The difficulty is general, and bound to a very simple fact : when z is 
complex (iL e. not known to be real), z^(1/3) denotes *some* quantity whose 
cube is z, with no way to specify which. This ambiguity is lifted when z is 
known real, where one can specify that, by *convention*, z^(1/3) (or 
$\sqrt[3]{z}$) is *the* real quantity whose cube is z (which is unique). 
This is similar (but not identical) to the *convention* that makes z^(1/2) 
(or $\sqrt{z}$) the positive quantity whose square is the *positive* real 
z, convention which breaks for z not known real and nonnegative...

In the same vein, I found that Sage was perfectly able to work out the 
roots of the "depressed" cubic x^3+p*x+q (variable and coefficients not 
known real) by Viete's substitution, but was unable to show that the two 
sets of solutions obtained were equivalent (assuming, of course, $\p\neq 
0$, $q\neq 0$ : it's perfectly obvious *numerically*, but I have been 
unable to convince Sage to state that those two roots sets were identical.

Emmanuel Charpentier

Le dimanche 13 août 2017 05:09:30 UTC+2, Nils Bruin a écrit :
> On Saturday, August 12, 2017 at 4:50:28 PM UTC-7, rjf wrote:
>> seems to me that asking for the "sign" of b^(1/3) in the complex domain
>> is nonsense.  There are 3 cube roots.  Let q be one of them; it doesn't
>> matter which.
>> then  - (1+sqrt(3)*i)*q/2    are the other two.  Yes, two.  because there
>> are two sqrt(3).
> I think the right pedantry here is to remark that there are *only* two 
> (not four) other roots, since, while there are two sqrt(3)'s and two i's, 
> the independent choices happen to lead to only two different elements of C. 
> It's a good argument for *not* writing (1+sqrt(3)*i)/2 for a primitive cube 
> root of unity.

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