anonymous
  • anonymous
use the demoivre s 'theorem to find the answer [1-/3i/1+/3i]9 1- underroot 3i divided by 1+underroot 3i
Mathematics
katieb
  • katieb
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dumbcow
  • dumbcow
\[[\frac{1-\sqrt{3}i}{1+\sqrt{3}i}]^{9}\] is that correct
anonymous
  • anonymous
yes it is
anonymous
  • anonymous
plz solve...

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dumbcow
  • 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
anonymous
  • anonymous
thanx...:)

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