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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