[SciPy-dev] [Numpy-discussion] Should numpy.sqrt(-1) return 1j rather than nan?

[Fernando Perez]

[ Moved over from the numpy list at Travis' request ]

On 10/12/06, Scott Sinclair <sinclaird at ukzn.ac.za> wrote:

> As far as I can tell this is exactly what happens. Consider the issue
> under discussion...
>
> ----------------------------------
>  >>> import numpy as np
>  >>> np.sqrt(-1)
> -1.#IND
>  >>> np.sqrt(-1+0j)
> 1j
>  >>> a = complex(-1)
>  >>> np.sqrt(a)
> 1j
>  >>> import scipy as sp
>  >>> sp.sqrt(-1)
> -1.#IND
>  >>> np.sqrt(-1+0j)
> 1j
>  >>> sp.sqrt(a)
> 1j
>  >>> np.__version__
> '1.0rc1'
>  >>> sp.__version__
> '0.5.1'
>  >>>
>
> ----------------------------------
>
> I'm sure that this hasn't changed in the development versions.

Check again (I just rebuilt from SVN right now):

In [1]: import numpy as n, scipy as s

In [2]: n.sqrt(-1)
Out[2]: nan

In [3]: s.sqrt(-1)
Out[3]: 1j

In [4]: n.__version__,s.__version__
Out[4]: ('1.0.dev3315', '0.5.2.dev2264')


Travis gave all the details on this particular issue, so I won't rehash them.

Cheers,

f