Question

i ^2 = -1,
i ^ i = ? i ^2 = -1,
i ^ i = ? @Mathematics

Answer 1

i^i = \left( e^{i (2k \pi + \pi/2)} \right)^i = e^{i^2 (2k \pi + \pi/2)} = e^{- (2k \pi + \pi/2)} where k \in \mathbb{Z}, the set of integers.

The principal value (for k=0) is e − π / 2 or approximately 0.207879576350761908546955465465465..

Answer 2

lol, do you understand what did you copy and paste?

Answer 3

Using Euler's Formula:\[e^{i\theta}=\cos\theta+i \sin\theta\]let theta be pi/2.\[e^{i\frac{\pi}{2}}=i\]Raise both sides to the i-th power:\[\left(e^{i\frac{\pi}{2}}\right)^i=i^i\iff i^i=e^{i^2\frac{\pi}{2}}=e^{-\frac{\pi}{2}}\]