N-th roots of complex numbers?



On 10/2/2013 12:06 AM, Robert Dodier wrote:
>
> I think it's a bug in solve that the result is ambiguous.
I disagree.

Starting in high school we have had ambiguous results from the quadratic 
formula.

  Recall that the discriminant appears
inside a square-root  :   +- sqrt(b^2-4ac).
Which square root of b^2-4ac is that?   Remember that in general b^2-4ac 
could be equal to
um,  what was that example?  8i ?
Now it doesn't  matter WHICH root is chosen because the other one will 
be chosen by
the negative sign.  So it doesn't matter that it is ambiguous, because 
we are offering the
set of two solutions, and the set is unique.
> If we can't even solve a simple equation unambiguously,
You are missing the point.  We have solved the equation and offered all 
the roots.
What happens NEXT is not a problem with the solve program, but possibly 
an issue
that subsequent processing  illegitimately separates the multiple 
solutions and tries
to somehow intuit or label which is the "principle" root.
> who knows what
> will happen in complicated problems.
Probably some programs will go off in the wrong direction. Candidates 
for doing
so include  the  change-of-variables program and definite integration.

There are or were CAS that provide slots for bounding boxes around 
algebraic numbers so
you CAN distinguish the members of a  RootOf expression.  Not clear that 
this would work
for expressions retaining extra parameters.

Trying to fix SOLVE when it is not wrong would be a problem. I suppose 
you could always
have algebraic root sets be returned as RootOf  expressions, in which 
case the recipient of
that result would be unable to extract just one root without some real 
effort.  That result, once
extracted, could then be illegitimately passed along as say "the" 
solution, leading to the
same situation,  but then at least you could not blame that basic 
version of SOLVE.

RJF