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

$\omega^3=e^{3i\theta}=1\implies3i\theta=\ln1=0+2ni\pi\implies\theta=\dfrac{2n\pi}3$

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.


$n=1\implies\omega=\dfrac{2i\pi}3=\cos\dfrac{2\pi}3+i\sin\dfrac{2\pi}3$
$n=2\implies\omega^2=e^{4i\pi/3}=\cos\dfrac{4\pi}3+i\sin\dfrac{4\pi}3$

Recall that $\cos(2\pi-x)=\cos x$, so that $\cos\dfrac{4\pi}3=\cos\dfrac{2\pi}3$. On the other hand, $\sin(2\pi-x)=-\sin x$, so $\sin\dfrac{4\pi}3=-\sin\dfrac{2\pi}3$. 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

$\mathrm{Re}(\omega)=\cos\dfrac{2\pi}3=-\dfrac12$

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 $\omega\neq1$, we must have $1+\omega+\omega^2=0$.