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The much anticipated math.SE community blog will $\tiny\mathrm{hopefully}$ contain a contribution from Alex Becker with the topic The Complex Real Roots of $x^3-3x+1$, which I'm really looking forward to reading. It will cover the fact that the roots of certain cubic polynomials are real, but can be expressed with radicals only if involving imaginary numbers (Casus irreductibilis). This is new to me and it really caught my eye!

As an example, let's use the polynomial mentioned above. The zeroes of $x^3-3x+1=0$ can be found on Wolfram Alpha. $$x=-\frac12\left(1-i\sqrt3\right)\sqrt[3]{\frac12\left(-1+i\sqrt3\right)}-\frac{1+i\sqrt3}{2^{2/3}\sqrt[3]{-1+i\sqrt3}} \approx -1.8794$$ $$x=-\frac{1-i\sqrt3}{2^{2/3}\sqrt[3]{-1+i\sqrt3}}-\frac12\sqrt[3]{\frac12\left(-1+i\sqrt3\right)}\left(1+\sqrt3\right) \approx 0.34730$$ $$x=\frac1{\sqrt[3]{\frac12\left(-1+i\sqrt3\right)}}+\sqrt[3]{\frac12\left(-1+i\sqrt3\right)} \approx 1.5321$$ There are all real numbers, but need imaginary numbers to be expressed by radicals.

To my question(s):

  • How do you evaluate such numbers to achieve a real approximation? If a complicated complex number as one of the above is thrown at you without knowing where it came from, would you need to first find the polynomial for which it is a root and then use numerical methods to extract an approximation of the root?
  • Can you determine if a radical expression involving imaginary numbers is real without going through the whole process of evaluating (approximating) it?
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The problem of determining whether an expression is purely real or complex is generally undecidable. This is an easy corollary of Richardson's theorem, which says that it is impossible to decide if two real expressions $x$ and $y$ are equal. Assuming Richardson's theorem, note that $(x-y)i$ is real if and only if $x=y$.

Even if you evaluate the expression numerically, that won't necessarily decide it as a purely real expression can lead to number like $10.0^{-16}i$, which might or might not be zero.

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I do not know the structure of the general case, but all three solutions of the stated problem have the form

$$x=zw+\frac{\bar{z}}{w},\quad z,w\in\mathbb{C}\tag{1}$$

with $$w=\sqrt[3]{\frac12\left(-1+i\sqrt3\right)}$$

Note that $|w|=1$. Equation (1) can be rewritten as

$$x=\frac{zw^2+\bar{z}}{w}=\frac{|w|^2zw+\bar{z}\bar{w}}{|w|^2}=2\Re\{zw\}$$

where the last equality follows from $|w|=1$. So all solutions can be easily shown to be real-valued for any $z$ as long as $|w|=1$ is satisfied.

Let $u=\frac12(-1+i\sqrt{3})$. Then the complex variable $z$ for the three solutions is given by

$$z=u \quad(x\approx −1.8794)\\ z=\bar{u}\quad(x\approx 0.3473)\\ z=1\quad(x\approx 1.5321)$$

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  • $\begingroup$ Thanks, but this doesn't really address the general case in my first question, does it? Also, how would you go about to find the real part of $z$? WA suggests finding a root of a cubic polynomial numerically, which is connected to my first question. $\endgroup$
    – Daniel R
    Commented Apr 28, 2014 at 12:58
  • $\begingroup$ I'm not sure how you define the "general case". My answer applies to all numbers which can be expressed as $1/z+z$, and then the above holds. If there is a general form for such numbers you could include it in your question. As for finding the real part of $z$, if you don't mind evaluating a cosine function, what else is the problem? $\endgroup$
    – Matt L.
    Commented Apr 28, 2014 at 13:25
  • $\begingroup$ Check out the other solutions in the WA link for other forms of a number of this kind. By the general case, I guess I mean any (complicated) complex number. $\endgroup$
    – Daniel R
    Commented Apr 28, 2014 at 13:39
  • $\begingroup$ @DanielR I included the form of the other two solutions in my answer. I do not know what the "general case" would be. $\endgroup$
    – Matt L.
    Commented Apr 28, 2014 at 15:59

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