The problem with (-1)^(1/6), is that as a complex number it has six
possible values. However, rectform returns just one of them:
rectform((-1)^(1/6)) -> sqrt(3)/2+i/2
There is clearly a discrepancy between the displayed solution and the
"actual" value which Maxima is using internally; or as John says, rectform
just picks one value. Using de Moivre's theorem, since -1 = exp(pi*i), we
would have, exp(i*pi/6) as one of the values, and maybe it is this one
which rectform chooses.
It seems to me that using rectform is dodgy unless you are simply
converting from the polar form of a particular complex number. Otherwise
you can't be quite sure of what you might obtain.
On Tue, Oct 1, 2013 at 7:08 PM, John Lapeyre <lapeyre.math122a at gmail.com>wrote:
>
> On 10/01/2013 09:06 AM, Robert Dodier wrote:
> > On Mon, Sep 30, 2013 at 10:08 PM, Alasdair McAndrew <amca01 at gmail.com><amca01 at gmail.com>wrote:
> > Dunno what's going on. With Maxima 5.31.1 + Clisp + Linux, I get:
>
>
> It also works for me with linux + (5.28.0 gcl and 5.31.1 sbcl)
>
>
>
> > (%i2) solve(z^3 = 8*%i,z);
> > (%o2) [z = (-1)^(1/6)*sqrt(3)*%i-(-1)^(1/6),
> > z = -(-1)^(1/6)*sqrt(3)*%i-(-1)^(1/6),z = 2*(-1)^(1/6)]
> > (%i3) rectform(%);
> > (%o3) [z = %i-sqrt(3),z = -2*%i,z = %i+sqrt(3)]
>
> I see three possible interpretations of this example
> (are there more ?) :
>
> 1. (-1)^(1/6) means one particular root of z^6=-1
> (e.g. the principal root). solve gave the
> correct answer. But perhaps solve should have
> applied rectform itself.
>
> 2. the meaning of (-1)^(1/6) depends on the context.
> eg if it appears at the top-level of expressions
> in a list of three elements, then it means
> the same root in each case, maybe a specific
> root. This is obviously problematic.
>
> 3. (-1)^(1/6) means any one of the the six roots,
> or maybe all of them. It's up to the user to
> decide. In this case, solve gave an incorrect answer
> and this is a bug in solve.
>
> --John
>
>
>
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