z^3= i/8

解

z^{3} = \frac{i}{8}

z = \frac{3}{4} + i \frac{1}{4}, z = - \frac{3}{4} + i \frac{1}{4}, z = - i \frac{1}{2}

解答ステップ

z^{3} = \frac{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 = \frac{i}{8}

∣ a ∣ = \frac{1}{8}

ar g (a) = \frac{π}{2}

z = 3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 0 π}{3})), z = 3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 1 π}{3})), z = 3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}))

簡素化

3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 0 π}{3})) : \frac{3}{4} + i \frac{1}{4}

簡素化

3 \frac{1}{8} (cos (\frac{\frac{π}{2} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 1 π}{3})) : - \frac{3}{4} + i \frac{1}{4}

簡素化

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

z = \frac{3}{4} + i \frac{1}{4}, z = - \frac{3}{4} + i \frac{1}{4}, z = - i \frac{1}{2}