Find all real values $a$ and $b$ such that $a+ib=i^{i^{i}}$? Find all real values $a$ and $b$ such that  $a+ib=i^{i^{i}}$ ?
My Try :
using the fact that $z^w=e^{w \log z}$, First I compute $i^i$ 
$$i^i=e^{i \log i}= e^{i (\log |i| + i arg (i))}=e^{- \frac{\pi}{2} - 2k \pi} \ \ \ \ k \in \mathbb{Z}$$
Now 
$$i^{i^{i}} = \left( e^{- \frac{\pi}{2} - 2k \pi} \right)^i = e^{i (\log |e^{- \frac{\pi}{2} - 2k \pi}| + i arg (e^{- \frac{\pi}{2} - 2k \pi}))} =  e^{i (\log |e^{- \frac{\pi}{2} - 2k \pi}| + i 2m\pi )} 
= e^{ (- \frac{\pi}{2} i - 2k \pi i - 2m\pi )} = -i e^{-2m\pi} $$
$m \in \mathbb{Z}$.
Therefore $a=0$ and $b=-e^{-2m\pi} $
Because We raised to $i$ two times I thought I might end up with two arbitrary factors $m$ and $k$, but one of them has to disappear. Is my solution correct ? Can we possibly have more than one arbitrary factor in the final result ?
Thanks. 
 A: Your first step is good, but
$$i^{i^i}=i^{(i^i)}$$
so you second step should be the other way around.
$$(i^i)^i=i^{i\cdot i}=i^{-1}=-i$$
(This is not the final result; it's just an example of the calculation of the value of the other (wrong) way of interpretation.)
A: $i=e^{i\pi/2}$, so $\displaystyle i^i=(e^{i\pi/2})^i=e^{-\pi/2}$ and $$i^{i^i}=i^{(i^i)}=(e^{i\pi/2})^{e^{-\pi/2}}=e^{i\left(\frac\pi2 e^{-\pi/2}\right)}=\cos\left(\frac\pi2 e^{-\pi/2}\right)+i\sin\left(\frac\pi2 e^{-\pi/2}\right).$$
As pointed out in the comments, I should probably make an assumption explicit: I am using the standard convention that writes $\alpha\ne0$ as $|\alpha|e^{i\theta}$ for some $\theta\in(-\pi,\pi]$ when evaluating $\alpha^\beta$. Without an explicit description of what is meant by $\alpha^\beta$ or, if we do not follow the convention, we can end up with different answers, since $i=e^{i5\pi/2}=e^{-3\pi/2}=\dots$, and different choices here lead to different  values of $i^{e^{-\pi/2}}$. 
I see the question has changed and now asks for all possible values of the expression: $i=e^{i(4a+1)\pi/2}$ for some integer $a$, so $i^i$ can be any number of the form $e^{-(4a+1)\pi/2}=e^{(4b-1)\pi/2}$ for some integer $b$. Finally, $\displaystyle i^{i^i}$ could be any number of the form $$\cos\left((4c+1)\frac\pi2e^{(4b-1)\pi/2}\right)+i\sin\left((4c+1)\frac\pi2e^{(4b-1)\pi/2}\right)$$ for any integers $b,c$.
A: I don't think any one of these answers is complete. First, it is universally understood that $i^{i^i}$ means only $i^{(i^i)}$ and never $(i^i)^i$. So we must first find all values $v$ of $i^i$, then find all values of $i^v$.
We compute $v = i^i = (e^{\pi i/2+2\pi ik})^i = e^{-\pi/2-2\pi k}$ for each integer $k$ (there are multiple values for this expression). Note these are all real numbers.
For each of these, we compute $i^v = (e^{\pi i/2+2\pi im})^v =
(e^{\pi i/2+2\pi im})^{e^{-\pi/2-2\pi k}} = e^{i\cdot\frac{\pi}{2}(4m+1)e^{-\frac{\pi}{2}(4k+1)}}$ for all integers $k$ and $m$. These can be written as $\cos \theta_{mk} + i \sin \theta_{mk}$, where $\theta_{mk} = \frac{\pi}{2}(4m+1)e^{-\frac{\pi}{2}(4k+1)}$.
I hope I don't have a typo in all of that. It's a doubly indexed family of points on the unit circle.
A: $i^{i}=e^{-\frac{1}{2}\pi+k2\pi}$ is correct. 
Then: 
$i^{i^{i}}=i^{e^{-\frac{1}{2}\pi+k2\pi}}=e^{\left(e^{-\frac{1}{2}\pi+k2\pi}\right)\ln i}=e^{\left(e^{-\frac{1}{2}\pi+k2\pi}\right)i\left(\frac{1}{2}\pi+m2\pi\right)}$
This results in: 
$i^{i^{i}}=e^{i\left(\frac{1}{2}\pi+m2\pi\right)\left(e^{-\frac{1}{2}\pi+k2\pi}\right)}=\cos\left[\left(\frac{1}{2}\pi+m2\pi\right)\left(e^{-\frac{1}{2}\pi+k2\pi}\right)\right]+i\sin\left[\left(\frac{1}{2}\pi+m2\pi\right)\left(e^{-\frac{1}{2}\pi+k2\pi}\right)\right]$
Here $k$ and $m$ are integers. 
A: If $\alpha \not = 0$ then define $\alpha^{\beta} = e^{\beta \log \alpha} $. But to define $\log \alpha$ there is ambiguity in how we choose to define a logarithm. We can define the principal logarithm by $\log \alpha = \log |\alpha| + i\arg \alpha$ where $\alpha \in (-\pi,\pi]$. Thus, all other possible logarithms are given by $\log \alpha = \log |\alpha| + i\arg \alpha + 2\pi i n$. 
Now $\log i = \log |i| + i\arg i + 2\pi i n = \pi i + 2\pi i n = \pi i (2n+1)$. Thus, 
$$ i^i = e^{i\log i} = e^{-\pi(2n+1)} $$ 
Thus, 
$$i^{i^i} = e^{e^{-\pi(2n+1)}\log i} = e^{e^{-\pi(2n+1)}\pi i (2m+1)} = \cos \left( e^{-\pi(2n+1)}\pi(2m+1)\right) + i \sin\left( e^{-\pi(2n+1)}\pi(2m+1)\right)   $$
My answer differs from what some people wrote above, I am too lazy to check the small details, just rather to show you the idea of one way of doing these problems. 
Note: I am not a fan of using the exponent multiplication rule $(a^b)^c = a^{bc}$ involving complex numbers. Because it is not true. Of course, here we are considering all possible exponents so it is safe to use it but in general it is not true and can lead to mistakes if one is not careful with it.  
