Re: HSC 2013 4U Marathon
let \alpha be a root of z^n-1=0
Lets form a polynomial with roots, 1-1, 1-z_1, 1-z_2, ....
z=1-\alpha \ \ \alpha = 1-z
Therefore the polynomial with roots:
0, (1-z_1), (1-z_2) ... , (1-z_{n-1})
is
(1-z)^n - 1 = 0
z^n - nz^{n-1}+ \dots + (-1)^{n-1} n z = 0...