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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
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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[[\frac{1\sqrt{3}i}{1+\sqrt{3}i}]^{9}\] is that correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first 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
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