Question

Find \((-1+\sqrt{2} i)^3\) using DeMoivre's Theorem.

Answer 1

first job is to write this in polar (trig) form
do you know how to do that?

Answer 2

yea i got\[\sqrt{3}^3(cos(-0.955)+i~sin(-0.955))\]butI dont know if im supposed to add \(\pi\) or \(2\pi\)

Answer 3

@satellite73

Answer 4

ok lets back up a second

Answer 5

\(r\left(\cos(\theta)+i\sin(\theta)\right)\) is the form you want, and so you need \(r\) and \(\theta\)

Answer 6

\(r\) is easiest. if you have \(a+bi\) then \(r=\sqrt{a^2+b^2}\)

Answer 7

\[r=\sqrt{(-1)^2+(\sqrt{2})^2}\]\[r=\sqrt{3}\]\[\theta=tan^{-1}\left({\sqrt{2}\over-1}\right)\]\[\theta\approx-0.955\]

Answer 8

in your case you have \(-1+\sqrt{2}i\) and so \(r=\sqrt{(-1)^2+\sqrt{2}^2}=\sqrt{1+2}=\sqrt{3}\)

Answer 9

ok good for \(r\) lose the calculator for \(\theta\) it is not helping you

Answer 10

if you were in quadrant 1, then you can use \(\theta=\tan^{-1}(\frac{b}{a})\) but you are in quadrant 3

Answer 11

how do i know that im in quadrant 3?

Answer 12

that's quadrant 2..

Answer 13

i lied, you are in quadrant 2

Answer 14

but in any case \(\tan^{-1}(x)\) only gives answers in quad 1 or 4

Answer 15

right so i would add \(\pi\) to to \(-0.955\) and get \(2.186\) then

Answer 16

which is in quadrant 2

Answer 17

yeah i guess that is what you need, right

Answer 18

now you can multiply that angle by 3, and cube \(\sqrt{3}\) to get your answer

Answer 19

just for my own information, you are sure this is not \(-1+\sqrt{3}i\) right? that would be much easier

Answer 20

when you multiply that by 3 .. you get 6.559 which means i would subtract \(2\pi\) to get the angle within the range \(0\le\theta\le2\pi\) and that would be \(0.276\)

Answer 21

no it is what i posted, though it would be easier if it was
so is that right?

Answer 22

@satellite73