Question

How do you use De Moivre's theorem to solve the equation below?

(`sqrt(3)-i)^5`

Answer 1

`sqrt3 - i = 2 (sqrt(3)/2 - 1/2 i) = 2 (cos(-pi/6) + i sin(-pi/6))`

Therefore,
`(sqrt3 - i)^5 = 2^5 (cos(-pi/6) + i sin(-pi/6))^5 = 2^5 (cos(-(5pi)/6) + i sin(-(5pi)/6)) = 32 (-sqrt3/2 - 1/2i) = -16sqrt3-16i`