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

解

z^{3} = (3 + i)

z = 32 cos (\frac{π}{18}) + 32 i sin (\frac{π}{18}), z = 32 cos (\frac{13 π}{18}) + 32 i sin (\frac{13 π}{18}), z = 32 cos (\frac{25 π}{18}) + 32 i sin (\frac{25 π}{18})

解答ステップ

z^{3} = (3 + i)

括弧を削除する:

(a) = a

z^{3} = 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 = 3 + i

∣ a ∣ = 2

ar g (a) = \frac{π}{6}

z = 32 (cos (\frac{\frac{π}{6} + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{6} + 2 \cdot 0 π}{3})), z = 32 (cos (\frac{\frac{π}{6} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{6} + 2 \cdot 1 π}{3})), z = 32 (cos (\frac{\frac{π}{6} + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{6} + 2 \cdot 2 π}{3}))

簡素化

32 (cos (\frac{\frac{π}{6} + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{6} + 2 \cdot 0 π}{3})) : 32 cos (\frac{π}{18}) + 32 i sin (\frac{π}{18})

簡素化

32 (cos (\frac{\frac{π}{6} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{6} + 2 \cdot 1 π}{3})) : 32 cos (\frac{13 π}{18}) + 32 i sin (\frac{13 π}{18})

簡素化

32 (cos (\frac{\frac{π}{6} + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{6} + 2 \cdot 2 π}{3})) : 32 cos (\frac{25 π}{18}) + 32 i sin (\frac{25 π}{18})

z = 32 cos (\frac{π}{18}) + 32 i sin (\frac{π}{18}), z = 32 cos (\frac{13 π}{18}) + 32 i sin (\frac{13 π}{18}), z = 32 cos (\frac{25 π}{18}) + 32 i sin (\frac{25 π}{18})