jmc

From the equation $\omega +\frac{1}{\omega }=1,$ ${\omega }^2+1=\omega ,$ so $${\omega }^2-\omega +1=0.$$Then $(\omega +1)({\omega }^2-\omega +1)=0,$ which expands as ${\omega }^3+1=0.$ Hence, ${\omega }^3=-1.$

We divide into cases where $n$ is of the form $3k,$ $3k+1,$ and $3k+2.$

If $n=3k,$ then $${\omega }^n+\frac{1}{{\omega }^n}={\omega }^{3k}+\frac{1}{{\omega }^{3k}}=({\omega }^3{)}^k+\frac{1}{({\omega }^3{)}^k}=(-1{)}^k+\frac{1}{(-1{)}^k}.$$If $k$ is even, then this becomes 2, and if $k$ is odd, then this becomes $-2.$

If $n=3k+1,$ then $$\begin{array}{rl}{\omega }^n+\frac{1}{{\omega }^n}& ={\omega }^{3k+1}+\frac{1}{{\omega }^{3k+1}}=({\omega }^3{)}^k\omega +\frac{1}{({\omega }^3{)}^k\omega }\\ & =(-1{)}^k\omega +\frac{1}{(-1{)}^k\omega }\\ & =(-1{)}^k\frac{{\omega }^2+1}{\omega }\\ & =(-1{)}^k\frac{-\omega }{\omega }\\ & =(-1{)}^k.\end{array}$$This can be $1$ or $-1$.

And if $n=3k+2,$ then $$\begin{array}{rl}{\omega }^n+\frac{1}{{\omega }^n}& ={\omega }^{3k+2}+\frac{1}{{\omega }^{3k+2}}=({\omega }^3{)}^k{\omega }^2+\frac{1}{({\omega }^3{)}^k{\omega }^2}\\ & =(-1{)}^k{\omega }^2+\frac{1}{(-1{)}^k{\omega }^2}\\ & =(-1{)}^k\frac{{\omega }^4+1}{{\omega }^2}\\ & =(-1{)}^k\frac{-\omega +1}{{\omega }^2}\\ & =(-1{)}^k\frac{-{\omega }^2}{{\omega }^2}\\ & =-(-1{)}^k.\end{array}$$This can be $1$ or $-1$.

Hence, the possible values of ${\omega }^n+\frac{1}{{\omega }^n}$ are $\boxed{-2,-1,1,2}.$