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Beliebt Algebra >

z^4-i=0

Lösung

z^{4} - i = 0

Lösung

z = \frac{2 + 2}{2} + i \frac{2 - 2}{2}, z = - \frac{2 - 2}{2} + i \frac{2 + 2}{2}, z = - \frac{2 + 2}{2} - i \frac{2 - 2}{2}, z = \frac{2 - 2}{2} - i \frac{2 + 2}{2}

Schritte zur Lösung

z^{4} - i = 0

Verschiebe

i

auf die rechte Seite

z^{4} = 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 = 4, a = i

∣ a ∣ = 1

ar g (a) = \frac{π}{2}

z = 41 (cos (\frac{\frac{π}{2} + 2 \cdot 0 π}{4}) + i sin (\frac{\frac{π}{2} + 2 \cdot 0 π}{4})), z = 41 (cos (\frac{\frac{π}{2} + 2 \cdot 1 π}{4}) + i sin (\frac{\frac{π}{2} + 2 \cdot 1 π}{4})), z = 41 (cos (\frac{\frac{π}{2} + 2 \cdot 2 π}{4}) + i sin (\frac{\frac{π}{2} + 2 \cdot 2 π}{4})), z = 41 (cos (\frac{\frac{π}{2} + 2 \cdot 3 π}{4}) + i sin (\frac{\frac{π}{2} + 2 \cdot 3 π}{4}))

Vereinfache

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

Vereinfache

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

Vereinfache

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

Vereinfache

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

z = \frac{2 + 2}{2} + i \frac{2 - 2}{2}, z = - \frac{2 - 2}{2} + i \frac{2 + 2}{2}, z = - \frac{2 + 2}{2} - i \frac{2 - 2}{2}, z = \frac{2 - 2}{2} - i \frac{2 + 2}{2}

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