${\left[\left({ }^{\mathrm{n}}{\mathrm{C}}_0{+}^{\mathrm{n}}{\mathrm{C}}_3+\cdots \right)-(1/2)\left({ }^{\mathrm{n}}{\mathrm{C}}_1{+}^{\mathrm{n}}{\mathrm{C}}_2{+}^{\mathrm{n}}{\mathrm{C}}_4{+}^{\mathrm{n}}{\mathrm{C}}_5+\cdots \right)\right]}^2+(3/4){\left({ }^n{C}_1{-}^n{C}_2{+}^n{C}_4{-}^n{C}_5+\cdots \right)}^2=$

$$\begin{gathered} (1+\mathrm{\omega }{)}^{\mathrm{n}}{=}^{\mathrm{n}}{\mathrm{C}}_0{+}^{\mathrm{n}}{\mathrm{C}}_1\mathrm{\omega }+\cdots \\ =\left({ }^{\mathrm{n}}{\mathrm{C}}_0{+}^{\mathrm{n}}{\mathrm{C}}_3+\cdots \right)+\left({ }^{\mathrm{n}}{\mathrm{C}}_1{+}^{\mathrm{n}}{\mathrm{C}}_4+\cdots \right)\left(\frac{-1+\sqrt{3}\mathrm{i}}{2}\right)+\left({ }^{\mathrm{n}}{\mathrm{C}}_2{+}^{\mathrm{n}}{\mathrm{C}}_5+\cdots \right)\left(\frac{-1-\sqrt{3}\mathrm{i}}{2}\right) \\ =\left({ }^{\mathrm{n}}{\mathrm{C}}_0{+}^{\mathrm{n}}{\mathrm{C}}_3+\cdots \right)-\frac{1}{2}\left({ }^{\mathrm{n}}{\mathrm{C}}_1{+}^{\mathrm{n}}{\mathrm{C}}_2{+}^{\mathrm{n}}{\mathrm{C}}_4{+}^{\mathrm{n}}{\mathrm{C}}_5\dots \right)+\frac{\mathrm{i}\sqrt{3}}{2}\left({ }^{\mathrm{n}}{\mathrm{C}}_1{-}^{\mathrm{n}}{\mathrm{C}}_2{+}^{\mathrm{n}}{\mathrm{C}}_4{-}^{\mathrm{n}}{\mathrm{C}}_5+\cdots \right) \end{gathered}$$

Equating the modulus, we get $\left|{\left(-{\mathrm{\omega }}^2\right)}^{\mathrm{n}}\right|=1$.   ( If z = x +iy then ${\left|\mathrm{z}\right|}^2={\mathrm{x}}^2+{\mathrm{y}}^2$ )