Let #epsilon_1#, #epsilon_2#, #epsilon_3# be the three cube roots of unity.
DeMoivre's Theorem states that, for any integer #n# and complex #x#,
#(cosx+isinx)^n = cosnx+isinnx#
Where #i# is the imaginary unit with the property that #i^2=-1#.
Let's assume that the numbers #epsilon_k# for #k=1,2,3# are complex numbers with Polar form:
#epsilon_k = r(cosx_k+isinx_k)#
For some #x# in terms of #k#. By definition,
#epsilon_k^3=1#, #forall k in {1,2,3}#
#:. r^3(cosx_k+isinx_k)^3= 1#
#:. r^3(cos3x_k+isin3x_k)=1#
Since #epsilon_k# are roots of one, #r# is constant between them.
We already know one root is #epsilon_1 = 1#, the real root. Hence,
#{(r=1),(cos3x_1=1),(sin3x_1=0) :} <=> 3x_1=2pi#
While #x_1# is a particular value of #"arg"(epsilon_k)#, it tells us a lot about the other arguements and roots.
We know that
#{(cos(2mpi)=1),(sin(2mpi)=0) :}#
for any integer #m#.
#=> {(cos(3*(2mpi)/3)=1),(sin(3*(2mpi)/3)=0) :}#
This seemingly answers our question:
#x_k=(2kpi)/3#
For #k in {1,2,3}#. While we defined #k# to in this set, why can't it be higher than #3#? Wouldn't it mean there are infinitely many roots of unity for any #n#?
Let #k=4#. Then:
#x_k = (8pi)/3 = 2pi+(2pi)/3#
As #2pi# is a period of the trigonometric functions, this means that it just resets. This is equivalent for any #k in ZZ >3#. It's just going to repeat with a period of #3#, meaning there truly are only three roots of unity of order #3#.
Finally,
#:. epsilon_k = cos((2kpi)/3)+isin((2kpi)/3)#, #k in {1,2,3}#
is a cube root of unity.
If we do calculate these, we have
#{(epsilon_1 = 1),(epsilon_2 =-1/2+isqrt3/2),(epsilon_3=-1/2-isqrt3/2) :}#
