zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的代数 >

z^3+1-i=0

解答

z^{3} + 1 - i = 0

解答

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

求解步骤

z^{3} + 1 - i = 0

将

i

到右边

z^{3} + 1 = i

将

1

到右边

z^{3} = i - 1

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 = i - 1

∣ a ∣ = 2

ar g (a) = - \frac{π}{4} + π

z = 32 (cos (\frac{- \frac{π}{4} + π + 2 \cdot 0 π}{3}) + i sin (\frac{- \frac{π}{4} + π + 2 \cdot 0 π}{3})), z = 32 (cos (\frac{- \frac{π}{4} + π + 2 \cdot 1 π}{3}) + i sin (\frac{- \frac{π}{4} + π + 2 \cdot 1 π}{3})), z = 32 (cos (\frac{- \frac{π}{4} + π + 2 \cdot 2 π}{3}) + i sin (\frac{- \frac{π}{4} + π + 2 \cdot 2 π}{3}))

化简

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

化简

32 (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 )}{4} + i \frac{2 ^{\frac{2}{3}} ( - 1 + 3 )}{4}

化简

32 (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 )}{4} - i \frac{2 ^{\frac{2}{3}} ( 1 + 3 )}{4}

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

流行的例子

2 x^{5} + 5 x = 2215

12x^3-26x^2-12x=0

12 x^{3} - 26 x^{2} - 12 x = 0

25x^4+9x^2-16=0

25 x^{4} + 9 x^{2} - 16 = 0

x^{6} = 23

x^{6} = 2 x