$e^{5 \frac{2 \pi i}{6}}+e^{\frac{2 \pi i}{6}}=1$

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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.

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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 $\cos(2\pi/6)=\frac12$.

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Because $(\mathbb{Z} / 6 \mathbb{Z},+) \to (E_6,\cdot)$ with $E_6 = \{ e^{\frac{2 \pi i k}{5}} : k = 0,1,\dots,5 \} $ is a group isomorphism, the statement follows because $5+1=6=0$ in $\mathbb{Z} / 6 \mathbb{Z}$.