Exercise [05.13]

ZZR Puig · 08 Jun 2008, 18:20

This one is also very easy. We had $\log i=\frac{\pi}{2}i+2n\pi i$ , so:

$i^i=e^{i\log{i}}=e^{i(\frac{\pi}{2}i+2n\pi i)}=e^{i^2(\frac{\pi}{2}+2n\pi )}=e^{-\frac{\pi}{2}(4n+1)}$

Obviously

can be either a positive or negative integer, as it refers to the number of full ( $2\pi$ ) rotations on the complex plane, which are equivalent regardless of their direction.

Exercise [05.13]

Who is online