Should numpy.sqrt(-1) return 1j rather than?nan?

Stefan van der Walt stefan at
Thu Oct 12 11:14:50 CDT 2006

On Thu, Oct 12, 2006 at 10:53:12AM -0400, Alan G Isaac wrote:
> On Thu, 12 Oct 2006, Stefan van der Walt apparently wrote: 
> > I tried to explain the argument at 
> > 
> Helpful.  But you start off by saying:
>         In mathematics, the above assumption is true -- that 
>         the square root of -1 is 1j.
> Since square root is (here) a function, and part of the 
> function definition is the domain and codomain, this 
> statement is not correct.  If the codomain is real numbers, 
> the domain must correspondingly be (a subset of) the 
> nonnegative reals.  The nan output is the result of an 
> **invalid** input.
> So the question is really: "Why is a negative real number an 
> invalid input in this implementation", or "Why in this 
> implementation is the type of the output restricted by the 
> type of the input?" You get a good start on answering that.

You are more than welcome to change the wiki if you can think of a
simpler way to explain the problem... In fact, I would encourage that
-- I just put it there as a starting point.


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