i^i

Date: 04/03/97 at 21:13:32
From: Anonymous
Subject: i^i

What is i to the power of i -- i^i?

Thanks,
Guillemo

Date: 04/04/97 at 05:42:23
From: Doctor Mitteldorf
Subject: Re: i^i

Dear Guillemo,

     One of the exciting things you discover when you explore complex
numbers is that expressions like this have meaning!  Expanding our
number system from the reals to the complex numbers wouldn't have made
much sense if we LOST some of the ability to do arithmetic operations.
After all, we did it in order to GAIN the ability to take the square
root of -1.  A general theorem assures you that arithmetic operations
on imaginary and complex numbers are all meaningful and well-defined -
with exceptions like ln(0) and 1/0.

     The Rosetta stone is the Euler equation
               e^ix = cos(x) + i sin(x)

You can verify that this equation makes a sensible definition by 
expanding the two sides as Taylor series in x.  You can also 
differentiate both sides and see that the answer is self-consistent.
Thirdly, you can use the formula for cos(2x) and sin(2x) to show
that the right side has the property you expect from an exponential,
so that e^i(2x) = (e^ix)^2.

    Once you believe the Euler equation, you've got a pretty
easy time of it.  If you let x=pi/2, then the real part is zero

       e^i (pi/2) = cos(pi/2) + i sin(pi/2) = i

     So i^i = (e^i*pi/2)^i = e^(i*i*pi/2) = e^(-pi/2).

    You may find it odd that not only is i^i well-defined, it's real!

-Doctor Mitteldorf,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/