z^3=4+4isqrt(3)

Lösung

z^{3} = 4 + 4 i 3

z = 2 cos (\frac{π}{9}) + 2 i sin (\frac{π}{9}), z = 2 cos (\frac{7 π}{9}) + 2 i sin (\frac{7 π}{9}), z = 2 cos (\frac{13 π}{9}) + 2 i sin (\frac{13 π}{9})

Schritte zur Lösung

z^{3} = 4 + 4 i 3

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 = 4 + 4 i 3

∣ a ∣ = 8

ar g (a) = \frac{π}{3}

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

Vereinfache

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

Vereinfache

38 (cos (\frac{\frac{π}{3} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{3} + 2 \cdot 1 π}{3})) : 2 cos (\frac{7 π}{9}) + 2 i sin (\frac{7 π}{9})

Vereinfache

38 (cos (\frac{\frac{π}{3} + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{3} + 2 \cdot 2 π}{3})) : 2 cos (\frac{13 π}{9}) + 2 i sin (\frac{13 π}{9})

z = 2 cos (\frac{π}{9}) + 2 i sin (\frac{π}{9}), z = 2 cos (\frac{7 π}{9}) + 2 i sin (\frac{7 π}{9}), z = 2 cos (\frac{13 π}{9}) + 2 i sin (\frac{13 π}{9})

(5 - 2 x)^{3} = - 1

x^{4} + x^{3} - 8 x^{2} - 9 x - 9 = 0

(x - 3) (x - 1)^{2} (x^{2} + 4 x + 13) = 0

- 2 x^{3} + x^{2} + 13 x + 6 = 0

x x^{2} = 4