Re: Is there a simpler way to get all 8 gaussian integers with same norml2() ?

["Ruud H.G. van Tol" <rvtol@isolution.nl>]

On 2024-11-03 21:39, Bill Allombert wrote:
> On Sun, Nov 03, 2024 at 04:00:16PM +0100, hermann@stamm-wilbrandt.de wrote:
> > ? id(c)=c;
> > ? neg(x)=-x;
> > ? flip(c)=I*conj(c);
> > ? c=7+4*I;vecsort([f(g(h(c)))|f<-[id,conj];g<-[id,neg];h<-[id,flip]])
> > [-7 - 4*I, -7 + 4*I, -4 - 7*I, -4 + 7*I, 4 - 7*I, 4 + 7*I, 7 - 4*I, 7 + 4*I]
> I suggest
>
> [w*z | w<-powers(I,3); z<-[c,conj(c)]]

Another one:

vecsort(concat(c*=[-1,1,-I,I], conj(c)),,8)

(assuming that duplicates are undesired)

-- Ruud