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...

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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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