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人気のある代数 >

z^5+z=0

解

z^{5} + z = 0

解

z = 0, z = - \frac{2}{2} + i \frac{2}{2}, z = - \frac{2}{2} - i \frac{2}{2}, z = \frac{2}{2} + i \frac{2}{2}, z = \frac{2}{2} - i \frac{2}{2}

解答ステップ

z^{5} + z = 0

因数

z^{5} + z : z (z^{2} + 2 z + 1) (z^{2} - 2 z + 1)

z (z^{2} + 2 z + 1) (z^{2} - 2 z + 1) = 0

零因子の原則を使用:

ab = 0

ならば

a = 0

または

b = 0

z = 0 or z^{2} + 2 z + 1 = 0 or z^{2} - 2 z + 1 = 0

解く

z^{2} + 2 z + 1 = 0 : z = - \frac{2}{2} + i \frac{2}{2}, z = - \frac{2}{2} - i \frac{2}{2}

解く

z^{2} - 2 z + 1 = 0 : z = \frac{2}{2} + i \frac{2}{2}, z = \frac{2}{2} - i \frac{2}{2}

解答は

z = 0, z = - \frac{2}{2} + i \frac{2}{2}, z = - \frac{2}{2} - i \frac{2}{2}, z = \frac{2}{2} + i \frac{2}{2}, z = \frac{2}{2} - i \frac{2}{2}

グラフ

人気の例

5x^3+5102x+2015=0

5 x^{3} + 5102 x + 2015 = 0

5 x - 6 = 2 x^{3} - 3 x^{2}

(2+b)(11b-2b^2)+189=0

(2 + b) (11 b - 2 b^{2}) + 189 = 0

solvefor x,x^4y^3-2xy^4=xy^3

so l v e f or x, x^{4} y^{3} - 2 x y^{4} = x y^{3}

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