Re: Desired behaviour ?

[Bill Allombert <Bill.Allombert@math.u-bordeaux1.fr>]

On Sat, Mar 10, 2007 at 05:34:22PM +0100, Loic Grenie wrote:
> 
>     I'm just wondering if there is a strong reason why 1.*I*I has an
>   imaginary part (equal to 0., but it's there). For instance neither I*I*1.
>   nor 1.*(I*I) have any imaginary part.

My understanding is that I*I should return -1+0*I but wrongly return
-1. So 1.*I*I should have an imaginary part (along with I*I*1. and
1.*(I*I) of course).

The rationale is that arithmetic operation should preserve the
definition domain so that domain detection work:

Compare:
factor(x^2-(-1+0*I))
factor(x^2-I*I)
both should returns
[x - I 1]

[x + I 1]

(The second doesn't)

Cheers,
Bill.