iz^3+8=0

解

i z^{3} + 8 = 0

z = 3 + i, z = - 3 + i, z = - 2 i

解答ステップ

i z^{3} + 8 = 0

8

を右側に移動します

i z^{3} = - 8

以下で両辺を割る

i

z^{3} = 8 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 = 8 i

∣ a ∣ = 8

ar g (a) = \frac{π}{2}

z = 38 (cos (\frac{\frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 0 π}{3})), z = 38 (cos (\frac{\frac{π}{2} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 1 π}{3})), z = 38 (cos (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}))

簡素化

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

簡素化

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

簡素化

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

z = 3 + i, z = - 3 + i, z = - 2 i