The real odd one out

note that from De Moivre Theorem $\left ( cos(\theta)+i.sin(\theta) \right )^{n} = \left ( e^{i\theta} \right )^{n} = e^{i.n\theta} = \left ( cos(n\theta)+i.sin(n\theta) \right )$

$\,\,\,\,\,\,\, \,\,\,\, \bullet \,\,\,\, i^{i}$

from Moivre Identity we know

$\displaystyle e^{\pi i+2k\pi i} = -1$ for some integers $k$

then

$\displaystyle e^{ \frac{\pi i}{2}+k\pi i} = (-1)^{\frac{1}{2}} = i$ for some integers $k$

$\displaystyle i^{i} = e^{\frac{-\pi}{2} + k.\pi }$ for some integers $k$

So $i^{i}$ is real number.

$\,\,\,\,\,\,\, \,\,\,\, \bullet \,\,\,\, i \left ( e^{-ix} -e^{ix} \right )$

$e^{-ix} = cos(-x)+i sin(-x) = cos(x) - i sin(x)$

So

$e^{-ix} - e^{ix} = -2i sin(x)$

$i \left ( e^{-ix} -e^{ix} \right ) = 2sin(x)$

So $i \left ( e^{-ix} -e^{ix} \right )$ is real number

$\,\,\,\,\,\,\, \,\,\,\, \bullet \,\,\,\, e^{\pi i}$

$e^{\pi i} = cos(\pi)+i sin(\pi) = -1 +0 = -1$

So $e^{\pi i}$ is real number

$\,\,\,\,\,\,\, \,\,\,\, \bullet \,\,\,\, \left ( 3-2i \right ) \sqrt{5+12i}$

$\left ( 3-2i \right ) \sqrt{5+12i} = \sqrt{\left ( 3-2i \right )^{2} } \sqrt{5+12i}$

$= \sqrt{5-12i}\sqrt{5+12i} = \sqrt{|25-144|} = \sqrt{169} = 13$

So $\left ( 3-2i \right ) \sqrt{5+12i}$ is real number

$\,\,\,\,\,\,\, \,\,\,\, \bullet \,\,\,\, ln\left ( cos^{-1} (e.i) \right )$

let $p = ln\left ( cos^{-1} (e.i) \right )$

$e^{p} = cos^{-1} (e.i)$

use this identity $cos^{-1}(x) = - i.ln \left ( x \pm \sqrt{x^{2}-1} \right )$

then $cos^{-1}(e.i) = -i ln \left ( e.i \pm \sqrt{-e^{2}-1} \right ) = -i ln \left ( e \pm \sqrt{e^{2}+1} \right ) -i ln(i)$

$= \frac{\pi}{2} -i ln \left ( e \pm \sqrt{e^{2}+1} \right )$

for simplify the equation let $m = e \pm \sqrt{e^{2}+1}$

then let again $k.e^{i \theta } = k.cos(\theta)+i.k.sin(\theta) = \frac{\pi}{2} - i.ln(m)$

we have $k.cos(\theta) = \frac{\pi}{2}$ and $k.sin(\theta) = - ln(m)$

then $k = \sqrt{ \frac{\pi^{2}}{4} +ln^{2}(m) }$ and $\theta = tg^{-1}\left ( \frac{-2 ln(m)}{\pi} \right )$

so $e^{p} = cos^{-1}(e.i) = k.e^{\theta i}$ then $p = ln(k) + \theta i$

$p = ln \left ( \sqrt{ \frac{\pi^{2}}{4} +ln^{2}(m) } \right ) + i.tg^{-1}\left ( \frac{-2 ln \left ( e \pm \sqrt{e^{2}+1} \right ) }{\pi} \right )$

because $p$ can be written as $a+b.i$ where $a,b \in \mathbb{R}$ and $b \ne 0$

we conclude that $p$ is complex number and not real number.