z^3=(1-sqrt(3)i)

Lösung

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

z = 32 cos (\frac{π}{9}) - 32 i sin (\frac{π}{9}), z = 32 cos (\frac{5 π}{9}) + 32 i sin (\frac{5 π}{9}), z = 32 cos (\frac{11 π}{9}) + 32 i sin (\frac{11 π}{9})

Schritte zur Lösung

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

Entferne die Klammern:

(a) = a

z^{3} = 1 - 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 = 1 - 3 i

∣ a ∣ = 2

ar g (a) = - \frac{π}{3}

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

Vereinfache

32 (cos (\frac{- \frac{π}{3} + 2 \cdot 0 π}{3}) + i sin (\frac{- \frac{π}{3} + 2 \cdot 0 π}{3})) : 32 cos (\frac{π}{9}) - 32 i sin (\frac{π}{9})

Vereinfache

32 (cos (\frac{- \frac{π}{3} + 2 \cdot 1 π}{3}) + i sin (\frac{- \frac{π}{3} + 2 \cdot 1 π}{3})) : 32 cos (\frac{5 π}{9}) + 32 i sin (\frac{5 π}{9})

Vereinfache

32 (cos (\frac{- \frac{π}{3} + 2 \cdot 2 π}{3}) + i sin (\frac{- \frac{π}{3} + 2 \cdot 2 π}{3})) : 32 cos (\frac{11 π}{9}) + 32 i sin (\frac{11 π}{9})

z = 32 cos (\frac{π}{9}) - 32 i sin (\frac{π}{9}), z = 32 cos (\frac{5 π}{9}) + 32 i sin (\frac{5 π}{9}), z = 32 cos (\frac{11 π}{9}) + 32 i sin (\frac{11 π}{9})

roo t s x^{4} - 4 x^{3} + 8 x^{2} - 16 x + 16

x^{3} - 4 x - 8 = 0

2 x^{3} - 3 x^{2} - 32 x + 48 = 0

x^{4} + 6 x^{3} + 18 x^{2} + 30 x + 25 = 0

z^{3} - 6 z - 4 = 0