z^3=-8i

Lösung

z^{3} = - 8 i

z = 3 - i, z = 2 i, z = - 3 - i

Schritte zur Lösung

z^{3} = - 8 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 = - 8 i

∣ a ∣ = 8

ar g (a) = - \frac{π}{2}

z = 38 (cos (\frac{- \frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{- \frac{π}{2} + 2 \cdot 0 π}{3})), z = 38 (cos (\frac{- \frac{π}{2} + 2 \cdot 1 π}{3}) + i sin (\frac{- \frac{π}{2} + 2 \cdot 1 π}{3})), z = 38 (cos (\frac{- \frac{π}{2} + 2 \cdot 2 π}{3}) + i sin (\frac{- \frac{π}{2} + 2 \cdot 2 π}{3}))

Vereinfache

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

Vereinfache

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

Vereinfache

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

z = 3 - i, z = 2 i, z = - 3 - i

(d^{3} + 1)^{3} (d^{2} + d + 1) = 0

x^{6} + 11 x^{4} + 19 x^{2} - 91 = 0

x^{2} (x^{2} - 10) + 9 = (x + 1) \cdot (x - 1)

x^{4} - 20 x^{2} + 36 = 0

3 + 3 x^{3} = 9