Question

How do you solve `z^4+z^2+1 = 0` ?

Answer 1

`"See explanation"`

Explanation:

`"first put " y = z^2 "."`
`"so we have"`
`y^2 + y + 1 = 0`
`"and this is a quadratic equation"`
`"the discriminant is "1^2 - 4*1 = -3"`
`"so we have no real solutions, only complex ones"`
`y_{1,2} = (-1 pm sqrt(3) i)/2`
`=> z = pm sqrt (-1 pm sqrt(3) i)/ sqrt(2)`

Answer 2

`z = +-1/2+-sqrt(3)/2i`

Explanation:

Given:

`z^4+z^2+1 = 0`

We can complete the square to find:

`0 = z^4+z^2+1`

`color(white)(0) = (z^2+1)^2-z^2`

`color(white)(0) = ((z^2+1)-z)((z^2+1)+z)`

`color(white)(0) = (z^2-z+1)(z^2+z+1)`

Then we can find the zeros of the quadratics by completing the square again:

`0 = 4(z^2-z+1)`

`color(white)(0) = 4z^2-4z+1+3`

`color(white)(0) = (2z-1)^2-(sqrt(3)i)^2`

`color(white)(0) = ((2z-1)-sqrt(3)i)((2z-1)+sqrt(3)i)`

`color(white)(0) = (2z-1-sqrt(3)i)(2z-1+sqrt(3)i)`

Hence:

`z = 1/2+-sqrt(3)/2i`

Similarly:

`0 = 4(z^2+z+1) = (2z+1-sqrt(3)i)(2z+1+sqrt(3)i)`

Hence:

`z = -1/2+-sqrt(3)/2i`

Alternatively, note that:

`(z^2-1)(z^4+z^2+1) = z^6-1`

Hence the zeros of `z^4+z^2+1` are the sixth roots of `1` apart from `+-1`.