Observe that
|√3 + i| = √((√3)² + 1²) = √4 = 2
and
arg(√3 + i) = arctan(1/√3) = π/6
so that
√3 + i = 2 exp(i π/6)
By de Moivre's theorem, we get
(√3 + i)⁶⁰ = 2⁶⁰ exp(i π/6 • 60) = 2⁶⁰ exp(i 10π) = 2⁶⁰
