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受欢迎的代数 >

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

解答

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

解答

z = 32 cos (\frac{2 π}{9}) + 32 i sin (\frac{2 π}{9}), z = 32 cos (\frac{8 π}{9}) + 32 i sin (\frac{8 π}{9}), z = 32 cos (\frac{14 π}{9}) + 32 i sin (\frac{14 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{8 π}{9}) + 32 i sin (\frac{8 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{14 π}{9}) + 32 i sin (\frac{14 π}{9})

z = 32 cos (\frac{2 π}{9}) + 32 i sin (\frac{2 π}{9}), z = 32 cos (\frac{8 π}{9}) + 32 i sin (\frac{8 π}{9}), z = 32 cos (\frac{14 π}{9}) + 32 i sin (\frac{14 π}{9})

流行的例子

x^3+6x^2-x-30=0

x^{3} + 6 x^{2} - x - 30 = 0

q^{3} - q^{2} - 30 q = 0

solvefor y,sqrt(x)sqrt(y)=sqrt(xy)

so l v e f or y, x y = x y

-2/(3x^{1/3)}=0

- \frac{2}{3 x ^{\frac{1}{3}}} = 0

3-sqrt(2x+1)=x-2

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