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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 $z \in \mathbb{C}$?

Quanto
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Novice
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    why not get the real root and then get the roots of quadratic equation hence formed? Probably because of this: https://i.sstatic.net/cSi2y.png – Anindya Prithvi Nov 21 '20 at 16:56
  • In the complex, a number has three cubic roots and Cardano's solutions are multiple, but only three combinations are valid. –  Nov 21 '20 at 17:02
  • @AnindyaPrithvi so it's not reasonable to calculate it. Is there another way? Or is it simply sufficient to say that "we ought to do it this way, but it's very complicated in evaluating"? – Novice Nov 21 '20 at 17:08
  • And you'll need Galois theory to understand the deep reason why the roots have an expression with $+,\times, \sqrt[n]{.}$ and complex roots of unity – reuns Nov 21 '20 at 17:10
  • See https://math.stackexchange.com/questions/2157643/how-can-i-solve-the-equation-x3-x-1-0 OR https://math.stackexchange.com/questions/2283046/what-are-the-roots-of-the-equation-8x3-6x1-0 OR https://math.stackexchange.com/questions/2203364/solve-the-following-equation-x3-3x-sqrtx2 – lab bhattacharjee Nov 21 '20 at 17:14
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    @AnindyaPrithvi Given the (real) cube roots $A,B$ such that the real root is $A+B,$ the other two roots are $A\omega + B \omega^2, $ and $A \omega^2 + B \omega,$ where $\omega = \frac{-1 + i \sqrt 3}{2}$ is a cube root of $1$ Note that $\omega^2 = \bar{\omega},$ so that these two roots are complex conjugates – Will Jagy Nov 21 '20 at 17:45
  • @Novice Buraian offers an answer without using Cardano although the complexity to solve it is a little higher. But....to find the real root you'll need either of the methods proposed. – Anindya Prithvi Nov 21 '20 at 18:25
  • @reuns Could you tell me more about this theory? – Novice Nov 21 '20 at 18:53

3 Answers3

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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 $$\omega^3 = 1 $$

Will Jagy
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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 $$x_{1,2}=\frac12\left( -a\pm i \sqrt{1-\frac3a}\right)$$

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

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)

Clemens Bartholdy
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