How many values does $i^{i^z}$ have? I came across this question in a written exam:  
For which complex $z$ does $i^{i^z}$ have finitely many values?
Is there a standard way in which this expression is defined? I tried to define it as the set $$\{e^{(i \pi/2 + 2 \pi i n)e^{z(i \pi/2 + 2 \pi i m)}}\}_{n,m \in \mathbb{Z}}$$ because of the expression $\log z = \log|z| + i \arg(z) + 2 \pi i k$ 
Is the question asking me to count the size of this set?
There are some answers which help me define $i^z \equiv \{e^{z i \pi/2 + z 2 \pi i k}\}_{k \in \mathbb{Z}}$ but it's not clear that this will now lead to a definition for $i^{i^z}$ unless we define what it means to exponentiate a set of complex numbers since the exponent is now a set. I think what makes sense is to take $\cup_k i^{e^{z i \pi/2 + z 2 \pi i k}} $ in which case I believe we get back the definition I made up $\{e^{(i \pi/2 + 2 \pi i n)e^{z(i \pi/2 + 2 \pi i m)}}\}_{n,m \in \mathbb{Z}}$
 A: My answer to Complex power of a complex number seems relevant here, because it explains in detail what $w^z$ means when $w$ and $z$ are complex, and also how many values $w^z$ represents. I am reproducing it below.

Let us find in general $w^z$ where $w$ and $z$ are complex.  This expression is by definition equal to $$\exp\{z\ln w\}$$ where $\ln w$ is one of the complex logarithms of $w$.  That is, $\ln w = w'$ where $$e^{w'} = w.$$  Suppose $w = re^{i\theta}$.  Then $$w' = \ln r + i\theta +2ik\pi$$ where $k$ is an arbitrary integer and the $\ln$ is the ordinary real-valued logarithm.  (Since $r\ge 0$ this is well-defined everywhere except for $r=0$, in which case we are dealing with $0^z$, which really is ambiguous.)
Putting this back into the original formula we have the answer, that $$\begin{align}
(re^{i\theta})^z & = \exp\{z (\ln r + i\theta + 2ik\pi)\}\tag{$\star$} \\
& =\exp\{z(\ln r + i\theta)\}\cdot \exp\{2ik\pi\cdot z\}
\end{align}$$
where $k$ is an integer.  

Now observe that although $(\star)$ seems to list an infinite number of solutions, they are not always distinct.  For example, when $z$ is a real integer,
the  second factor, $\exp\{2ik\pi\cdot z\}$, is $1$ for every choice of $k$, and so can be disregarded.
To find out how many values of $(\star)$ are distinct, one needs to ask about the values of $e^{2ik\pi \cdot z}$.  When $z$ has nonzero imaginary part,  or is a real irrational, these are all distinct and there are an infinite family of values of $w^z$, given by different choices of $k$.  But when $z$ is a real rational number with (lowest-terms) denominator $n$, there are exactly $n$ distinct values.
A: If $z = a + ib$ then 
$$i^z = e^{i\pi z/2} = e^{\frac{1}{2}(-b\pi + ia\pi)} = \frac{\cos \frac{a\pi}{2} + i\sin \frac{a\pi}{2}}{e^{b\pi/2}}$$
Therefore 
$$i^{i^z} = e^{(\cos \frac{a\pi}{2} + i\sin \frac{a\pi}{2})·e^{-b\pi/2}·\frac{i\pi}{2}} = \\e^{-\frac{\pi}{2}e^{-b\pi/2}\sin \frac{a\pi}{2}}·\left(\cos\left( \frac{\pi}{2}\cos\frac{a\pi}{2}e^\frac{-b\pi}{2}\right) + i\sin \left( \frac{\pi}{2}\cos\frac{a\pi}{2}e^\frac{-b\pi}{2}\right)\right)$$
So for $r_\theta$ to be expressed as $i^{i^z}$ then we need:
$$\ln \vert r\vert = -\frac{\pi}{2}e^\frac{\large -b\pi}{2}\sin\frac{a\pi}{2}$$
And
$$\theta = \frac{\pi}{2}e^\frac{\large -b\pi}{2}\cos\frac{a\pi}{2}$$
So given $z$ then the module and argument of $i^{i^z}$ are fixed.
In fact, I think $f(z)  = i^{i^z}$ bijects $\mathbb{C} \to \mathbb{C}-\lbrace 0\rbrace$ but I might be wrong.
A: Defining $$i^{i^z} \equiv \cup_{k\in\mathbb{Z}} i^{e^{i \pi/2 + 2 \pi i k}}$$  $$= \cup_{k\in\mathbb{Z}} \{e^{(i \pi/2 + 2 \pi i n)e^{z(i \pi/2 + 2 \pi i k)}}\}_{n \in \mathbb{Z}}$$
after plugging in the definition for $i^{e^{i \pi/2 + 2 \pi i k}}$  
Fixing some $k$, it's not hard to show that $ \{e^{(i \pi/2 + 2 \pi i n)e^{z(i \pi/2 + 2 \pi i k)}}\}_{n \in \mathbb{Z}} \subset i^{e^{i \pi/2 + 2 \pi i k}}$ is infinite cardinality unless $e^{z(i \pi/2 + 2 \pi i k)}$ is a rational number.
Writing $z = x+iy$ we get that $x$ must be even and $e^{-y(\pi/2 + 2 \pi k)}$ must be a rational number. But considering $k=0$ we get that $e^{-y(\pi/2)} = p/q$ for some integers $p,q$ and therefore $y= -2/\pi \log(p/q)$.  
Then $e^{z(i \pi/2 + 2 \pi i k)} = \pm e^{-y(\pi/2 + 2 \pi k)} = \pm e^ {\log(p/q) (1 + 4k)} = \pm (p/q)^{1+4k}$  
Now we have shown  $\{e^{(i \pi/2 + 2 \pi i n)e^{z(i \pi/2 + 2 \pi i k)}}\}_{n \in \mathbb{Z}} $ must be of the form $\{e^{\pm (i \pi/2 + 2 \pi i n)(p/q)^{4k+1}}\}_{n \in \mathbb{Z}} $ if it is to be finite. So we are reduced to searching for conditions on $p,q$ for which the set $\{e^{\pm (i \pi/2 + 2 \pi i n)(p/q)^{4k+1}}\}_{n,k \in \mathbb{Z}}$ is finite.  
It is necessary that $|p|>|q|$ and sufficient that $|q|=1$. If I can only show that it is necessary for $|q|=1$ then I am done.
