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