In reply to jonny taylor:
> I happen to remember (as he does) that e^{i pi} = -1, or to write it in a less familiar, but equally correct way, (e^pi}^i = -1. This implies that an ith root of -1 is actually e^pi.
> I don£t know if there are other roots (in the same way that there are two square roots of a number, a positive and a negative). That might be one for a mathematican not a physicist!
It is fairly easy to see that any number of the form e^kpi where k is an odd number (or the negative of an odd number) is an ith root of -1:
(e^kpi)^i
=e^kipi
=(e^ipi)k
=(-1)^k
Which is -1 when k is odd.
Obviously when k is even we have an ith root of 1
Note that I have not proved that these are the only ith roots of -1 ans 1............. but I suspect they are!
Edit: proved it now I think!
Post edited at 18:05