show that given one nth root of z the others are obtained by

show that, given one nth root of z, the others are obtained by multiplying it by the nth roots of unity. (z is complex number)

Solution

An nth root of unity is a complex number satisfying the equation

z^ n 1 = 0.

Obviously, by the fundamental theorem of algebra, there are n nth roots of unity.

Additionally, since 1 = e^(2i), we can write an nth root as = e^(2i/n).

All other nth roots are given by the integer powers of this root between 0 and n 1: the roots are 1, , ^2 , . . . , ^(n1) .

By factoring the equation z^n 1 = 0, we can obtain a useful identity:

z^n 1 = (z 1)(z^(n1) + z^(n2) + · · · + z + 1) = 0

This means that all nth roots of unity other than 1 satisfy the equation 1 + + ^2 + · · · + ^(n1) = 0

One other nice property of roots of unity is that they are periodic, which is due to the fact that ^n = 1. If we have a power of that is greater than n, it is equal to to that power mod n.

For example, if is a third root of unity, then

^5 + ^4 + 1 = ^2 + + 1 = 0.