Negative Numbers to Powers

Date: 11/23/2001 at 16:04:43
From: Filoxus Maximus
Subject: Powering negative numbers

If u is an irrational number and x is a negative number, what is x^u?

How do I determine even whether that number is positive or negative?
It isn't a problem when x is positive, but when x is negative the 
situation is a bit complicated. 

In the end: how much is (-e)^(-e)?

Date: 11/24/2001 at 12:03:57
From: Doctor Rob
Subject: Re: Powering negative numbers

Thanks for writing to Ask Dr. Math, Filoxus.

Your question boils down to evaluating (-1)^u.  The way to try to do
this is to use Euler's Equation:

  http://www.mathforum.org/dr.math/faq/faq.euler.equation.html   

If you do this, you'll find that, for every integer k,

   -1 = cos(Pi*[2*k+1]) + i*sin(Pi*[2*k+1]),
      = e^(i*Pi*[2*k+1]),
   (-1)^u = [e^(i*Pi*[2*k+1])]^u,
          = e^(i*Pi*u*[2*k+1]),
          = cos(Pi*u*[2*k+1]) + i*sin(Pi*u*[2*k+1]).

Since u is irrational, this means that (-1)^u has infinitely many
different values, one for each integer k. If u were rational, there
would only be finitely many different values, and one could pick out
of this finite set a value for (-1)^u which was the simplest: either a 
real one or a complex one with the smallest positive argument. In ,the 
case where u is irrational, there is no convenient way to pick a
"simplest" value, since the infinite set of values form a dense subset 
of the unit circle in the complex plane, none of which is real.

If you let k = 0, you get one of the values,

   (-e)^(-e) = e^(-e)*(-1)^(-e),
             = e^(-e)*(cos[-Pi*e]+i*sin[-Pi*e]),
             = e^(-e)*(cos[Pi*e]-i*sin[Pi*e]),

which I leave to you to evaluate numerically. If you let k = -1, you 
get the complex conjugate of the above answer. If you let k = 1, you 
get a different value,

   (-e)^(-e) = e^(-e)*(-1)^(-e),
             = e^(-e)*(cos[3*Pi*e]-i*sin[3*Pi*e]).

If you let k have other values, you get other different answers. All 
have the property that when raised to the power (-1/e), they give the 
answer -e.

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/