De Moivre's theorem can be summarized by
#(e^(ix))^n = (e^(i nx))#
Using Euler's theorem (#e^(ix) = cos(x) + i sin(x)#), we get
#[cos(x) + isin(x)]^n = cos(nx) + i sin(nx)#
which is the normal way to write the theorem.
We can use this to find the cube roots of unity. Let #n = 3#. We want the above to equal 1, i.e.
#cos(3x) + i sin(3x) = 1 = [cos(x) + i sin(x)]^3 = z^3#
We want to find #z# such that this is true.
It's clear that #3x# has to equal #0, 2pi, or 4pi# because that's when the sine term is eliminated and the cosine term gives us 1.
These three therefore have
# x in {0, (2pi)/3, (4pi)/3}#
Obviously, we can also use #6pi, 8pi, ...#, but these divided by 3 give us the same angles, just adding a #2pi#.
With these angles, we calculate #e^(ix)#:
#e^(i cdot 0) = 1 # which we totally expect: #1^3 = 1#
#e^(i cdot 2pi / 3) = cos((2pi)/3) + i sin((2pi)/3) = -1/2 + i sqrt(3)/2#
#e^(i cdot 4pi / 3) = cos((4pi)/3) + i sin((4pi)/3) = -1/2 - i sqrt(3)/2 #
The last two are complex conjugates, a necessary attribute of the roots of unity (For odd powers, we have the number 1 and a bunch of conjugate pairs. For even powers, we have #pm 1# and a bunch of conjugate pairs.)
