Steve S. answered 03/21/14
Tutor
5
(3)
Tutoring in Precalculus, Trig, and Differential Calculus
z^5 = (√3+i)^5
tan(θ) = 1/√(3) => θ = 30° from Special Triangle
r = √((√3)^2+1^2) = 2
z = 2 e^(i 30°)
z^5 = (2 e^(i 30°))^5 = 32 e^(i 150°)
z^5 =32(cos(150°) + i sin(150°) )
z^5 = 32(-√(3)/2 + 1/2 i) = -16√(3) + 16 i