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受欢迎的代数 >

iz^3+8=0

解答

i z^{3} + 8 = 0

解答

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

求解步骤

i z^{3} + 8 = 0

将

8

到右边

i z^{3} = - 8

两边除以

i

z^{3} = 8 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 = 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}))

化简

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

化简

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

化简

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

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