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

[Travis Oliphant]

Sven Schreiber wrote:

>>This is user adjustable.  You change the error mode to raise on 
>>'invalid' instead of pass silently which is now the default.
>>
>>-Travis
>>
>>    
>>
>
>Could you please explain how this adjustment is done, or point to the
>relevant documentation.
>  
>

numpy.sqrt(-1)

old = seterr(invalid='raise')
numpy.sqrt(-1)  # should raise an error

seterr(**old)  # restores error-modes for current thread
numpy.sqrt(-1)





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