z^3=64i

解

z^{3} = 64 i

z = 23 + 2 i, z = - 23 + 2 i, z = - 4 i

解答ステップ

z^{3} = 64 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 = 64 i

∣ a ∣ = 64

ar g (a) = \frac{π}{2}

z = 364 (cos (\frac{\frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 0 π}{3})), z = 364 (cos (\frac{\frac{π}{2} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 1 π}{3})), z = 364 (cos (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}))

簡素化

364 (cos (\frac{\frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 0 π}{3})) : 23 + 2 i

簡素化

364 (cos (\frac{\frac{π}{2} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 1 π}{3})) : - 23 + 2 i

簡素化

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

z = 23 + 2 i, z = - 23 + 2 i, z = - 4 i