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

Out=
-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/
>
>

```

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