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{13 π}{18}) + 32 i sin (\frac{13 π}{18}), z = 32 cos (\frac{25 π}{18}) + 32 i sin (\frac{25 π}{18})

Schritte zur Lösung

z^{3} = (3 + i)

Entferne die Klammern:

(a) = a

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{13 π}{18}) + 32 i sin (\frac{13 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{25 π}{18}) + 32 i sin (\frac{25 π}{18})

z = 32 cos (\frac{π}{18}) + 32 i sin (\frac{π}{18}), z = 32 cos (\frac{13 π}{18}) + 32 i sin (\frac{13 π}{18}), z = 32 cos (\frac{25 π}{18}) + 32 i sin (\frac{25 π}{18})

m^{3} = 1

roo t s r^{3} - 3 r^{2} + 3 r - 1

x^{3} - 3 x^{2} - 40 x = 0

(x^{2} + 25)^{2} = 0

so l v e f or y, x^{3} + y^{3} = 0