Raymond Toy <toy@rtp.ericsson.se> writes:
> Currently, maxima says
>
> (C21) solve(x^4+1=0,x);
>
> 1/4 1/4 1/4 1/4
> (D21) [x = (- 1) %I, x = - (- 1) , x = - (- 1) %I, x = (- 1) ]
>
> which is right, but I expected the answer to be powers of
> (1+%i)/sqrt(2). I could add the rule
>
> tellsimp((-1)^(1/4), (1+%i)/sqrt(2))
>
> to get what I want.
Hm. At some level Maxima agrees that (-1)^(1/4)=1/sqrt(2) + 1/sqrt(2)
(C1) (-1)^(1/4);
1/4
(D1) (- 1)
(C2) bfloat(%);
(D2) 7.071067811865475B-1 %I + 7.071067811865475B-1
And of course, even if Maxima were sensitive about branches of the
fourth root, using (1+i)/sqrt(2) in the solution above would give all
four solutions.
(This reminds me, for some reason, of the old demonstrations that 1=-1,
1 = sqrt(1) = sqrt( -1 * -1) = sqrt(-1) * sqrt(-1) = i * i = -1
The mistake being, of course, that in general sqrt(a*b) isn't the same as
sqrt(a)*sqrt(b), unless you're willing to change which branch of the
square root you take.)
Jay