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 $\omega =\frac{-1+i\sqrt{3}}{2}.$ Then $|a\omega +b{|}^2=1.$ Also, $$\begin{array}{rl}|a\omega +b{|}^2& ={\left|a\cdot \frac{-1+i\sqrt{3}}{2}+b\right|}^2\\ & ={\left|-\frac{1}{2}a+b+\frac{\sqrt{3}}{2}ai\right|}^2\\ & ={\left(-\frac{1}{2}a+b\right)}^2+{\left(\frac{\sqrt{3}}{2}a\right)}^2\\ & =\frac{1}{4}{a}^2-ab+b^2+\frac{3}{4}{a}^2\\ & =a^2-ab+b^2.\end{array}$$Thus, we want to find integers $a$ and $b$ so that $a^2-ab+b^2=1.$ Note that we derived this equation from the equation $${\left(-\frac{1}{2}a+b\right)}^2+{\left(\frac{\sqrt{3}}{2}a\right)}^2=1.$$Then $${\left(\frac{\sqrt{3}}{2}a\right)}^2\le 1,$$so $$\left|\frac{\sqrt{3}}{2}a\right|\le 1.$$Then $$|a|\le \frac{2}{\sqrt{3}}<2,$$so the only possible values of $a$ are $-1,$ $0,$ and $1.$

If $a=-1,$ then the equation $a^2-ab+b^2=1$ becomes $$b^2+b=0.$$The solutions are $b=-1$ and $b=0.$

If $a=0,$ then the equation $a^2-ab+b^2=1$ becomes $$b^2=1.$$The solutions are $b=-1$ and $b=1.$

If $a=1,$ then the equation $a^2-ab+b^2=1$ becomes $$b^2-b=0.$$The solutions are $b=0$ and $b=1.$

Therefore, the possible pairs $(a,b)$ are $(-1,-1),$ $(-1,0),$ $(0,-1),$ $(0,1),$ $(1,0),$ and $(1,1).$

We went with the value $\omega =\frac{-1+i\sqrt{3}}{2}.$ The other possible value of $\omega$ is $$\frac{-1-i\sqrt{3}}{2}=1-\omega ,$$so any number that can be represented in the form $a\omega +b$ can also be represented in this form with the other value of $\omega .$ (In other words, it doesn't which value of $\omega$ we use.)

Hence, there are $\boxed{6}$ possible pairs $(a,b).$

Note that the complex numbers of the form $a\omega +b$ form a triangular lattice in the complex plane. This makes it clear why there are six complex numbers that have absolute value 1.