Prove that $\left(\frac {-1+\sqrt {-3}}{2}\right)^n + \left(\frac {-1-\sqrt {-3}}{2}\right)^n$ follows this pattern 
Prove that:
  $$\left(\dfrac {-1+\sqrt {-3}}{2}\right)^n + \left(\dfrac {-1-\sqrt {-3}}{2}\right)^n=\begin{cases}
2, & \textrm { if } n \textrm { is a multiple of 3},\\
-1, & \textrm { if } n \textrm { is any other integer}
\end{cases}$$

My Attempt:
$$\dfrac {-1+\sqrt {-3}}{2}=\dfrac {-1+i\sqrt {3}}{2}$$ which is a complex cube root of unity. 
Let $\omega = \dfrac {-1+i\sqrt {3}}{2}$.
Similarly, $\omega^2=\dfrac {-1-i\sqrt {3}}{2}$
 A: You have an answer from @uswer8734617's comment. I would do it slightly differently. 
Note that $\overline{z^n}=\overline{z}^n$ for any complex number $z$. So 
$$
\omega^n+\overline{\omega}^n=\omega^n+\overline{\omega^n}=2Re(\omega^n)
$$
Now all you need to do is calculating $\omega^n$. 
But you have already done $n=1$, $n=2$. Calculate one more, you would fine the pattern: $\omega^3=\omega\cdot\omega^2=1$. 
A: Alternative solution: the two complex numbers $\frac{-1\pm\sqrt{3}i}{2}$ are the roots of $x^2+x+1=0.$ Consider a recursive formula
$$a_{n+2}+a_{n+1}+a_n=0,$$
with $a_0=2$ and $a_1=-1$ agreeing with $\left(\frac{-1+\sqrt{3}i}{2}\right)^n+\left(\frac{-1-\sqrt{3}i}{2}\right)^n$ at $n=0$ and $1$. Then from the characteristic root method, $a_n=\left(\frac{-1+\sqrt{3}i}{2}\right)^n+\left(\frac{-1-\sqrt{3}i}{2}\right)^n$. We can then prove the periodic-$3$ pattern using mathematical induction. 
A: $\{1,\omega,\omega^2\}$, as a group with respect to $\cdot$, is isomorphic to $\left(\mathbb{Z}/(3\mathbb{Z}),+\right)$.
It follows that for any $n\in\mathbb{Z}$ the power sum
$$ 1^n +\omega^n +\omega^{2n} $$
equals $3$ if $n\equiv 0\pmod{3}$ and zero otherwise, i.e.
$$ \frac{1^n+\omega^n+\omega^{2n}}{3} = \mathbb{1}_{\equiv 0\pmod{3}}(n)$$
which is the driving principle of the discrete Fourier transform / multisection technique for computing objects like 
$$\sum_{\substack{0\leq k\leq n\\ k\equiv 1\!\!\pmod{\!\!3}}}\!\!\!\!\binom{n}{k}\,2^k,$$
for instance.
A: Using How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i*\sin(\varphi)$?,
$$-\dfrac12+\dfrac{\sqrt3i}2=\cos120^\circ+i\sin120^\circ=e^{i120^\circ}$$
$$\left(-\dfrac12+\dfrac{\sqrt3i}2\right)^n=\cos(120^\circ n)+i\sin(120^\circ n)$$
Similarly, 
$$-\dfrac12-\dfrac{\sqrt3i}2=\cos(-120^\circ)+i(-\sin120^\circ)=e^{-i120^\circ}$$
$$\left(-\dfrac12-\dfrac{\sqrt3i}2\right)^n=\cos(-120^\circ n)+i\sin(-120^\circ n)=\cos(120^\circ n)-i\sin(120^\circ n)$$
$$\left(-\dfrac12+\dfrac{\sqrt3i}2\right)^n+\left(-\dfrac12-\dfrac{\sqrt3i}2\right)^n=2\cos(120^\circ n)$$
Now $n$ can be written as one of $3m,3m+1,3m+2$ where $m$ is any integer.
