Question

If `kappa = cos((2pi)/9)+i sin((2pi)/9)` (i.e. the primitive Complex `9`th root of `1`) then how do you express `kappa^8` in the form `a+b kappa + c kappa^2 + d kappa^3 + e kappa^4 + f kappa^5` where `a, b, c, d, e, f` are rational?

Answer 1

`kappa^8 = -kappa^5-kappa^2`

Explanation:

Consider:

`kappa = cos((2pi)/9) + i sin((2pi)/9)`

Using de Moivre's formula:

`(cos(x) + i sin(x))^n = cos(nx) + i sin(nx)`

it should be fairly clear that `kappa` satisfies:

`kappa^9 = 1`

That is: `kappa` is a zero of the polynomial:

`x^9-1`

What may be less obvious is that this is not the simplest polynomial for which `kappa` is a zero.

We find:

`x^9-1 = (x-1)(x^2+x+1)(x^6+x^3+1)`

`kappa` is not a zero of `(x-1)`, nor of `(x^2+x+1)`, so it must be a zero of:

`x^6+x^3+1`

This is the minimum polynomial for `kappa` over the rational numbers.
This polynomial is known as a cyclotomic polynomial, specifically `Phi_9`

So if we can express:

`kappa^8 = q_kappa(kappa^6+kappa^3+1) + r_kappa`

For some quotient polynomial `q_kappa` and remainder polynomial `r_kappa` where `r_kappa` is of degree `5` or lower, then:

`kappa^8 = r_kappa`

since `(kappa^6+kappa^3+1) = 0`

So all we have to do is long divide `kappa^8` by `kappa^6+kappa^3+1` and identify the remainder.

Actually the long division terminates after the first term of the quotient, since:

`kappa^8 - kappa^2(kappa^6+kappa^3+1) = color(red)(cancel(color(black)(kappa^8))) - color(red)(cancel(color(black)(kappa^8)))-kappa^5-kappa^2`

`color(white)(kappa^8 - kappa^2(kappa^6+kappa^3+1)) = -kappa^5-kappa^2`

which is already of degree `5`.

So:

`kappa^8 = -kappa^5-kappa^2`