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Re: DirectedInfinity[1 + I], why does it get replaced by (1 + I)/(sqrt(2) ?


Evidently, the resulting complex number inside the DirectedInfinity 
expression has complex modulus 1, so that number is suitable to 
represent a "pure direction", without a magnitude.  If you think of a 
complex number as a vector in the real plane, then you're getting a unit 
vector with the same direction as the original vector.

Of course it would have been nice to have this properly documented!

Matt wrote:
> Hello,
>   This isn't particularly important probably, however, I am trying to
> learn as much about Mathematica as possible, and I thought this might
> shed some light on a 'Why Mathematica does this or that' principle.
> 
>   I'm working my way through 'The Mathematica Guidebook for
> Programming' and on page 177, he gives an example as follows:
> 
> In[39]:= DirectedInfinity[1 + I] DirectedInfinity[I]
> Out[39]:= DirectedInfinity[-(1 - I)/sqrt(2)]
> 
> That puzzled me a bit, so I decided to see what Mathematica would do
> with just the first part:
> 
> In[40]:= DirectedInfinity[1 + I]
> Out[40]:= DirectedInfinity[(1 + I)/sqrt(2)]
> 
> I realize that (1 + I)/sqrt(2) is in the same direction as (1 + I), but
> why did Mathematica change it to the more 'strange' form?
> 
> Thanks,
> 
> Matt
> 
> 

-- 
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.
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University of Massachusetts                413 545-2859 (W)
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Amherst, MA 01003-9305


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