Question #d99ea

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Apr 15, 2017

Answer:

#2i#, #sqrt(3)-i#, #-sqrt(3)-i#

Explanation:

The primitive complex cube root of #1# is:

#omega = -1/2+sqrt(3)/2i#

If #alpha# is one root of #z^3=-8i# then the other two are #omega alpha# and #omega^2 alpha#.

Note that:

#(2i)^3 = 2^3i^3 = -8i#

So #2i# is one root.

So the other roots are:

#omega * 2i = (-1/2+sqrt(3)/2i) 2i = -sqrt(3)-i#

#omega^2 * 2i = (-1/2-sqrt(3)/2i) 2i = sqrt(3)-i#

Here are the three roots in the complex plane, plotted with the circle #abs(z) = 2#:

graph{(x^2+(y-2)^2-0.01)((x-sqrt(3))^2+(y+1)^2-0.01)((x+sqrt(3))^2+(y+1)^2-0.01)(x^2+y^2-4) = 0 [-5, 5, -2.5, 2.5]}

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