Question

How do you find all the real and complex roots of `z^3 +1 = 0`?

Answer 1

Factor then solve the remaining quadratic using the quadratic formula to find:

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

Explanation:

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

Solve `z^2-z+1 = 0` using the quadratic formula with `a=1`, `b=-1` and `c=1`:

`z = (-b+-sqrt(b^2-4ac))/(2a) = (1+-sqrt(-3))/2 = 1/2+-sqrt(3)/2i`

Hence `z^3+1 = 0` has roots `z=-1`, `z=1/2+-sqrt(3)/2i`