Evaluate and comment:

f = i^i

From the polar form of i, we know the natural logarithm of i is i*pi/2. The natural log of i^i is therefore i*i*pi/2 = -pi/2, and the corresponding value (antilog) is e^(-pi/2) or about 0.207879576350762.

Posted by Charlie on 2024-10-30 17:47:33