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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