Question

What is the value of $i^i$

Where $i=\sqrt{-1}$

Answer 1

Using Euler's formula
$e^{ix}=\cos x+i\sin x$
Using $x=\frac{\pi }{2}$
we get $e^{i\frac{\pi }{2}}=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}$
Now raising both sides by $i^{th}$ power , we get
$e^{i^2\left(\frac{\pi }{2}\right)}=i^i$
$i^2=-1$
$i^i=e^{-\frac{\pi }{2}}$