z^3=(-1+sqrt(3)i)

解

z^{3} = (- 1 + 3 i)

z = 32 cos (\frac{2 π}{9}) + 32 i sin (\frac{2 π}{9}), z = 32 cos (\frac{8 π}{9}) + 32 i sin (\frac{8 π}{9}), z = 32 cos (\frac{14 π}{9}) + 32 i sin (\frac{14 π}{9})

解答ステップ

z^{3} = (- 1 + 3 i)

括弧を削除する:

(a) = a

z^{3} = - 1 + 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 = - 1 + 3 i

∣ a ∣ = 2

ar g (a) = - \frac{π}{3} + π

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

簡素化

32 (cos (\frac{- \frac{π}{3} + π + 2 \cdot 0 π}{3}) + i sin (\frac{- \frac{π}{3} + π + 2 \cdot 0 π}{3})) : 32 cos (\frac{2 π}{9}) + 32 i sin (\frac{2 π}{9})

簡素化

32 (cos (\frac{- \frac{π}{3} + π + 2 \cdot 1 π}{3}) + i sin (\frac{- \frac{π}{3} + π + 2 \cdot 1 π}{3})) : 32 cos (\frac{8 π}{9}) + 32 i sin (\frac{8 π}{9})

簡素化

32 (cos (\frac{- \frac{π}{3} + π + 2 \cdot 2 π}{3}) + i sin (\frac{- \frac{π}{3} + π + 2 \cdot 2 π}{3})) : 32 cos (\frac{14 π}{9}) + 32 i sin (\frac{14 π}{9})

z = 32 cos (\frac{2 π}{9}) + 32 i sin (\frac{2 π}{9}), z = 32 cos (\frac{8 π}{9}) + 32 i sin (\frac{8 π}{9}), z = 32 cos (\frac{14 π}{9}) + 32 i sin (\frac{14 π}{9})