Pre-Calculus

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Could someone help me with this-we learned this today in school but I'm not getting it now that I'm home with my homework-I have lots of these to do can someone help with this one and then maybe it'll click-thank you
Find (sqrt3/2 -1/2i)^4 by using DeMoivre's Theorem
I have to convert to polar, use the theorem and put it back into rectangular form

  • Pre-Calculus -

    tan(t) = y/x = (-1/2) /(sqrt(3)/2) = -1/sqrt(3)

    so, t = -pi/6

    r^2 = x^2 + y^2 = 3/4 + 1/4 = 1
    r = 1

    so, point z = (1,-pi/6)

    z^4 = 1^4 cis 4*(-pi/6) = 4 cis -2pi/3

    x = r cos -2pi/3 = 1 * -1/2
    y = r sin -2pi/3 = 1 * -sqrt(3)/2

    z^4 = (-1/2, -sqrt(3)/2)

    Just for confirmation,

    (sqrt(3)/2 - i/2)^4 =

    9/16 - 4*3*sqrt(3)/(8*2)i + 6*3/4*1/4 i^2 - 4*sqrt(3)/2 * 1/8 i^3 + 1/16 i^4

    = 9/16 - 3sqrt(3)/4 i - 18/16 + 1/4 sqrt(3) i + 1/16

    -8/16 - sqrt(3)/2 i

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