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

[Robert Israel]

On Thu, 5 Aug 2010, Charles Greathouse wrote:

> (e i)^(pi i) = e^(pi i) * i^(pi i) = -1 * exp(pi i log i) = -exp(pi i
> * pi/2 i) = -exp(-pi^2/2) = -0.00719188335582636560780136639...
>
> So yes, no imaginary part.

One does have to be careful about branches of multivalued complex 
functions.  By definition (e i)^(pi i) is
exp(pi i log(e i)) [using any of the branches of log] 
= exp(pi i (1 + pi i/2 + 2 n pi i)) [ for any integer n ]
= - exp(- pi^2/2 - 2 n pi^2)
So a more complete answer is: yes, no imaginary part in any of its
branches.

Robert Israel                                israel at math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada