Re: imag () function detail regarding -0.0000

[Jordi Gutiérrez Hermoso]

On 11 September 2012 20:00, Daniel J Sebald <address@hidden> wrote:
> octave:59> [1,-1,-0]'*i
> ans =
>
>    0 + 1i
>   -0 - 1i
>   -0 - 0i
>
> Is this proper behavior?  I would think

There is no actual "i" in Octave. Instead, "i" is a function that
returns the equivalent of std::complex<double>(0.0, 1.0). You can see
this behaviour in other ways, e.g. inf*i giving NaNs.

- Jordi G. H.