z^3+1=e^{pi/2 i}

Lösung

z^{3} + 1 = e^{\frac{π}{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} + 1 = e^{\frac{π}{2} i}

Verschiebe

1

auf die rechte Seite

z^{3} = i - 1

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 = i - 1

∣ 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}

x^{4} - 5 x^{2} - 50 = 0

t = - 3.9 (t)^{2} - (t)^{4}

so l v e f or x, y = q^{2} + (a x)^{3}

q^{3} - 10 q^{2} - 5808 = 0

1.05 = (1 + \frac{x}{12})^{12}