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, $\omega^2=\dfrac {-1-i\sqrt {3}}{2}$

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.
