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Re: DirectedInfinity[1 + I], why does it get replaced by (1 + I)/(sqrt(2) ?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg55842] Re: [mg55802] DirectedInfinity[1 + I], why does it get replaced by (1 + I)/(sqrt(2) ?
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Thu, 7 Apr 2005 07:15:02 -0400 (EDT)
*References*: <200504060711.DAA13652@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
On 6 Apr 2005, at 09:11, 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
>
Because (1 + I)/Sqrt[2] is a complex number of modulus 1, or a "unit
vector", and unit vectors are normally used to indicate direction
(there is a unique unit vector in a given direction).
Andrzej Kozlowski
Chiba, Japan
http://www.akikoz.net/andrzej/index.html
http://www.mimuw.edu.pl/~akoz/
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