z^3= 1/8 i

Lösung

z^{3} = \frac{1}{8} i

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

Schritte zur Lösung

z^{3} = \frac{1}{8} 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 = \frac{1}{8} i

∣ a ∣ = \frac{1}{8}

ar g (a) = \frac{π}{2}

z = 3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 0 π}{3})), z = 3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 1 π}{3})), z = 3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}))

Vereinfache

3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 0 π}{3})) : \frac{3}{4} + i \frac{1}{4}

Vereinfache

3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 1 π}{3})) : - \frac{3}{4} + i \frac{1}{4}

Vereinfache

3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 2 π}{3})) : - i \frac{1}{2}

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

49 x^{4} - 25 x^{2} = 0

x^{(4)} - 4 x^{(2)} + 3 = 0

r^{4} + 9 r^{2} = 0

x (x - 2)^{2} - 3 = 7

2 x^{3} - 15 x^{2} + 27 x - 10 = 0