jmc

Let the pure imaginary root be $ki,$ where $k$ is real, so $$-k^2+ki+\omega =0.$$Thus, $\omega =k^2-ki.$ Then $\overline{\omega }=k^2+ki,$ so $$1=|\omega {|}^2=\omega \overline{\omega }=(k^2-ki)(k^2+ki)=k^4+k^2.$$Then $k^4+k^2-1=0.$ By the quadratic formula, $$k^2=\frac{-1\pm \sqrt{5}}{2}.$$Since $k$ is real, $$k^2=\frac{-1+\sqrt{5}}{2}.$$Therefore, $$\omega +\overline{\omega }=k^2-ki+k^2+ki=2k^2=\boxed{\sqrt{5}-1}.$$