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=16sqrt(3)+16i

解答

z^{5} = 163 + 16 i

解答

z = 2 cos (\frac{π}{30}) + 2 i sin (\frac{π}{30}), z = 2 cos (\frac{13 π}{30}) + 2 i sin (\frac{13 π}{30}), z = - 3 + i, z = 2 cos (\frac{37 π}{30}) + 2 i sin (\frac{37 π}{30}), z = 2 cos (\frac{49 π}{30}) + 2 i sin (\frac{49 π}{30})

求解步骤

z^{5} = 163 + 16 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 = 163 + 16 i

∣ a ∣ = 32

ar g (a) = \frac{π}{6}

z = 532 (cos (\frac{\frac{π}{6} + 2 \cdot 0 π}{5}) + i sin (\frac{\frac{π}{6} + 2 \cdot 0 π}{5})), z = 532 (cos (\frac{\frac{π}{6} + 2 \cdot 1 π}{5}) + i sin (\frac{\frac{π}{6} + 2 \cdot 1 π}{5})), z = 532 (cos (\frac{\frac{π}{6} + 2 \cdot 2 π}{5}) + i sin (\frac{\frac{π}{6} + 2 \cdot 2 π}{5})), z = 532 (cos (\frac{\frac{π}{6} + 2 \cdot 3 π}{5}) + i sin (\frac{\frac{π}{6} + 2 \cdot 3 π}{5})), z = 532 (cos (\frac{\frac{π}{6} + 2 \cdot 4 π}{5}) + i sin (\frac{\frac{π}{6} + 2 \cdot 4 π}{5}))

化简

532 (cos (\frac{\frac{π}{6} + 2 \cdot 0 π}{5}) + i sin (\frac{\frac{π}{6} + 2 \cdot 0 π}{5})) : 2 cos (\frac{π}{30}) + 2 i sin (\frac{π}{30})

化简

532 (cos (\frac{\frac{π}{6} + 2 \cdot 1 π}{5}) + i sin (\frac{\frac{π}{6} + 2 \cdot 1 π}{5})) : 2 cos (\frac{13 π}{30}) + 2 i sin (\frac{13 π}{30})

化简

532 (cos (\frac{\frac{π}{6} + 2 \cdot 2 π}{5}) + i sin (\frac{\frac{π}{6} + 2 \cdot 2 π}{5})) : - 3 + i

化简

532 (cos (\frac{\frac{π}{6} + 2 \cdot 3 π}{5}) + i sin (\frac{\frac{π}{6} + 2 \cdot 3 π}{5})) : 2 cos (\frac{37 π}{30}) + 2 i sin (\frac{37 π}{30})

化简

532 (cos (\frac{\frac{π}{6} + 2 \cdot 4 π}{5}) + i sin (\frac{\frac{π}{6} + 2 \cdot 4 π}{5})) : 2 cos (\frac{49 π}{30}) + 2 i sin (\frac{49 π}{30})

z = 2 cos (\frac{π}{30}) + 2 i sin (\frac{π}{30}), z = 2 cos (\frac{13 π}{30}) + 2 i sin (\frac{13 π}{30}), z = - 3 + i, z = 2 cos (\frac{37 π}{30}) + 2 i sin (\frac{37 π}{30}), z = 2 cos (\frac{49 π}{30}) + 2 i sin (\frac{49 π}{30})

流行的例子

15 = k^{3}

12y^2-13y^3=2y^2-2y

12 y^{2} - 13 y^{3} = 2 y^{2} - 2 y

8k-48-8k^8=k+15k

8 k - 48 - 8 k^{8} = k + 15 k

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

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

2 x^{3} - 6 x^{2} - 8 x = 0