see what to do now?

Answer 4

you would then divide 27 by 3? and 279 by 3? this is where i get very confused!

Answer 5

You can't apply De Moivre's Theorem to the cube root of 27.
\[27^{\frac{1}{3}}=\sqrt[3]{27}=?\]

Answer 6

that would be 3 ?

Answer 7

That's correct! So now we have:
\[3\times (\cos 279+i \sin 279)^{\frac{1}{3}}\]
and can apply De Moivre's Theorem.

Answer 8

Okay, so where do i got from there?

Answer 9

\[3\times (\cos 279+i \sin 279)^{\frac{1}{3}}=3(\cos \frac{279}{3}+i \sin \frac{279}{3})\]

Answer 10

279/3 is 93, so thats my cubed root?

Answer 11

The cube root is 3(cos 93 + i sin 93)

Answer 12

ohhh okay, thats my final form! I need to do some more practicing! Thank you so much!!!

Answer 13

You're Welcome :)