z^3=1+i

Lösung

z^{3} = 1 + i

z = \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}

Schritte zur Lösung

z^{3} = 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}

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

Vereinfache

32 (cos (\frac{\frac{π}{4} + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{4} + 2 \cdot 0 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} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{4} + 2 \cdot 1 π}{3})) : - \frac{2 ^{\frac{2}{3}}}{2} + i \frac{2 ^{\frac{2}{3}}}{2}

Vereinfache

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

x^{3} = 36 x

25 x^{4} - 241 x^{2} + 144 = 0

w^{3} = - i

(x - 1)^{2} (x^{2} + 1) = 0

(x + 2)^{4} = x^{4} + 16