From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: i^i
Date: 16 Feb 1998 02:02:14 GMT

In article <6c5qtm$1od$1@excalibur.flash.net>,
Brandon Berg <bberg@flash.net> wrote:
>Why does i^i=e^(-pi/2)?

It doesn't. Exponentiation of complex numbers is undefined, in general.
You can discuss principal values, I suppose, but arguably  i^i  is
	i^i = (e^((2*n+1/2)*pi*i)^i   for any integer  n
	    = e^((2*n+1/2)*pi*(-1))
that is, it is indeed  e^(-pi/2), or that value times any power of
e^(2 pi).

dave