Answer:
The number [tex]i^i[/tex] is a real number.
Step-by-step explanation:
First, let us recall the exponential form of a complex number: if [tex]z\in\mathbb{C}[/tex] it can be written as
[tex]z=re^{i\phi}[/tex]
where [tex]r=|z|[/tex] and [tex]\phi[/tex] is the argument of [tex]z[/tex].
Also, let us recall Euler's formula: if x is a real number
[tex] e^{ix} = \cos x +i\sin x[/tex].
Using this, we have
[tex]e^{i\pi/2} = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = i[/tex].
So, if we elevate both sides to [tex]i[/tex]
[tex]\left(e^{i\pi/2}\right)^i = i^i[/tex].
But, [tex]\left(e^{i\pi/2}\right)^i = e^{i^2\pi/2} = e^{-\pi/2}[/tex].
Therefore, [tex]i^i = e^{-\pi/2}[/tex] which is a real number.
