How to solve $z^3+z+1=0$? I have troubles with solving an equation
$$z^3+z+1=0$$
in complex numbers. I know that in real numbers we can use Cardano's formula and it's clear to me. But what about the situation when $z \in \mathbb{C}$?
Given the (real) cube roots $A,B$ in Cardano's mthod, so that the real root of your cubic is $A+B,$ the other two roots are complex conjugates, given by
$$  A \omega + B \omega^2 $$
$$  A \omega^2 + B \omega $$
where
$$ \omega = \frac{-1 + i\sqrt 3}{2}$$
and
$$ \omega^2 = \bar{\omega}  $$
and $$\omega^3 = 1 $$
$$ z = re^{i \theta}$$
Or,
$$ r^3 \left[  e^{i3 \theta} \right] + re^{i \theta} + 1 =0$$
$$ r^3 \left[ \sum_{k=even}^{2}\binom{3}{k} \cos^k \theta (i)^{3-k} \sin^{3-k} \theta+ \sum_{k=odd}^{3} \binom{3}{k} \cos^k \theta (i)^{3-k} \sin^{3-k} \theta \right] + r e^{i \theta} + 1 =0$$
Equating real and imaginary part to zero,
$$ r^3 \left[ \sum_{k=odd}^3 \binom{3}{k} (i)^{3-k} \sin^{3-k} \theta \cos^k \theta  \right] + r \cos \theta + 1 = 0$$
Or,
$$ r^3 \left[ \sum_{k=0}^1 \binom{3}{2k+1}  (-1)^{1+k} \sin^{2-2k} \theta \cos^{2k+1} \theta  \right] + r \cos \theta + 1 = 0 \tag{1}$$
And,
$$  r^3 \left[\sum_{k=even}^2 \binom{3}{k} \cos^k \theta (i)^{3-k} \sin^{3-k} \theta  \right] + r i \sin \theta = 0$$
Or,
$$  r^2 \left[\sum_{k=0}^1 \binom{3}{2k} \cos^{2k} \theta (-1)^{k+1}  \sin^{3-2k} \theta  \right] +  \sin \theta = 0 \tag{2}$$
I'll leave you with solving the trignometric equation of (1) an d (2)