Studying for Pre Cal exam

Find the fourth roots of
− 1/2 + (square root)3/2 i

Write the roots in trigonometric form.

A -
w 1=cos(35°)+isin(35°)
w2 =cos(125°)+isin(125°)
w3 =cos(215°)+isin(215°)
w4 =cos(305°)+isin(305°)

B -
w1 =cos(40°)+isin(40°)
w2 =cos(130°)+isin(130°)
w3 =cos(220°)+isin(220°)
w4 =cos(310°)+isin(3105°)

C - w =cos(20°)+isin(20°) 1
w2 =cos(110°)+isin(110°)
w3 =cos(200°)+isin(200°)
w4 =cos(290°)+isin(290°)

D -
w1 =cos(30°)+isin(30°)
w2 =cos(120°)+isin(120°)
w3 =cos(210°)+isin(210°)
w4 =cos(300°)+isin(300°)

E -
w1 =cos(25°)+isin(25°)
w2 =cos(115°)+isin(115°)
w3 =cos(205°)+isin(205°)
w4 =cos(295°)+isin(295°)

I'm studying for my pre cal exam. Can you please help me with this question?

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  1. let z = ( -1/2 + √3/2 i)
    = (cos 120° + sin 120 i)

    z^(1/4) = cos (120/4) + sin(120/4) i
    = cos 30 + sin30 i ------> my primary solution.

    there will be 4 roots, and 360/4 = 90°
    by adding multiples of 90 to 30° we can get the other 3

    z^(1/3)
    = cos 30 + sin 30 i
    or
    = cos 120 + sin 120 i
    or
    cos 210 + sin 210 i
    or
    cos 300 + sin 300 i

    looks like D

    looks like you are reviewing De Moivre's Theorem.
    In general , in finding the nth root, there will be n such roots.
    Once you have the primary one, divide 360 by n, then repeatedly add 360/n degrees to get the others.
    Notice if you add one more at that point, it brings you back to your first solution.
    e.g. adding 90° to cos 300 + sin300i will yield
    cos 390 + sin 390 i
    which is
    cos 30 + sin 30 i by coterminal angles, 390 = 360 + 30

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