1i=e^\pi/2 i
i^i=(e^\pi/2 i)^i
=e^\pi/2i^2
e^-\pi /2
arg =0
5 Answers
66You should make it a habit to first search the site if the problems you are going to ask has been already discussed. This question for instance has been discussed a number of times earlier.
Anyway
i^i = (e^{i(2n\pi + \frac{\pi}{2})})^i=e^{i^2(2n\pi + \frac{\pi}{2})}=e^{-(2n\pi + \frac{\pi}{2})}
which is a positive real number. (Here n is an integer)
1thanx msp,was really careless didn't notice that there was no i in power
