Re: What is Infinity+Pi*I

• To: mathgroup at smc.vnet.net
• Subject: [mg65881] Re: [mg65860] What is Infinity+Pi*I
• From: "Erickson Paul-CPTP18" <Paul.Erickson at Motorola.com>
• Date: Thu, 20 Apr 2006 05:15:15 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```Ted,

OK, this is interesting why:

In[1]:=
Exp[ Infinity +  Ï? I ]

Out[1]=
â??

In[2]:=
Limit[ Exp[ x +  Ï? I ] , x \[Rule] Infinity ]

Out[2]=
-â??

The reason is, of course,

In[3]:=
Infinity +  Ï? I

Out[3]=
â??

Because any finite (even complex finite) value is insignificant against infinity, while any infinite (even though large) maintains the imaginary part. Therefore the limit is evaluated differently than the direct.

In[5]:=
10^10000  +  Ï? I//Short

Out[5]//Short=
1000000000000000000000\[LeftSkeleton]9957\[RightSkeleton]\
0000000000000000000000+\[LeftSkeleton]1\[RightSkeleton]

In[6]:=
%[[2]]

Out[6]=
\[ImaginaryI] Ï?

Therefore, we have the interesting result:

In[8]:=
Exp[ Ï? I ] Exp[ Infinity ] â?  Exp[ Infinity +  Ï? I ]

Out[8]=
True

Whereas:

In[9]:=
Limit[ Exp[ Ï? I ] Exp[ x ] == Exp[ x +  Ï? I ], x \[Rule] â??]

Out[9]=
True

So by playing with the order of evaluation to delay Plus:

In[22]:=
SetAttributes[ Plus, HoldAll ]

In[23]:=
Infinity +  Ï? I

Out[23]=
Ï? \[ImaginaryI]+â??

In[25]:=
Exp[ Infinity +  Ï? I ] //Simplify

From In[25]:=
\!\(\*
RowBox[{\(Simplify::"infd"\), \(\(:\)\(\ \)\), "\<\"Expression \\!\\(-\
\[ExponentialE]\\^\\(\\(\\(
Ï?\\\\ \[ImaginaryI]\\)\\) + â?? - \\(\\(\[ImaginaryI]\\\\ Ï?\\)\\)\\)\\) \
simplified to \\!\\(-â??\\). \\!\\(\\*ButtonBox[\\\"Moreâ?¦\\\", \
ButtonData:>\\\"Simplify::infd\\\"]\\)\"\>"}]\)

Out[25]=
-â??

Now:

In[27]:=
Exp[ Ï? I ] Exp[ Infinity ] == Exp[ Infinity +  Ï? I ] // Simplify

Out[27]=
True

Paul

-----Original Message-----
From: ted.ersek at tqci.net [mailto:ted.ersek at tqci.net]
To: mathgroup at smc.vnet.net
Subject: [mg65881] [mg65860] What is Infinity+Pi*I

I am using Mathematica 4.1, but I suspect all versions do the same in this case.

In[1]:=
Infinity + Pi * I

Out[1]=
Infinity

I think it the input above should return itself.
Am I wrong here? If we do it my way the following would return
-Infinity. (*Negative Infinity*)

In[1]:=
E^( Infinity + Pi * I )

Out[1]=
Infinity

Either way it's an interesting example.

-------
Regards,
Ted Ersek

```

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