Express √3 + i in polar form:
|√3 + i| = √((√3)² + 1²) = √4 = 2
arg(√3 + i) = arctan(1/√3) = π/6
Then
√3 + i = 2 (cos(π/6) + i sin(π/6))
By DeMoivre's theorem,
(√3 + i)³ = (2 (cos(π/6) + i sin(π/6)))³
… = 2³ (cos(3 • π/6) + i sin(3 • π/6))
… = 8 (cos(π/2) + i sin(π/2))
… = 8i
