z^3+8i+8=0

解

z^{3} + 8 i + 8 = 0

z = 2^{\frac{2}{3}} - 2^{\frac{2}{3}} i, z = \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{2} + i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{2}, z = \frac{2 ^{\frac{2}{3}} ( - 1 - 3 )}{2} + i \frac{2 ^{\frac{2}{3}} ( 1 - 3 )}{2}

解答ステップ

z^{3} + 8 i + 8 = 0

8 i

を右側に移動します

z^{3} + 8 = - 8 i

8

を右側に移動します

z^{3} = - 8 i - 8

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 - 8

∣ a ∣ = 82

ar g (a) = \frac{π}{4} - π

z = 382 (cos (\frac{\frac{π}{4} - π + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{4} - π + 2 \cdot 0 π}{3})), z = 382 (cos (\frac{\frac{π}{4} - π + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{4} - π + 2 \cdot 1 π}{3})), z = 382 (cos (\frac{\frac{π}{4} - π + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{4} - π + 2 \cdot 2 π}{3}))

簡素化

382 (cos (\frac{\frac{π}{4} - π + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{4} - π + 2 \cdot 0 π}{3})) : 2^{\frac{2}{3}} - 2^{\frac{2}{3}} i

簡素化

382 (cos (\frac{\frac{π}{4} - π + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{4} - π + 2 \cdot 1 π}{3})) : \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{2} + i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{2}

簡素化

382 (cos (\frac{\frac{π}{4} - π + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{4} - π + 2 \cdot 2 π}{3})) : \frac{2 ^{\frac{2}{3}} ( - 1 - 3 )}{2} + i \frac{2 ^{\frac{2}{3}} ( 1 - 3 )}{2}

z = 2^{\frac{2}{3}} - 2^{\frac{2}{3}} i, z = \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{2} + i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{2}, z = \frac{2 ^{\frac{2}{3}} ( - 1 - 3 )}{2} + i \frac{2 ^{\frac{2}{3}} ( 1 - 3 )}{2}