Proof that $(e^{i\pi/3}+e^{-i\pi/3}) = 1$ I tried to  prove that 
$$\left(e^{i\pi/3}+e^{-i\pi/3}\right) = 1$$
I have geometric argument: 
$e^{i\pi/3}$ is the complex number of imaginary unit and the number $e^{- i\pi/3}$ also from imaginary unit, but above of real axis. From parallelogram law we have 1. 
This is also of complex conjugate. 
Of-course, the second argument is dividing this equation by 2 and argument of 
$$\frac{\left(e^{i\pi/3}+e^{-i\pi/3}\right)}{2} = \frac{1}{2}$$
$$\cos(\pi/3)= \frac{1}{2} $$
Do you have third argument of proof this? 
 A: If $x=\exp\left(\frac{\pi i}3\right)+\exp\left(-\frac{\pi i}3\right)$, then$$x^3=-1-1+3x=3x-2.$$But the equation $x^3=3x-2$ only has $2$ solutions: $1$ and $-2$. And since it is clear that both $\exp\left(\frac{\pi i}3\right)$ and $\exp\left(-\frac{\pi i}3\right)$ have absolute value equal to $1$ but one of them is $1$, it is clear that their sum is not $2$. So, $x=1$.
A: Note that \begin{align}(e^{i\pi/3}+e^{-i\pi/3})^3&=e^{i\pi}+3e^{i\pi/3}+3e^{-i\pi/3}+e^{-i\pi}\\&=-1+3(e^{i\pi/3}+e^{-i\pi/3})-1\end{align}$$\implies(e^{i\pi/3}+e^{-i\pi/3})^3-3(e^{i\pi/3}+e^{-i\pi/3})+2=0.$$ This cubic $x^3-3x+2=(x+2)(x-1)^2$ has two roots: $$e^{i\pi/3}+e^{-i\pi/3}=-2,1.$$ The former cannot be attained as $\pi/3$ is not a multiple of $\pi$.
A: It is $$e^{i\pi/3}=\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)$$
A: By symmetry for the centroid point, we have that
$$e^{i\pi/3}+e^{-i\pi/3}+e^{i\pi}=0 \implies e^{i\pi/3}+e^{-i\pi/3}=1$$
A: If $x^3=-1\iff x^3+1=0$
$x^3=e^{i(2n+1)\pi}$ where $n$ is any integer
$x_n=e^{i(2n+1)\pi/3}$ where $n=-1,0,1$
But using Vieta's formula, $$x_{-1}+x_0+x_1=-\dfrac03$$
