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

[Andrew N W Hone]

Hi - 

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 
complex case.

>From the basic rules of exponentiation,

(a b)^c = a^c b^c

means that 

(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 

and hence  

i^(pi i) =  e^( ( pi i  )(pi i /2 ) = e^( - pi^2 /2 ), 

so finally 

(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,
Andy 



________________________________________
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!

Zak


--- 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
> 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
> > >
> > > _______________________________________________
> > >
> > > Seqfan Mailing list - http://list.seqfan.eu/
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> >
> >
> >
> >
> >
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