(-1+i 3^1/2)^12 using moivre's theorem
so we have
(-1 + i√3)^12
let u = -1 + i√3
= √10(cos(2π/3) + i sin(2π/3))
u^12 = √10^12(cos 8π + i sin 8π)
= 10^6( 1 + i(0))
= 10^6 or 1,000,000
(-1 + i√3)^12
let u = -1 + i√3
= √10(cos(2π/3) + i sin(2π/3))
u^12 = √10^12(cos 8π + i sin 8π)
= 10^6( 1 + i(0))
= 10^6 or 1,000,000