(a+ib)^2=i

Lösung

(a + ib)^{2} = i

(a = \frac{1}{2}, a = - \frac{1}{2}, b = \frac{1}{2} b = - \frac{1}{2})

Schritte zur Lösung

(a + ib)^{2} = i

Schreibe

(a + ib)^{2}

um:

(a^{2} - b^{2}) + 2 iab

(a^{2} - b^{2}) + 2 iab = i

Schreibe

i

in der Standard komplexen Form um:

0 + i

(a^{2} - b^{2}) + 2 iab = 0 + i

Komplexe Zahlen können nur gleich sein wenn ihre realen und imaginären Teile gleich sind.Schreibe als Gleichungssystem um

[a^{2} - b^{2} = 0 2 ab = 1]

[a^{2} - b^{2} = 0 2 ab = 1] : (a = \frac{1}{2}, a = - \frac{1}{2}, b = \frac{1}{2} b = - \frac{1}{2})

(a = \frac{1}{2}, a = - \frac{1}{2}, b = \frac{1}{2} b = - \frac{1}{2})

x - 2 y i = - 1 + 5 i

\frac{2 + i}{z} = \frac{2 + 2 i}{z + i}

x + i y = 3 i - 4

(11 - 3 x) + (7 + 4 y) i = - 17 + i

a + bi + \frac{1}{a + bi} = 0