QUESTION: how to prove sum of complex numbers z is zero when z^3=1.
I think that what you want is to show that the sum of the roots is zero.
The cubic equation f(z) = z^3 - 1 = 0 has three roots, z1, z2, and z3 s.t. z1^3 =1, z2^3 = 1 and z3^3 = 1.
We wish to show that z1 + z2 + z3 = 0.
We have z^3 = 1
writing the complex number z^3 in polar form,
z^3 = 1 = cos(2n.pi) + i.sin(2n.pi) = 1.e^(i.(2n.pi))
Using z^3 = cos(2n.pi) + i.sin(2n.pi), then
z = cos(2n.pi/3) + i.sin(2n.pi/3), n = 0,1,2
n = 0: z1 = cos(0) + i.sin(0) = 1
n = 1: z2 = cos(2.pi/3) + i.sin(2,pi/3) = -1/2 + i.sqrt(3)/2
n = 2: z3 = cos(4.pi/3) + i.sin(4.pi/3) = -1/2 - i.sqrt(3)/2
Adding together the three roots,
z1 + z2 + z3 = 1 + (-1/2 + i.sqrt(3)/2) + (-1/2 - i.sqrt(3)/2)
z1 + z2 + z3 = (1 - 1/2 - 1/2) + i.(sqrt(3)/2 - sqrt(3)/2)
z1 + z2 + z3 = 0