Wie drückt man (1-i) ^ 3 in a + bi -Form aus?
Antworten:
(1-i)^3 = -2-2i
Erläuterung:
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