[seqfan] Re: Does (e i)^(pi i) have an imaginary part?

[Alonso Del Arte]

Thank you very much for the explanations. I've worried that my knowledge of
calculus and trigonometry is sorely lacking, but I see I really need to
brush up on logarithms.

And thanks to Zak for reminding me of the Im and Simplify functions. But it
leaves me wondering, why exactly is simplification necessary here? After
all, Im[E^(Pi I)] returns 0 without having to ask for a Simplify.

Al

On Thu, Aug 5, 2010 at 10:55 PM, zak seidov <zakseidov at yahoo.com> wrote:

> In Mathematica,
>
> In[4]:=
> Im[(E*I)^(Pi I)]//Simplify
>
> Out[4]=
> 0
>
> Zak
>
>
> --- On Thu, 8/5/10, Alonso Del Arte <alonso.delarte at gmail.com> wrote:
>
> > From: Alonso Del Arte <alonso.delarte at gmail.com>
> > Subject: [seqfan]  Does (e i)^(pi i) have an imaginary part?
> > To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
> > Date: Thursday, August 5, 2010, 6:34 PM
> > This may be a question with a very
> > obvious response, but it seems to be a
> > bit over my head. We all know e^(pi i) has no imaginary
> > part (its real part
> > famously being -1). What about (e i)^(pi i)?
> >
> > In Mathematica, I get a real part of
> > approx. -0.007191883355826365607801366396371202955362318
> > and an imaginary
> > part of 0. * 10^(-23) (if I ask for 20 decimal places
> > precision) or 0. *
> > 10^(-53) (if I ask for 50). Is the imaginary part so small
> > it's beyond
> > machine precision, or am I chasing a phantom?
> >
> > Al
> >
> > _______________________________________________
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> >
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