Question

Find all the complex roots. Write the answer in the indicated form. The complex cube roots of 27(cos 234° + i sin 234°) (polar form)
B. 3(cos 78° + i sin 78°), 3(cos 118° + i sin 118°), 3(cos 158° + i sin 158°)
C. -3(cos 78° + i sin 78°), 3(cos 118° + i sin 118°), -3(cos 158° + i sin 158°)
D. 3(cos 78° + i sin 78°), 3(cos198° + i sin 198°), 3(cos 318° + i sin 318°)

Answer 1

@bibby

Answer 2

\[\Large\rm \left[27\left(\cos234^o+\mathcal i \sin234^o\right)\right]^{1/3}\]So we need to apply our De Moivre's Theorem, yes? :)

Answer 3

We should think of our angle like this:\[\Large\rm \left[27\left(\cos\left(234^o+360^ok\right)+\mathcal i \sin\left(234^o+360^ok\right)\right)\right]^{1/3}\]We can get that same angle by spinning around.

Answer 4

And our roots will correspond to k=0,1,2.
The first three k values.
That will give us our three unique roots.

Answer 5

\[\Large\rm 3\left[\cos\left(234^o+360^ok\right)+\mathcal i \sin\left(234^o+360^ok\right)\right]^{1/3}\]Take the cube root of the 27.

Answer 6

Remember how to apply your De Moivre's Theorem?
We just multiply the angles by this exponent.\[\Large\rm =3\left[\cos\left(\frac{234^o+360^ok}{3}\right)+\mathcal i \sin\left(\frac{234^o+360^ok}{3}\right)\right]\]So we're multiplying by 1/3, or dividing by 3.

Answer 7

Plug in k=0 to get your first solution.
k=1 for the next,
and k=2 for the final solution.

Answer 8

Lady Skitttleeeee :U where you at
got it?

Answer 9

Where do I plug in k=0? I don't understand any of this, it really confuses me

Answer 10

So you're at some angle 234 degrees.

Answer 11

But we can also get to that same location by spinning around the circle, yah?