z^3=64i

Lösung

z^{3} = 64 i

z = 23 + 2 i, z = - 23 + 2 i, z = - 4 i

Schritte zur Lösung

z^{3} = 64 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 = 64 i

∣ a ∣ = 64

ar g (a) = \frac{π}{2}

z = 364 (cos (\frac{\frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 0 π}{3})), z = 364 (cos (\frac{\frac{π}{2} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 1 π}{3})), z = 364 (cos (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}))

Vereinfache

364 (cos (\frac{\frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 0 π}{3})) : 23 + 2 i

Vereinfache

364 (cos (\frac{\frac{π}{2} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 1 π}{3})) : - 23 + 2 i

Vereinfache

364 (cos (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 2 π}{3})) : - 4 i

z = 23 + 2 i, z = - 23 + 2 i, z = - 4 i

- 2 x^{3} - 9 x^{2} + 108 x - 10 = 0

3 + 5 x (2 x^{3})^{2} = 48

6 x^{2} - 11 x^{4} - 2 = 0

10 x^{4} - 3 x^{2} + 1 = 0

x^{3} + 7 x^{2} - 4 x - 28 = 0