Question

Find the cube roots of 27(cos 330° + i sin 330°

Answer 1

@agent0smith please help :(

Answer 2

@ranga

Answer 3

have they covered De Moivre's theorem yet?

Answer 4

Yes i just dont get it

Answer 5

\[\Large e^{ix} = \cos x + isin x\]

Answer 6

r^3(cos(3Θ + 360nº)+i sin(3Θ + 360nº))
z^3= 27(cos 330º + i sin 330º)
z1 = 3(cos 110º + i sin 110º)
z2 = 3(cos 230º + i sin 230º)
z3 = 3(cos 350º + i sin 350º)
i have that

Answer 7

So is what i have the answer or do i need to do more?

Answer 8

\[\Large (27(\cos 330 + isin 330))^{1/3} = (27e^{i330})^{1/3} = 3(\cos 110 + i \sin 110)\]

Answer 9

No that is it.

Answer 10

so should i earase all that i have

Answer 11

No, those are the steps.
We start with the Euler's formula: e^(ix) = cos(x) + isin(x)
Then to find cube root of 27(cos 330° + i sin 330°) we raise it to the power of 1/3: {27(cos 330° + i sin 330°)}^(1/3) = {27e^(330i)}^(1/3) = 3e^(110i) =
3(cos 110º + i sin 110º)

Answer 12

Thank you~