z^3=-i

Lösung

z^{3} = - i

z = \frac{3}{2} - i \frac{1}{2}, z = i, z = - \frac{3}{2} - i \frac{1}{2}

Schritte zur Lösung

z^{3} = - 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 = - i

∣ a ∣ = 1

ar g (a) = - \frac{π}{2}

z = 31 (cos (\frac{- \frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{- \frac{π}{2} + 2 \cdot 0 π}{3})), z = 31 (cos (\frac{- \frac{π}{2} + 2 \cdot 1 π}{3}) + i sin (\frac{- \frac{π}{2} + 2 \cdot 1 π}{3})), z = 31 (cos (\frac{- \frac{π}{2} + 2 \cdot 2 π}{3}) + i sin (\frac{- \frac{π}{2} + 2 \cdot 2 π}{3}))

Vereinfache

31 (cos (\frac{- \frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{- \frac{π}{2} + 2 \cdot 0 π}{3})) : \frac{3}{2} - i \frac{1}{2}

Vereinfache

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

Vereinfache

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

z = \frac{3}{2} - i \frac{1}{2}, z = i, z = - \frac{3}{2} - i \frac{1}{2}

x^{3} - 5 x^{2} - 6 x - 8 = 0

(x - 4)^{4} + (x - 6)^{4} = 34

x (x + 1)^{2} (x + 2) = 72

4 x^{3} - 6 x^{2} = 0

(x + 1) \cdot (x + 2)^{2} \cdot (x + 4) - 2 x^{2} = 0