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

• To: mathgroup at smc.vnet.net
• Subject: [mg55823] Re: [mg55802] DirectedInfinity[1 + I], why does it get replaced by (1 + I)/(sqrt(2) ?
• From: DrBob <drbob at bigfoot.com>
• Date: Thu, 7 Apr 2005 05:10:02 -0400 (EDT)
• References: <200504060711.DAA13652@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```> 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?

Because directions have unit length.

Norm[(1 + I)/Sqrt[2]]

1

DirectedInfinity[2 + I*3]
First[%]
Norm[%]

DirectedInfinity[(2 + 3*I)/Sqrt[13]]
(2 + 3*I)/Sqrt[13]
1

Bobby

On Wed, 6 Apr 2005 03:11:59 -0400 (EDT), Matt <anonmous69 at netscape.net> 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
>
>
>
>

--
DrBob at bigfoot.com

```

• Prev by Date: Re: DirectedInfinity[1 + I], why does it get replaced by (1 + I)/(sqrt(2) ?
• Next by Date: Re: How do I remove operator status?
• Previous by thread: DirectedInfinity[1 + I], why does it get replaced by (1 + I)/(sqrt(2) ?
• Next by thread: Re: DirectedInfinity[1 + I], why does it get replaced by (1 + I)/(sqrt(2) ?