Question

What is the `rootii`?

Answer 1

Using a similar question:
`root(n)(i)=cos(\pi/(2n))+isin(\pi/(2n))`

`root(i)(i)=cos(\pi/(2i))+isin(\pi/(2i))`

Using Euler's identity:
`e^(ix)=cosx+isinx`

We know that:
`e^(i*\pi/(2i))=cos(\pi/(2i))+isin(\pi/(2i))`

`cos(\pi/(2i))+isin(\pi/(2i))=e^(\pi/2)`

Therefore, `root(i)(i)=e^(\pi/2)`

This can be proved by calculating `(e^(\pi/2))^i` which should give `i`

Answer 2

`e^(pi/2+2npi)" "` for `n in ZZ`, with principal value `e^(pi/2)`

Explanation:

Note that:

`i = e^((pi/2+2npi)i)" "` for any `n in ZZ`

So:

`root(i)(i) = i^(1/i) = i^(-i) = (e^((pi/2+2npi)i))^(-i) = e^((pi/2+2npi)*(-i^2))= e^(pi/2+2npi)`

with principal value `e^(pi/2)` when `n=0`