z^3=sqrt(3)-i

Lösung

z^{3} = 3 - i

z = 32 cos (\frac{π}{18}) - 32 i sin (\frac{π}{18}), z = 32 cos (\frac{11 π}{18}) + 32 i sin (\frac{11 π}{18}), z = 32 cos (\frac{23 π}{18}) + 32 i sin (\frac{23 π}{18})

Schritte zur Lösung

z^{3} = 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 = 3 - i

∣ a ∣ = 2

ar g (a) = - \frac{π}{6}

z = 32 (cos (\frac{- \frac{π}{6} + 2 \cdot 0 π}{3}) + i sin (\frac{- \frac{π}{6} + 2 \cdot 0 π}{3})), z = 32 (cos (\frac{- \frac{π}{6} + 2 \cdot 1 π}{3}) + i sin (\frac{- \frac{π}{6} + 2 \cdot 1 π}{3})), z = 32 (cos (\frac{- \frac{π}{6} + 2 \cdot 2 π}{3}) + i sin (\frac{- \frac{π}{6} + 2 \cdot 2 π}{3}))

Vereinfache

32 (cos (\frac{- \frac{π}{6} + 2 \cdot 0 π}{3}) + i sin (\frac{- \frac{π}{6} + 2 \cdot 0 π}{3})) : 32 cos (\frac{π}{18}) - 32 i sin (\frac{π}{18})

Vereinfache

32 (cos (\frac{- \frac{π}{6} + 2 \cdot 1 π}{3}) + i sin (\frac{- \frac{π}{6} + 2 \cdot 1 π}{3})) : 32 cos (\frac{11 π}{18}) + 32 i sin (\frac{11 π}{18})

Vereinfache

32 (cos (\frac{- \frac{π}{6} + 2 \cdot 2 π}{3}) + i sin (\frac{- \frac{π}{6} + 2 \cdot 2 π}{3})) : 32 cos (\frac{23 π}{18}) + 32 i sin (\frac{23 π}{18})

z = 32 cos (\frac{π}{18}) - 32 i sin (\frac{π}{18}), z = 32 cos (\frac{11 π}{18}) + 32 i sin (\frac{11 π}{18}), z = 32 cos (\frac{23 π}{18}) + 32 i sin (\frac{23 π}{18})

5 X^{3} = - 45 X

2 x^{3} + 5 c - 3 = 0

3 x^{3} + 7 x^{2} - 33 x - 77 = 0

9 x^{4} + 49 = 58 x^{2}

x^{3} - 125 x^{2} + 3842 x - 16129 = 0