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Re: How to evaluate Exp[I Pi(1+x)]?

• To: mathgroup at smc.vnet.net
• Subject: [mg52770] Re: How to evaluate Exp[I Pi(1+x)]?
• From: "Dr. Wolfgang Hintze" <weh at snafu.de>
• Date: Mon, 13 Dec 2004 04:22:09 -0500 (EST)
• References: <200412100123.UAA18952@smc.vnet.net> <cpehuc$6hr$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

I remember that ComplexExpand[expr] assumes parameters in expr to be 
real. Hence the following was the shortest succesful expansion I found:

In[4]:=
FullSimplify[ComplexExpand[E^(I*Pi*(1 + x))]]

Out[4]=
-E^(I*Pi*x)

Simplify is not enough.

By the way, while doing much complex function work recently I devolped 
the habit to always use ComplexExpand as the first opration.
Otherwise even the simplest operations Re, Im, Abs don't work.

Regards,
Wolfgang

Andrzej Kozlowski wrote:

> On 10 Dec 2004, at 10:23, hello wrote:
> 
> 
>>Is there a way to evaluate Exp[I Pi (1+x)]? I am expecting to see the
>>result to be:
>>
>>-Exp[I Pi x]
>>
>>because Exp[I Pi] can be reduced to shorter form.
>>
>>
>>
>>
>>
> 
> Hello?
> 
> FullSimplify[ComplexExpand[Exp[I*Pi*(1 + x)]],
>    x $B":(B Reals]
> 
> 
> -E^(I*Pi*x)
> 
> 
> 
> Andrzej Kozlowski
> Chiba, Japan
> http://www.akikoz.net/~andrzej/
> http://www.mimuw.edu.pl/~akoz/
> 
> 

• Follow-Ups:
  • Re: Re: How to evaluate Exp[I Pi(1+x)]?
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>

• References:
  • How to evaluate Exp[I Pi(1+x)]?
    • From: hello@noreply.com (hello)