>>830
It's wrong. \eqn{\frac{-1+i\sqrt{3}}{2}} is an a complex cube root of 1 (as can be easily verified by multiplication) \[(\frac{-1+i\sqrt{3}}{2})^2019 = (\frac{-1+i\sqrt{3}}{2})^{3\times673} = ((\frac{-1+i\sqrt{3}}{2})^3)^673 = 1^673 =1\]. Thus the answer is 3