Use DeMoivre's Theorem to find the three cube roots of 1?

1 Answer
Jun 22, 2018

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) :}