### $i^i$ where $i=\sqrt {-1}$ is a real number. T/F

We have, $e^{i\theta}=cos\theta +i sin\theta$

Let, $\theta = 2n\pi + \frac {\pi}{2}$

Then, $e^{i.(2n\pi + \frac {\pi}{2})}=cos (2n\pi + \frac {\pi}{2}) +i sin (2n\pi + \frac {\pi}{2})$

$\Rightarrow e^{i.(4n+1) \frac {\pi}{2}}=0+i.1$

$\Rightarrow i.(4n+1)\frac {\pi}{2} = \ln i$

Let, $z=i^i$

$\Rightarrow \ln z = i.\ln i$

So, $\ln z = i.i(4n+1)\frac {\pi}{2}= i^2.(4n+1)\frac {\pi}{2} = -(4n+1)\frac {\pi}{2}$

$\Rightarrow z = e^{-(4n+1)\frac {\pi}{2}}$ = Real number

Hence, True.