Why is $e^{5 \frac{2 \pi i}{6}}+e^{\frac{2 \pi i}{6}}=1$?
I know $e^{5 \frac{2 \pi i}{6}} = e^{-\frac{\pi i}{3}}=\overline{e^{\frac{\pi i}{3}}}=\overline{e^{\frac{2 \pi i}{6}}}$, but I still don't see why the equation is true.
Why is $e^{5 \frac{2 \pi i}{6}}+e^{\frac{2 \pi i}{6}}=1$?
I know $e^{5 \frac{2 \pi i}{6}} = e^{-\frac{\pi i}{3}}=\overline{e^{\frac{\pi i}{3}}}=\overline{e^{\frac{2 \pi i}{6}}}$, but I still don't see why the equation is true.
Copyright © 2021 JogjaFile Inc.
Hint.
Note that for any complex number $z$, $ z+\overline{z}=2\textrm{Re}(z) $. Also $\textrm{Re}(e^{i\theta})=\cos(\theta)$ for $\theta\in\mathbb{R}$.
Now observe that $\cos(2\pi/6)=\frac12$.