How do you express #(1-i)^3# in #a+bi# form?
1 Answer
May 7, 2017
Explanation:
Method 1 - direct evaluation
#(1-i)^3 = (1-i)(1-i)(1-i)#
#color(white)((i-i)^3) = (1-2i+i^2)(1-i)#
#color(white)((i-i)^3) = (-2i)(1-i)#
#color(white)((i-i)^3) = -2i+2i^2#
#color(white)((i-i)^3) = -2-2i#
Method 2 - binomial expansion then simplification
#(1-i)^3 = 1^3+3(1^2)(-i)+3(1)(-i)^2+(-i)^3#
#color(white)((1-i)^3) = 1-3i-3+i#
#color(white)((1-i)^3) = -2-2i#
Method 3 - de Moivre
#(1-i)^3 = (sqrt(2)(cos(-pi/4)+i sin(-pi/4)))^3#
#color(white)((1-i)^3) = (sqrt(2))^3(cos(-(3pi)/4)+i sin(-(3pi)/4))#
#color(white)((1-i)^3) = 2sqrt(2)(-sqrt(2)/2-i sqrt(2)/2)#
#color(white)((1-i)^3) = -2-2i#