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.