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I unable to resolve the following equation (I don't have any idea how even start resolving this equation): $$8z^3-12z^2+6z-1-i=0$$

M.Naszko
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  • $(8)z^3-(12)z^2+(6)z+(-1-i)z^0=0$. This is just a cubic. Solve it like you would any other cubic. The only challenge is to recognize and keep track of that you might need both the real and imaginary components written together for each of the coefficients and that here the $-1$ and the $-i$ were both parts of the constant part of the polynomial. – JMoravitz Nov 04 '22 at 12:06
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  • You will have heard of the "quadratic formula"... you know, the $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ thing? Well, there's a version for cubics and quartics as well. (No, there is not one for quintics or higher). Now... granted... not that many people bother learning or memorizing the cubic formula, it is long and tedious (the quartic is even worse), and just knowing it exists at all is more than enough most of the time. The fact you are talking about complex numbers is irrelevant. It is the same formula. – JMoravitz Nov 04 '22 at 12:10
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    There is one very simple root – Claude Leibovici Nov 04 '22 at 12:13
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    Does this answer your question? Is there anything like “cubic formula”? – JMoravitz Nov 04 '22 at 12:15
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    @ClaudeLeibovici I would not expect any of these roots to be spotted by inspection by a beginner. Which root are you suggesting is so "simple"? $\frac{1}{2}-\frac{i}{2}$? If you could spot that, then you are quite impressive at this... but I for one struggle trying to evaluate the polynomial with that value in my head quickly, much less plucking that out of thin air ahead of time. – JMoravitz Nov 04 '22 at 12:18
  • @JMoravitz. This is the "advantage" of being blind. I memorize almost everything (not the quartic). Cheers :-) – Claude Leibovici Nov 04 '22 at 12:22
  • @Claude Leibovici : Waouh ! $\frac12(1-i)$ is not what is called a "very simple root", even in France, n'est-ce pas ? :) :) – Stéphane Jaouen Nov 04 '22 at 12:54
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    @StéphaneJaouen. In Pau, it is ! Back to serious, since $\pm i$ is not a root, neither $\pm 1\pm i$, what do you want to try ? Cheers and nice to meet you. – Claude Leibovici Nov 04 '22 at 13:05
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    @StéphaneJaouen You might try the rational root theorem in the Gaussian integers. It suggests Claude's root as a possible answer. – B. Goddard Nov 04 '22 at 13:42

2 Answers2

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Let $w=2z$. Then your equation is

$$w^3-3w^2+3w-1 = i.$$

So

$$(w-1)^3 = i.$$

So $w-1$ is a cube root of $i$. So the three roots are $z=(t+1)/2$ where $t$ is a cube root of $i$.

B. Goddard
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So, $C:=\frac12(1-i)$ is a very very simple root. Then, $$8z^3-12z^2+6z-1-i=0$$

$$\iff$$

$$(z-C)(4z^2-2(i+2)z+i)=0$$

Stéphane Jaouen
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