2022-03-25

Given a complex number $\omega =\mathrm{cos}\left(\frac{2\pi }{5}\right)+i\mathrm{sin}\left(\frac{2\pi }{5}\right)$

Wilson Rivas

Step 1
You seem to have written a perfectly valid proof. Just note that the general case is also true (and pretty important as well). Let $n\in \mathbb{N}$, and consider the complex nth root of 1, i.e., the complex number
${\omega }_{n}={e}^{\frac{2\pi t}{n}}=\mathrm{cos}\left(\frac{2\pi }{n}\right)+i\mathrm{sin}\left(\frac{2\pi }{n}\right)$
that satisfies the equation
${z}^{n}=1$
Of course there are n solutions to this equation, we take ${\omega }_{n}$ as the one with the minimal argument (except from 1). The calculation of ${\omega }_{n}^{n}$ is as you have done - using de Moivre formula,
${\omega }_{n}^{n}={\left({e}^{\frac{2\pi t}{n}}\right)}^{n}={e}^{2\pi }=1$
By the sum of a geometric sequence (which is also true for complex numbers), we know that
$1+z+{z}^{2}+\cdots +{z}^{n-1}=\frac{{z}^{n}-1}{z-1}$
and therefore plugging in $z={\omega }_{n}$ yields zero in the RHS, and the sum you were requested to calculate in the LHS.

Do you have a similar question?