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)