What are the solutions to the equation z^5+z=0 ?

The solutions to the equation z^5+z=0 are

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z^5+z=0

z^{5} + z = 0

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}

Solution steps

z^{5} + z = 0

Factor

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

Solve

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

The solutions are

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}

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

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

(3 x - 2)^{5} = 24

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

x, x^{4} y^{3} - 2 x y^{4} = x y^{3}

What are the solutions to the equation z^5+z=0 ?
The solutions to the equation z^5+z=0 are
Find the zeros of z^5+z=0
The zeros of z^5+z=0 are