Prove that i to the power i is a real no.
Answers
Write i=eπ2i, then ii=(eπ2i)i=e−π2∈R. Be careful though, taking complex powers is more... complex... than it may appear on first sight − see here for more info.
In particular, it's not well-defined (until we make some choice that makes it well-defined); we could just have well written i=e5π2i and obtained ii=e−5π2. But ii can't be equal to both e−π2 and e−5π2 can it?
Despite the lack-of-well-defined-ness, though, ii is always real, no matter which 'ith power of i' we decide to take.
More depth: If z,α∈C then we can define
zα=exp(αlogz)
where expw is defined in some independent manner, e.g. by its power series. The complex logarithm is defined by
logz=log|z|+iargz
and therefore depends on our choice of range of argument. If we fix a range of argument, though, then zα becomes well-defined.
Now, here, z=i and so logi=iargi, so
ii=exp(i⋅iargi)=exp(−argi)
so no matter what we choose for our range of range of argument, we always have ii∈R.
Good luck