Question

use the demoivre s 'theorem to find the answer
[1-/3i/1+/3i]9
1- underroot 3i divided by 1+underroot 3i

Answer 1

\[[\frac{1-\sqrt{3}i}{1+\sqrt{3}i}]^{9}\]
is that correct

Answer 2

yes it is

Answer 3

plz solve...

Answer 4

first multiply by conjugate of denominator on top and bottom
\[\frac{1-\sqrt{3}i}{1+\sqrt{3}i}*\frac{1-\sqrt{3}i}{1-\sqrt{3}i}\]

\[ = -[\frac{1}{2} + \frac{\sqrt{3}}{2}i]\]
replace with trig functions
1/2 = cos(pi/3)
sqrt3/2 = sin(pi/3)
\[= -(\cos \pi/3 + i \sin \pi/3)^{9}\]
use de Moivre theorem
\[= -(\cos 3\pi + i \sin 3\pi)\]
cos 3pi = -1, sin 3pi = 0
= 1

Answer 5

thanx...:)