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QUESTION

# How do you get the complex cube root of 8?

The cube roots of 8 are 2, 2omega and 2omega^2 where omega=-1/2+sqrt(3)/2 i is the primitive Complex cube root of 1.

Here are the cube roots of 8 plotted in the Complex plane on the circle of radius 2:

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

They can be written as:

2(cos(0)+i sin(0)) = 2

2(cos((2pi)/3) + i sin((2pi)/3)) = -1 + sqrt(3)i = 2omega

2(cos((4pi)/3) + i sin((4pi)/3)) = -1 - sqrt(3)i = 2omega^2

One way of finding these cube roots of 8 is to find all of the roots of x^3-8 = 0.

x^3-8 = (x-2)(x^2+2x+4)

The quadratic factor can be solved using :

x = (-b +-sqrt(b^2-4ac))/(2a)

= (-2+-sqrt(2^2-(4xx1xx4)))/(2*1)

=(-2+-sqrt(-12))/2

=-1+-sqrt(3)i