z^3=-1-i

Lösung

z^{3} = - 1 - i

z = \frac{2 ^{\frac{2}{3}}}{2} - i \frac{2 ^{\frac{2}{3}}}{2}, z = \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4}, z = - \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4} - i \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4}

Schritte zur Lösung

z^{3} = - 1 - i

Für

z^{n} = a

sind die Lösungen

z_{k} = n ∣ a ∣ (cos (\frac{ar g ( a ) + 2 kπ}{n}) + i sin (\frac{ar g ( a ) + 2 kπ}{n})),

k = 0, 1, \dots, n - 1

Für

n = 3, a = - 1 - i

∣ a ∣ = 2

ar g (a) = \frac{π}{4} - π

z = 32 (cos (\frac{\frac{π}{4} - π + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{4} - π + 2 \cdot 0 π}{3})), z = 32 (cos (\frac{\frac{π}{4} - π + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{4} - π + 2 \cdot 1 π}{3})), z = 32 (cos (\frac{\frac{π}{4} - π + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{4} - π + 2 \cdot 2 π}{3}))

Vereinfache

32 (cos (\frac{\frac{π}{4} - π + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{4} - π + 2 \cdot 0 π}{3})) : \frac{2 ^{\frac{2}{3}}}{2} - i \frac{2 ^{\frac{2}{3}}}{2}

Vereinfache

32 (cos (\frac{\frac{π}{4} - π + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{4} - π + 2 \cdot 1 π}{3})) : \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4}

Vereinfache

32 (cos (\frac{\frac{π}{4} - π + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{4} - π + 2 \cdot 2 π}{3})) : - \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4} - i \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4}

z = \frac{2 ^{\frac{2}{3}}}{2} - i \frac{2 ^{\frac{2}{3}}}{2}, z = \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4}, z = - \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4} - i \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4}

x^{2} (x + 3) = 0

so l v e f or y, (y - x)^{3} = C x

x^{3} + 3 x^{2} - 36 x - 108 = 0

17 + x^{3} = x^{2} + 17

4 x^{3} = 32