Evaluate the sum of two complex conjugates to the power of $n$ Please help me out with the following problem for I'm stuck and don't even have any idea on how to proceed.
Find all integers $n$ such that 
$$\left(\frac {-1 +i{\sqrt3} } {2}\right)^n+\left(\frac {-1 -i{\sqrt3} } {2}\right)^n=2$$

Hint:  let $\displaystyle z_{1,2}=\frac {-1 \pm i{\sqrt3} } {2}\,$, then $z_1+z_2=-1$ and $z_1z_2=1\,$, so $z_{1,2}$ are the roots of $z^2+z+1\,$ which implies that they are roots of $z^3-1=(z-1)(z^2+z+1)$ i.e $z_1$ and $z_2 = \overline{z_1}$ are the complex cube roots of unity, therefore $z_1^n = z_1^{n \bmod 3}\,$.

[ EDIT ]  expanded hint: let $\displaystyle z_{1}=\frac {-1 + i{\sqrt3} } {2}\,$, then $\displaystyle z_{2}=\frac {-1 - i{\sqrt3} } {2}=\overline{z_1}\,$, and the sum $S_n=z_1^n+z_2^n=z_1^n+\bar z_1^n=z_1^n+\overline{z_1^n}=2 \operatorname{Re}(z_1^n)$. Working out the calculations for $n=1,2,3\,$:


*

*$\displaystyle \;S_1=2 \operatorname{Re}(z_1)=-1$

*$\displaystyle \;z_1^2=\frac{(-1+i\sqrt{3})^2}{4}=\frac{(-1)^2-2 \cdot 1 \cdot i \sqrt{3} + i^2 \left(\sqrt{3}\right)^2}{4}=\frac{1-2 \sqrt{3} i-3}{4}=\frac{-1 - i \sqrt{3}}{2}\,$, therefore $S_2=2 \operatorname{Re}(z_1^2) = -1$

*$\displaystyle \;z_1^3 = z_1 \cdot z_1^2=\frac {(-1 + i{\sqrt3})(-1 - i{\sqrt3}) } {4} = \frac{(-1)^2 - i^2 \left(\sqrt{3}\right)^2}{4}=\frac{1+3}{4}=1\,$, therefore $S_3=2 \operatorname{Re}(z_1^3)=2$
Let $n=3k+m$ where $m = n \bmod 3 \in \{0,1,2\}\,$, then using the just established fact that $z_1^3=1$ it follows that  $z_1^n=z_1^{3k+m}=\left(z_1^3\right)^k \cdot z_1^m = 1 \cdot z_1^m=z_1^m\,$. Therefore, the sum $S_n=2 \operatorname{Re}(z_1^n)$ always takes one of the values $-1$ or $2\,$, depending on the remainder $n \bmod 3\,$. In particular, $S_n$ is $2$ iff $n$ is a multiple of $3\,$, and $-1$ otherwise.
