[seqfan] Re: Does (e i)^(pi i) have an imaginary part?
Andrew N W Hone
A.N.W.Hone at kent.ac.uk
Sun Aug 8 18:25:41 CEST 2010
It's best to learn how to unravel this without a computer if you can, as with mathematica or maple
one can always worry about whether zero really means zero!
I would suggest reading an introductory text on complex analysis: the complex exponential is usually one of
the first examples of analytic functions in any such book.
The multi-valuedness of the logarithm may take a bit longer to understand than the exponential.
However, if you know how the ordinary (real) exp and log functions work, then you should quickly get the hang of the
>From the basic rules of exponentiation,
(a b)^c = a^c b^c
(e i)^(pi i) = e^(pi i) i^(pi i) = - i^(pi i).
Also, exp and log are inverses of each other, and one of the basic rules of logs is that
log (a^b) = b log a,
so this means
i^(pi i) = e^log ( i^(pi i) ) = e^( pi i log i )
Now i = e^( pi i /2 ) so one branch of the logarithm (the principal branch) gives
log i = pi i /2
i^(pi i) = e^( ( pi i )(pi i /2 ) = e^( - pi^2 /2 ),
(e i)^(pi i) = - e^( - pi^2 /2 ),
which is real and about -0.007, agreeing with M-ma below.
However, bear in mind that the function i^z is multi-valued due to the multi-valuedness of log, so
there are an infinity of different answers.
All the best,
From: seqfan-bounces at list.seqfan.eu [seqfan-bounces at list.seqfan.eu] On Behalf Of zak seidov [zakseidov at yahoo.com]
Sent: 06 August 2010 07:04
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Does (e i)^(pi i) have an imaginary part?
Because I guess
(e i)^(pi i) is a slightly more complex case and
Mathematica needs Simplify to reduce it.
In even more comlex cases you need FullSimplify!
--- On Fri, 8/6/10, Alonso Del Arte <alonso.delarte at gmail.com> wrote:
> From: Alonso Del Arte <alonso.delarte at gmail.com>
> Subject: [seqfan] Re: Does (e i)^(pi i) have an imaginary part?
> To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
> Date: Friday, August 6, 2010, 1:56 AM
> 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
> On Thu, Aug 5, 2010 at 10:55 PM, zak seidov <zakseidov at yahoo.com>
> > In Mathematica,
> > In:=
> > Im[(E*I)^(Pi I)]//Simplify
> > Out=
> > 0
> > Zak
> > --- On Thu, 8/5/10, Alonso Del Arte <alonso.delarte at gmail.com>
> > > 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
> > > part (its real part
> > > famously being -1). What about (e i)^(pi i)?
> > >
> > > In Mathematica, I get a real part of
> > > approx.
> > > and an imaginary
> > > part of 0. * 10^(-23) (if I ask for 20 decimal
> > > 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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