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^5=i

解答

z^{5} = i

解答

z = \frac{2 5 + 5}{4} + i \frac{2 3 - 5}{4}, z = i, z = - \frac{2 5 + 5}{4} + i \frac{2 3 - 5}{4}, z = - \frac{10 - 2 5}{4} + i \frac{- 1 - 5}{4}, z = \frac{10 - 2 5}{4} + i \frac{- 1 - 5}{4}

求解步骤

z^{5} = 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 = 5, a = i

∣ a ∣ = 1

ar g (a) = \frac{π}{2}

z = 51 (cos (\frac{\frac{π}{2} + 2 \cdot 0 π}{5}) + i sin (\frac{\frac{π}{2} + 2 \cdot 0 π}{5})), z = 51 (cos (\frac{\frac{π}{2} + 2 \cdot 1 π}{5}) + i sin (\frac{\frac{π}{2} + 2 \cdot 1 π}{5})), z = 51 (cos (\frac{\frac{π}{2} + 2 \cdot 2 π}{5}) + i sin (\frac{\frac{π}{2} + 2 \cdot 2 π}{5})), z = 51 (cos (\frac{\frac{π}{2} + 2 \cdot 3 π}{5}) + i sin (\frac{\frac{π}{2} + 2 \cdot 3 π}{5})), z = 51 (cos (\frac{\frac{π}{2} + 2 \cdot 4 π}{5}) + i sin (\frac{\frac{π}{2} + 2 \cdot 4 π}{5}))

化简

51 (cos (\frac{\frac{π}{2} + 2 \cdot 0 π}{5}) + i sin (\frac{\frac{π}{2} + 2 \cdot 0 π}{5})) : \frac{2 5 + 5}{4} + i \frac{2 3 - 5}{4}

化简

51 (cos (\frac{\frac{π}{2} + 2 \cdot 1 π}{5}) + i sin (\frac{\frac{π}{2} + 2 \cdot 1 π}{5})) : i

化简

51 (cos (\frac{\frac{π}{2} + 2 \cdot 2 π}{5}) + i sin (\frac{\frac{π}{2} + 2 \cdot 2 π}{5})) : - \frac{2 5 + 5}{4} + i \frac{2 3 - 5}{4}

化简

51 (cos (\frac{\frac{π}{2} + 2 \cdot 3 π}{5}) + i sin (\frac{\frac{π}{2} + 2 \cdot 3 π}{5})) : - \frac{10 - 2 5}{4} + i \frac{- 1 - 5}{4}

化简

51 (cos (\frac{\frac{π}{2} + 2 \cdot 4 π}{5}) + i sin (\frac{\frac{π}{2} + 2 \cdot 4 π}{5})) : \frac{10 - 2 5}{4} + i \frac{- 1 - 5}{4}

z = \frac{2 5 + 5}{4} + i \frac{2 3 - 5}{4}, z = i, z = - \frac{2 5 + 5}{4} + i \frac{2 3 - 5}{4}, z = - \frac{10 - 2 5}{4} + i \frac{- 1 - 5}{4}, z = \frac{10 - 2 5}{4} + i \frac{- 1 - 5}{4}

流行的例子

2 x^{3} = 16

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

x^{4} - i = 0

4 x = x^{3}

x^{3} - 9 x^{2} = 0