21The roots of the equation Z^n-1 = 0 , are given by U_n = \left \{ 1,\epsilon,\epsilon^2,...,\epsilon^{n-1}\right \}, \epsilon^r = e^{\frac {2r\pi}{n}}, r \in \left \{ 0,...,n-1 \right \}. The root \epsilon ^r \in U_n is called primitive if for all positive integers m<n we have (\epsilon ^r)^m \ne 1.
Its not tough to prove that Primitive roots of Z^n-1 = 0 are given by \epsilon _k = e^{\frac{2k\pi}{n}} where k\in \left \{ 1,...,n \right \}, \gcd(k,n)= 1
And its also easy to prove that if Z_c \in U_n is a primitive root of unity ,Then the roots of Z^n-1 = 0 are given by Z_c ^r, Z_c^{r+1},..., Z_c^{r+n-1}. Where r is an arbitrary positive integer.
Now for the solution see http://www.targetiit.com/iit-jee-forum/posts/complex-nos-18416.html