It is required to determine the possible values for $1+\omega^n+\omega^{2n}$ where $n$ is a natural number and $\omega = -\frac{1}{2}-\frac{\sqrt{3}}{2}i$. The following identities are given; $\omega^{4}=\omega$ and $\omega^{2}+\omega+1=0$
My attempt:
Consider the expression for even $n$, $n=2k$ $$ 1+\omega^{n}+\omega^{2n}=1+\omega^{2k}+\omega^{4k}$$ $$ 1+\omega^{n}+\omega^{2n}=1+\omega^{k}+\omega^{2k}$$
Now for odd $n$, let $n=2m+1$ $$1+\omega^{n}+\omega^{2n}=1+\omega^{2m+1}+\omega^{4m+2}$$ $$1+\omega^{n}+\omega^{2n}=1+\omega^{m+2}+\omega^{2m+1}$$
I'm not sure where to go from here. I have just be testing some values for $n$ on my caluclator and it seems that the value of the expression is independent of whehter $n$ is even or odd so there must be a different method for solving this.
Hint:
As $w^3=1,$ the length of the period of $w^n$ is $3$
For any integer $m,n$ can be one of $3m+1,3m+2, 3m$
Case$\#1:$
If $n=3m+1\equiv1\pmod3, 2n=2(3m+1)\equiv2\pmod3$
Case$\#2:$
If $n=3m+2\equiv2\pmod3, 2n=\cdots\equiv1\pmod3$
Case$\#3:$
What if $n=3m?$