i to the i ?

Let $z = x + iy$ .

Then

$e^z = e^{x+iy} = e^xe^{iy}$

$= e^x (\cos y + i \sin y)$ (by applying Euler's formula)

This will equal $i$ only when $x=0$ and $y = 2n\pi + \frac{\pi}{2}$

Now then, if $e^z = i$ then $e^{iz} = i^i$ (So $e^{iz}$ will be a possible solution to $i^i$ )

In other words, solutions of $i^i$ will all be of the form:

$e^{iz}$

Where $z = i(\frac{\pi}{2} + 2n\pi)$

So, they will be of the form:

$e^{-(\frac{\pi}{2} + 2n\pi)}$

for all integers $n$ .

The only answer for which this equations fits, is $\boxed{e^{-\frac{\pi}{2}}}$

We know by euler's form that

$e^{i \dfrac{\pi}{2}} = i$ where $i = \sqrt{-1}$

So we get, for $i^i= e^{(i \times i \times \dfrac{\pi}{2})}$ = $e^{\dfrac{-\pi}{2}}$

$\displaystyle e^{i \pi} = -1 ~~\boxed{\text{most beautiful and famous equation}} \\ \displaystyle e^{i \pi \times i} = -1^{i} \\ \displaystyle e^{-1 \times \pi} = -1^{i} \\ \displaystyle e^{\dfrac{-\pi}{2}} = -1^{\dfrac{i}{2}} \\ \displaystyle e^{-\dfrac{\pi}{2}} = (-1^{\dfrac{1}{2}})^{i} \\ \displaystyle \boxed{\therefore i^i = e^{-\dfrac{\pi}{2}}}$