anonymous 5 years ago use the demoivre s 'theorem to find the answer [1-/3i/1+/3i]9 1- underroot 3i divided by 1+underroot 3i

1. dumbcow

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

2. anonymous

yes it is

3. anonymous

plz solve...

4. dumbcow

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

5. anonymous

thanx...:)