Question

Find the indicated power using De Moivre's Theorem. (Express your fully simplified answer in the form
a + bi.) [(Sqrt(3)-i]^-10

Answer 1

\[\Large\rm \left(\sqrt3-1\mathcal i\right)^{-10}\]Factor a 2 out of each term,\[\Large\rm 2^{-10}\left(\frac{\sqrt3}{2}-\frac{1}{2}\mathcal i\right)^{-10}\]Then relate it to your cosine and sine,\[\Large\rm \cos \theta=\frac{\sqrt3}{2}, \qquad\qquad\qquad \sin \theta=-\frac{1}{2}\]This is in the the uhhhh 4th quadrant, yes?
Cosine positive, Sine negative.

I believe the angle we're looking for is \(\Large\rm-\dfrac{\pi}{6}\)

Answer 2

\[\Large\rm 2^{-10}\left(\cos \left[-\frac{\pi}{6}\right]+\mathcal i\sin \left[-\frac{\pi}{6}\right]\right)^{-10}\]

Answer 3

Then we can finally apply De Moivre's Theorem, yes?\[\Large\rm 2^{-10}\left(\cos \left[\frac{10\pi}{6}\right]+\mathcal i\sin \left[\frac{10\pi}{6}\right]\right)\]

Answer 4

Well that's not in a+bi form, is that why maybe?

Answer 5

maybe I tried working it all out and i got -5/32 + 10 Sqrt3 /2

Answer 6

Angle simplifies to 5pi/3.
So we get uhhhhhh, something like,\[\Large\rm 2^{-10}\left(\frac{1}{2}-\frac{\sqrt3}{2}\mathcal i\right)\]If we write that in a+bi form (distributing the 2), we get,\[\Large\rm \frac{1}{2^{11}}-\frac{\sqrt3}{2^{11}}\mathcal i\]Like that maybe?
Did you forgot the 2^(-10) or something? :o
These numbers look a bit different than what you came up with, hmmm.