Find the value of $\space\large i^{i^i}$?

Is $\large i^{i^i}$ real ? How to find it?

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$i^i=e^{i\log i}$

Now on principal branch,using $i=e^{i\pi/2}\implies \log i=i\pi/2$ gives $i^i=e^{-\pi/2}$

Therefore, $i^{i^i}=i^{e^{-\pi/2}}=e^{e^{-\pi/2}\log i}=e^{i(\pi e^{-\pi/2})/2}=\cos\left(\pi \frac{e^{-\pi/2}}{2}\right)+i\sin\left(\pi \frac{e^{-\pi/2}}{2}\right)$

and hence its imaginary part is $\neq 0$ as $ \frac{e^{-\pi/2}}{2}$ is not an integer.

Complex powers may have more than one value. In our case
$$ i^i=e^{i \log i}=\exp \left(i \left(\ln(1)+i \frac{\pi}{2}+2\pi k i\right)\right)=e^{-\frac{\pi}{2}+2\pi k} $$ where $k$ is an integer. Thus $$i^{i^i}=e^{i^i \log(i)}=\exp\left(e^{-\frac{\pi}{2}+2\pi k}\cdot\left(i \frac{\pi}{2}+2\pi l i\right)\right), $$

which is $e$ to an imaginary power. It is therefore a point on the unit circle, but it can never be chosen real.

Wolfram Alpha gives the answer to be:

$0.94715899… + 0.320764449… i$.

Therefore it is not a real number, as the answer has an imaginary component to it.

We have $i^i=(e^{i\pi /2})^i=e^{-\pi /2}$. Then, $$i^{i^i}=i^{e^{-\pi /2}}=(e^{-i\pi /2})^{e^{-\pi /2}}=e^{-i\pi e^{-\pi /2}/2}=\cos (\pi e^{-\pi /2}/2)-i\sin(\pi e^{-\pi /2}/2) $$ Now $\sin(\pi e^{-\pi /2}/2)$ is non-zero, since $e^{-\pi /2}/2$ is not an integer.

First find all the values of $i^i$. Then, if $\beta$ is one of those values, a possible value of $i^{i^i}$ is $(e^{i\pi/2})^{\beta}$. Or $(e^{-3i\pi/2})^{\beta}$.