Question 1133322
Assume the cube root of 1 is z, that is,
:
z = 1^(1/3)
:
now cube both sides of the =
:
z^3 = 1
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rewrite as
:
z^3 -1 = 0
:
this cubic can be factored 
:
(z-1)(z^2 +z +1) = 0
:
use quadratic formula
:
z = -1 +square root(1^2 -4**1)/2*1 = -(1/2) +i * square root(3)/2
:
z = -1 -square root(1^2 -4**1)/2*1 = -(1/2) -i * square root(3)/2 
:
z = 1
:
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The 3 cubic roots of unity (1 + 0i) in complex standard form are
:
1+0i, (-1+i*square root(3))/2, (-1-i*square root(3))/2
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