Question

The sum of cube roots of unity (1+w+w^2) is

Answer 1

Let
`\omega=e^(i\theta)`. By DeMoivre's theorem, we have


`\omega^3=e^(3i\theta)=1\implies3i\theta=\ln1=0+2ni\pi\implies\theta=\frac{2n\pi}3`

where
`n` is any integer, but we can capture all the cube roots of unity by selecting
`n=0,1,2`, and in particular when
`n=0` we get 1.



`n=1\implies\omega=\frac{2i\pi}3=\cos\frac{2\pi}3+i\sin\frac{2\pi}3`

`n=2\implies\omega^2=e^(4i\pi/3)=\cos\frac{4\pi}3+i\sin\frac{4\pi}3`

Recall that
`\cos(2\pi-x)=\cos x`, so that
`\cos\frac{4\pi}3=\cos\frac{2\pi}3`. On the other hand,
`\sin(2\pi-x)=-\sin x`, so
`\sin\frac{4\pi}3=-\sin\frac{2\pi}3`. So in
`\omega+\omega^2`, we find the imaginary parts cancel, leaving us with


`1+\omega+\omega^2=1+2\mathrm{Re}(\omega)`

for which we have


`\mathrm{Re}(\omega)=\cos\frac{2\pi}3=-\frac12`

so the sum is
`1+\omega+\omega^2=0`.

Another way of computing the sum is to notice that


`(1-\omega)(1+\omega+\omega^2)=1-\omega^3`

for any
`\omega`. We assumed that
`\omega^3=1`, so the RHS is 0. Then for
`\omega\\eq1`, we must have
`1+\omega+\omega^2=0`.