How do you use DeMoivre's theorem to simplify #(-1/2-sqrt3/2i)^3#?
1 Answer
Sep 29, 2016
Explanation:
de Moivre's theorem tells us that:
#(cos theta + i sin theta)^n = cos n theta + i sin n theta#
So we find:
#(-1/2-sqrt(3)/2i)^3 = (cos (-(2pi)/3) + i sin (-(2pi)/3))^3#
#color(white)((-1/2-sqrt(3)/2i)^3) = cos (-2pi) + i sin (-2pi)#
#color(white)((-1/2-sqrt(3)/2i)^3) = 1 + i * 0#
#color(white)((-1/2-sqrt(3)/2i)^3) = 1#
