z^3+i=0

解

z^{3} + i = 0

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

解答ステップ

z^{3} + i = 0

i

を右側に移動します

z^{3} = - i

z^{n} = a

の場合, 解は

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

以下のため:

n = 3, a = - i

∣ a ∣ = 1

ar g (a) = - \frac{π}{2}

z = 31 (cos (\frac{- \frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{- \frac{π}{2} + 2 \cdot 0 π}{3})), z = 31 (cos (\frac{- \frac{π}{2} + 2 \cdot 1 π}{3}) + i sin (\frac{- \frac{π}{2} + 2 \cdot 1 π}{3})), z = 31 (cos (\frac{- \frac{π}{2} + 2 \cdot 2 π}{3}) + i sin (\frac{- \frac{π}{2} + 2 \cdot 2 π}{3}))

簡素化

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

簡素化

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

簡素化

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

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