z^3=(-1-sqrt(3)i)

解答

z^{3} = (- 1 - 3 i)

z = 32 cos (\frac{2 π}{9}) - 32 i sin (\frac{2 π}{9}), z = 32 cos (\frac{4 π}{9}) + 32 i sin (\frac{4 π}{9}), z = 32 cos (\frac{10 π}{9}) + 32 i sin (\frac{10 π}{9})

求解步骤

z^{3} = (- 1 - 3 i)

去除括号:

(a) = a

z^{3} = - 1 - 3 i

For

z^{n} = a

the solutions are

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

对于

n = 3, a = - 1 - 3 i

∣ a ∣ = 2

ar g (a) = \frac{π}{3} - π

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

化简

32 (cos (\frac{\frac{π}{3} - π + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{3} - π + 2 \cdot 0 π}{3})) : 32 cos (\frac{2 π}{9}) - 32 i sin (\frac{2 π}{9})

化简

32 (cos (\frac{\frac{π}{3} - π + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{3} - π + 2 \cdot 1 π}{3})) : 32 cos (\frac{4 π}{9}) + 32 i sin (\frac{4 π}{9})

化简

32 (cos (\frac{\frac{π}{3} - π + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{3} - π + 2 \cdot 2 π}{3})) : 32 cos (\frac{10 π}{9}) + 32 i sin (\frac{10 π}{9})

z = 32 cos (\frac{2 π}{9}) - 32 i sin (\frac{2 π}{9}), z = 32 cos (\frac{4 π}{9}) + 32 i sin (\frac{4 π}{9}), z = 32 cos (\frac{10 π}{9}) + 32 i sin (\frac{10 π}{9})

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