In the plane of complex numbers, 1 has the modulus and the argument values ,, ,....
Taking the cube root of the complex number 1, you have the modulus (positive value) and
three argument values , and .
It gives three values of the cube root of the complex number 1:
1, and .
Figure shows complex (real) number 1 in red and two other roots in blue.
Note that these tree points are vertices of the equilateral triangle.
Figure. Cube roots of the complex number 1
You can put this solution on YOUR website! Assume the cube root of 1 is z, that is,
:
z = 1^(1/3)
:
now cube both sides of the =
:
z^3 = 1
:
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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