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/

**References**:**DirectedInfinity[1 + I], why does it get replaced by (1 + I)/(sqrt(2) ?***From:*"Matt" <anonmous69@netscape.net>