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Beliebt Algebra >

z^5=1-i

Lösung

z^{5} = 1 - i

Lösung

z = \frac{2 ^{\frac{3}{5}} 4 + 10 + 2 5}{4} - i \frac{2 ^{\frac{3}{5}} 4 - 10 + 2 5}{4}, z = \frac{2 ^{\frac{3}{5}} 4 - 10 - 2 5}{4} + i \frac{2 ^{\frac{3}{5}} 4 + 10 - 2 5}{4}, z = - \frac{2 ^{\frac{3}{5}}}{2} + i \frac{2 ^{\frac{3}{5}}}{2}, z = - \frac{2 ^{\frac{3}{5}} 4 + 10 - 2 5}{4} - i \frac{2 ^{\frac{3}{5}} 4 - 10 - 2 5}{4}, z = \frac{2 ^{\frac{3}{5}} 4 - 10 + 2 5}{4} - i \frac{2 ^{\frac{3}{5}} 4 + 10 + 2 5}{4}

Schritte zur Lösung

z^{5} = 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 = 5, a = 1 - i

∣ a ∣ = 2

ar g (a) = - \frac{π}{4}

z = 52 (cos (\frac{- \frac{π}{4} + 2 \cdot 0 π}{5}) + i sin (\frac{- \frac{π}{4} + 2 \cdot 0 π}{5})), z = 52 (cos (\frac{- \frac{π}{4} + 2 \cdot 1 π}{5}) + i sin (\frac{- \frac{π}{4} + 2 \cdot 1 π}{5})), z = 52 (cos (\frac{- \frac{π}{4} + 2 \cdot 2 π}{5}) + i sin (\frac{- \frac{π}{4} + 2 \cdot 2 π}{5})), z = 52 (cos (\frac{- \frac{π}{4} + 2 \cdot 3 π}{5}) + i sin (\frac{- \frac{π}{4} + 2 \cdot 3 π}{5})), z = 52 (cos (\frac{- \frac{π}{4} + 2 \cdot 4 π}{5}) + i sin (\frac{- \frac{π}{4} + 2 \cdot 4 π}{5}))

Vereinfache

52 (cos (\frac{- \frac{π}{4} + 2 \cdot 0 π}{5}) + i sin (\frac{- \frac{π}{4} + 2 \cdot 0 π}{5})) : \frac{2 ^{\frac{3}{5}} 4 + 10 + 2 5}{4} - i \frac{2 ^{\frac{3}{5}} 4 - 10 + 2 5}{4}

Vereinfache

52 (cos (\frac{- \frac{π}{4} + 2 \cdot 1 π}{5}) + i sin (\frac{- \frac{π}{4} + 2 \cdot 1 π}{5})) : \frac{2 ^{\frac{3}{5}} 4 - 10 - 2 5}{4} + i \frac{2 ^{\frac{3}{5}} 4 + 10 - 2 5}{4}

Vereinfache

52 (cos (\frac{- \frac{π}{4} + 2 \cdot 2 π}{5}) + i sin (\frac{- \frac{π}{4} + 2 \cdot 2 π}{5})) : - \frac{2 ^{\frac{3}{5}}}{2} + i \frac{2 ^{\frac{3}{5}}}{2}

Vereinfache

52 (cos (\frac{- \frac{π}{4} + 2 \cdot 3 π}{5}) + i sin (\frac{- \frac{π}{4} + 2 \cdot 3 π}{5})) : - \frac{2 ^{\frac{3}{5}} 4 + 10 - 2 5}{4} - i \frac{2 ^{\frac{3}{5}} 4 - 10 - 2 5}{4}

Vereinfache

52 (cos (\frac{- \frac{π}{4} + 2 \cdot 4 π}{5}) + i sin (\frac{- \frac{π}{4} + 2 \cdot 4 π}{5})) : \frac{2 ^{\frac{3}{5}} 4 - 10 + 2 5}{4} - i \frac{2 ^{\frac{3}{5}} 4 + 10 + 2 5}{4}

z = \frac{2 ^{\frac{3}{5}} 4 + 10 + 2 5}{4} - i \frac{2 ^{\frac{3}{5}} 4 - 10 + 2 5}{4}, z = \frac{2 ^{\frac{3}{5}} 4 - 10 - 2 5}{4} + i \frac{2 ^{\frac{3}{5}} 4 + 10 - 2 5}{4}, z = - \frac{2 ^{\frac{3}{5}}}{2} + i \frac{2 ^{\frac{3}{5}}}{2}, z = - \frac{2 ^{\frac{3}{5}} 4 + 10 - 2 5}{4} - i \frac{2 ^{\frac{3}{5}} 4 - 10 - 2 5}{4}, z = \frac{2 ^{\frac{3}{5}} 4 - 10 + 2 5}{4} - i \frac{2 ^{\frac{3}{5}} 4 + 10 + 2 5}{4}

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