jmc

We have that $z^3-1=0,$ which factors as $(z-1)(z^2+z+1)=0.$ Since $\omega$ is not real, $\omega$ satisfies $${\omega }^2+\omega +1=0.$$By the quadratic formula, $$\omega =\frac{-1\pm i\sqrt{3}}{2}.$$Let $$\alpha =1+\omega =\frac{1\pm i\sqrt{3}}{2}.$$For $\alpha =\frac{1+i\sqrt{3}}{2},$ $$\begin{array}{rl}{\alpha }^2& =\frac{(1+i\sqrt{3}{)}^2}{2^2}=\frac{1+2i\sqrt{3}-3}{4}=\frac{-2+2i\sqrt{3}}{4}=\frac{-1+i\sqrt{3}}{2},\\ {\alpha }^3& =\alpha \cdot {\alpha }^2=\frac{1+i\sqrt{3}}{2}\cdot \frac{-1+i\sqrt{3}}{2}=\frac{-1^2+(i\sqrt{3}{)}^2}{4}=\frac{-1-3}{4}=-1,\\ {\alpha }^4& =\alpha \cdot {\alpha }^3=\frac{-1-i\sqrt{3}}{2},\\ {\alpha }^5& ={\alpha }^2\cdot {\alpha }^3=\frac{1-i\sqrt{3}}{2},\\ {\alpha }^6& =({\alpha }^3{)}^2=1.\end{array}$$After that, the powers of $\alpha$ repeat in a cycle of 6. The same occurs when $\alpha =\frac{1-i\sqrt{3}}{2},$ and the powers of $\frac{1-i\sqrt{3}}{2}$ achieve the same values as the powers of $\frac{1+i\sqrt{3}}{2},$ so there are $\boxed{6}$ different possible values of ${\alpha }^n.$