Wie drückt man # (1-i) ^ 3 # in # a + bi # -Form aus?

Methode 1 - direkte Auswertung

#(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#

#color(white)()#
Methode 2 - Binomialerweiterung, dann Vereinfachung

#(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#

#color(white)()#
Methode 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#