<p>i^i
is multivalued . . .
if i = e^(i* pi/2)
------take the complex plane, make polar (polar complex plane), r =1, theta = 90 degrees . . . also, remembering e^(i<em>theta) = cos(theta) + i</em>sin(theta) and for cartesian z = x + i*y (transform cartesian to polar)</p>
<p>then
i^i = (e^i*(pi/2))^i
=e^(-pi/2)</p>
<p>but then, pi/2 is effectively equal to 5pi/2 or -9pi/2 or something else . . .
so it's multivalued
that's probably not perfectly clear, sorry</p>