Answer :
You could just use the binomial theorem and expand it out then simplify it
BUT ... the chances of messing up are pretty good ... so I think it's way easier to use De Moivre's Theorem to do it
Drawing it on an argand diagram -√3 + i is in the 2nd quadrant
mod (-√3 + i) = √(3 + 1) = 2
arg (-√3 + i) = π - arctan (1 / √3) = π - (π/6) = 5π/6
so (-√3 + i)^6 = {2 [cos (5π/6) + i sin (5π/6)]}^6
= 2^6 [cos (5π) + i sin (5π)] ... [using De Moivre's theorem [r (cos θ + i sin θ]^n = r^n [cos (nθ) + i sin (nθ)]
= 64 [-1 + 0 i]
= -64
BUT ... the chances of messing up are pretty good ... so I think it's way easier to use De Moivre's Theorem to do it
Drawing it on an argand diagram -√3 + i is in the 2nd quadrant
mod (-√3 + i) = √(3 + 1) = 2
arg (-√3 + i) = π - arctan (1 / √3) = π - (π/6) = 5π/6
so (-√3 + i)^6 = {2 [cos (5π/6) + i sin (5π/6)]}^6
= 2^6 [cos (5π) + i sin (5π)] ... [using De Moivre's theorem [r (cos θ + i sin θ]^n = r^n [cos (nθ) + i sin (nθ)]
= 64 [-1 + 0 i]
= -64
