März 5, 2020

von Loutitia

Wie drückt man ${(1 - i)}^{3}$ $(1-i) ^ 3$ in $a + b i$ $a + bi$ -Form aus?

Antworten:

${(1 - i)}^{3} = - 2 - 2 i$ $(1-i)^3 = -2-2i$

Erläuterung:

Methode 1 - direkte Auswertung

${(1 - i)}^{3} = (1 - i) (1 - i) (1 - i)$ $(1-i)^3 = (1-i)(1-i)(1-i)$

${(i - i)}^{3} = (1 - 2 i + i^{2}) (1 - i)$ $color(white)((i-i)^3) = (1-2i+i^2)(1-i)$

${(i - i)}^{3} = (- 2 i) (1 - i)$ $color(white)((i-i)^3) = (-2i)(1-i)$

${(i - i)}^{3} = - 2 i + 2 i^{2}$ $color(white)((i-i)^3) = -2i+2i^2$

${(i - i)}^{3} = - 2 - 2 i$ $color(white)((i-i)^3) = -2-2i$

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

${(1 - i)}^{3} = 1^{3} + 3 (1^{2}) (- i) + 3 (1) {(- i)}^{2} + {(- i)}^{3}$ $(1-i)^3 = 1^3+3(1^2)(-i)+3(1)(-i)^2+(-i)^3$

${(1 - i)}^{3} = 1 - 3 i - 3 + i$ $color(white)((1-i)^3) = 1-3i-3+i$

${(1 - i)}^{3} = - 2 - 2 i$ $color(white)((1-i)^3) = -2-2i$

$color(white)()$
Methode 3 - de Moivre

${(1 - i)}^{3} = {(\sqrt{2} (cos (- \frac{π}{4}) + i sin (- \frac{π}{4})))}^{3}$ $(1-i)^3 = (sqrt(2)(cos(-pi/4)+i sin(-pi/4)))^3$

${(1 - i)}^{3} = {(\sqrt{2})}^{3} (cos (- \frac{3 π}{4}) + i sin (- \frac{3 π}{4}))$ $color(white)((1-i)^3) = (sqrt(2))^3(cos(-(3pi)/4)+i sin(-(3pi)/4))$

${(1 - i)}^{3} = 2 \sqrt{2} (- \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2})$ $color(white)((1-i)^3) = 2sqrt(2)(-sqrt(2)/2-i sqrt(2)/2)$

${(1 - i)}^{3} = - 2 - 2 i$ $color(white)((1-i)^3) = -2-2i$