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