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

• To: mathgroup at smc.vnet.net
• Subject: [mg55818] Re: [mg55802] DirectedInfinity[1 + I], why does it get replaced by (1 + I)/(sqrt(2) ?
• From: Murray Eisenberg <murray at math.umass.edu>
• Date: Thu, 7 Apr 2005 05:09:56 -0400 (EDT)
• Organization: Mathematics & Statistics, Univ. of Mass./Amherst
• References: <200504060711.DAA13652@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```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:= DirectedInfinity[1 + I] DirectedInfinity[I]
> Out:= 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:= DirectedInfinity[1 + I]
> Out:= 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.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305

```

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