Question

Help please:
(1-sqrt(3)*i)^5
using De Moivre's theorem

Answer 1

\[(1-\sqrt{3}i)^{5}\]

Answer 2

I think \(r=2\): \((\large1-\sqrt3 i)^5=[2(\frac{1}{2}-\frac{\sqrt3}{2}i)]^5=[2(cos\frac{\pi}{3}-isin\frac{\pi}{3})]^5\)

Answer 3

How did you get the 1/2 and \[\sqrt{3}/2\]

Answer 4

Take 2 outside: \(\large1+\sqrt3=2(\frac{1}{2}+\frac{\sqrt3}{2})\)

Answer 5

I mean -. Sorry

Answer 6

Why is r=2?

Answer 7

and how did you find \[\theta \]?

Answer 8

This is Moirve's theorem: \([r(cos~\theta + isin~\theta)]^n=r^n(cos~n\theta +isin~n\theta)\), right?
So you know why r = 2.
Now we have to find \(\theta\), knowing that \(cos~\theta=\frac{1}{2}\) and \(sin~\theta=\frac {\sqrt3}{2}\). Then \(\theta=\frac{\pi}{3}\)

Answer 9

so what would be the simplified answer?

Answer 10

So \([2(cos\frac{\pi}{3}+isin\frac{\pi}{3})]^5=2^5(cos\frac{5\pi}{3}+isin\frac{5\pi}{3})=...\)
Can you simplify that?

Answer 11

\[16-16\sqrt{3}i\]

Answer 12

But it's wrong

Answer 13

Oops. I am so sorry. It should be \(2^5(cos\frac{5\pi}{3}-isin\frac{5\pi}{3})\). My fault

Answer 14

Okay thank you so much!

Answer 15

You are welcome