nC0+nC3+⋯−(1/2) nC1+nC2+nC4+nC5+⋯2+(3/4) nC1−nC2+nC4−nC5+⋯2=

(1+ω)n=nC0+nC1ω+⋯            = nC0+nC3+⋯+ nC1+nC4+⋯−1+3i2+ nC2+nC5+⋯−1−3i2            = nC0+nC3+⋯−12 nC1+nC2+nC4+nC5…+i32 nC1−nC2+nC4−nC5+⋯Equating the modulus, we get −ω2n=1.   ( If z = x +iy then z2=x2+y2 )