How do I use DeMoivre's theorem to solve `z^3-1=0`?

If `z^3-1=0`, then we are looking for the cubic roots of unity, i.e. the numbers such that `z^3=1`.

If you're using complex numbers, then every polynomial equation of degree `k` yields exactly `k` solution. So, we're expecting to find three cubic roots.

De Moivre's theorem uses the fact that we can write any complex number as `\rho e^{i \theta}= \rho (\cos(\theta)+i\sin(\theta))`, and it states that, if
`z=\rho (\cos(\theta)+i\sin(\theta))`, then
`z^n = \rho^n (\cos(n \theta)+i\sin(n \theta))`

If you look at `1` as a complex number, then you have `\rho=1`, and `\theta=2\pi`. We are thus looking for three numbers such that `\rho^3=1`, and `3\theta=2\pi`.

Since `\rho` is a real number, the only solution to `\rho^3=1` is `\rho=1`. On the other hand, using the periodicity of the angles, we have that the three solutions for `\theta` are
`\theta_{1,2,3}=\frac{2k\pi}{3}`, for `k=0,1,2`.

This means that the three solutions are:

1. `\rho=1, \theta=0`, which is the real number `1`.
2. `\rho=1, \theta=\frac{2\pi}{3}`, which is the complex number `-1/2 + \sqrt{3}/2 i`
3. `\rho=1, \theta=\frac{4\pi}{3}`, which is the complex number `-1/2 - \sqrt{3}/2 i`

`z^3-1=0`

`z^3=1`

We know that any complex number, `a+bi`, can be written in modulus-argument form, `r(cosx+isinx)`, where `r=sqrt(a^2+b^2)` and `x` satisfies `sinx=b/r` and `cosx=a/r`.

`therefore 1 =1(cos0+isin0)`

So `z^3=cos(0+2kpi)+isin(0+2kpi) rarr` Since the solutions to trig equations aren't unique, we need to consider other possibilities.

`z=[cos(0+2kpi)+isin(0+2kpi)]^(1/3)`

Use de Moivre's theorem: `(cosx+isinx)^k=coskx+isinkx`

`z=cos(2/3kpi) +i sin(2/3kpi)`

Now we must consider every k such that `-pi< 2/3kpi ≤ pi`

`k = 0, z = 1`

`k = 1, z = cos(2/3pi) + isin(2/3pi) = -1/2+sqrt3/2 i`

`k= -1, z= cos(-2/3 pi) + i sin (-2/3pi) = -1/2 - sqrt3/2 i`

These values are called the cubic roots of unity and are usually written as `1, omega, omega^2`.

The fact that `omega' = omega^2` can easily be verified.