How to solve $z^3+z+1=0$?

Question

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 $z \in \mathbb{C}$?

Answer 1

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 $$\omega^3 = 1 $$

Answer 2

$$ z = re^{i \theta}$$

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)

Answer 3

The Cardano formula gives the real root $$a=\sqrt[3]{\frac{\sqrt{93}-9}{18}} - \sqrt[3]{\frac{\sqrt{93}+9}{18}} <0$$ Then, factorize the equation as $$x^3+x+1=(x-a)(x^2+ax-\frac1a)=0$$ where the quadratic factor gives the pair of complex roots $$x_{1,2}=\frac12\left( -a\pm i \sqrt{1-\frac3a}\right)$$