z^3=(-3+i)/(2+i)

Lösung

z^{3} = \frac{- 3 + i}{2 + i}

z = \frac{2 ^{\frac{2}{3}}}{2} + i \frac{2 ^{\frac{2}{3}}}{2}, z = \frac{2 ^{\frac{2}{3}} ( - 1 - 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4}, z = \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4} - i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4}

Schritte zur Lösung

z^{3} = \frac{- 3 + i}{2 + i}

Vereinfache

\frac{- 3 + i}{2 + i} : - 1 + 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 + i

∣ a ∣ = 2

ar g (a) = - \frac{π}{4} + π

z = 32 (cos (\frac{- \frac{π}{4} + π + 2 \cdot 0 π}{3}) + i sin (\frac{- \frac{π}{4} + π + 2 \cdot 0 π}{3})), z = 32 (cos (\frac{- \frac{π}{4} + π + 2 \cdot 1 π}{3}) + i sin (\frac{- \frac{π}{4} + π + 2 \cdot 1 π}{3})), z = 32 (cos (\frac{- \frac{π}{4} + π + 2 \cdot 2 π}{3}) + i sin (\frac{- \frac{π}{4} + π + 2 \cdot 2 π}{3}))

Vereinfache

32 (cos (\frac{- \frac{π}{4} + π + 2 \cdot 0 π}{3}) + i sin (\frac{- \frac{π}{4} + π + 2 \cdot 0 π}{3})) : \frac{2 ^{\frac{2}{3}}}{2} + i \frac{2 ^{\frac{2}{3}}}{2}

Vereinfache

32 (cos (\frac{- \frac{π}{4} + π + 2 \cdot 1 π}{3}) + i sin (\frac{- \frac{π}{4} + π + 2 \cdot 1 π}{3})) : \frac{2 ^{\frac{2}{3}} ( - 1 - 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4}

Vereinfache

32 (cos (\frac{- \frac{π}{4} + π + 2 \cdot 2 π}{3}) + i sin (\frac{- \frac{π}{4} + π + 2 \cdot 2 π}{3})) : \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4} - i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4}

z = \frac{2 ^{\frac{2}{3}}}{2} + i \frac{2 ^{\frac{2}{3}}}{2}, z = \frac{2 ^{\frac{2}{3}} ( - 1 - 3 )}{4} + i \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4}, z = \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4} - i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4}

so l v e f or z, A = \frac{p z ^{4}}{T}

2 x (4 x - 7) (x^{2} + 49) = 0

2 a^{4} - 11 a^{2} + 14 = 0

4 t - t^{3} = 0

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