<s>[INST] Complex number $i^{{i^{i^{.^{.^.}}}}}$ If $A+iB=i^{{i^{i^{.^{.^.}}}}}$
Principal  values only being considered, 
Prove that
(a)tan $ \frac {\pi}{2} $A= $\frac{B}{A}$
(b) $A^2 + B^2 = e^{-\pi B}$
I tried the concept A+iB= $y=i^y$ 
$i= e^{ \frac{i\pi}{2}}$
$\ln(A+iB)=i \frac{\pi}{2}$(A+iB)
After this step not able to proceed
 [/INST] Whatever the value of $x = i^{i^{i^\cdots}}$ (if it exists), it should satisfy the equation
$$
x = i^x.
$$
But what is the definition of $i^x$ ? Since it is ambiguous, we must loosen the constraint even further: Whatever the value of $x$, it should satisfy
$$
x = e^{x \log i} \text{ for } \textit{some} \text{ value of } \log i
$$
Now, the set of logarithms of $i$ (the set of $y$ such that $e^y = i$) is $\{2 \pi i k + \tfrac{\pi i}{2} \}_{k \in \mathbb{Z}}$.
So our condition is now
$$
x = \exp\left(2 \pi i k x + \frac{\pi i}{2} x\right) \tag{1}
$$
Letting $x = A + Bi$, we get
\begin{align*}
A + Bi
&= \exp\left[ (A + Bi)\left(2 \pi i k x + \frac{\pi i}{2} x\right)\right] \\
&= \exp\left[ (-B + Ai)\left(2 \pi k + \frac{\pi}{2} \right)\right] \\
&= \underbrace{\exp\left[ (-B)\left(2 \pi k + \frac{\pi}{2} \right)\right]}_{\text{magnitude}}
   \underbrace{\exp\left[ A \left(2 \pi k + \frac{\pi}{2} \right) i \right]}_{\text{direction}} \\
\end{align*}
Equate the magnitude squared of both sides and the slope (tangent of the angle) of both sides. For magnitude squared we get:
$$
A^2 + B^2 = e^{- \pi B  + 4 \pi k B}. \tag{2}
$$
For tangent of the angle we get
$$
\frac{B}{A} = \tan \left((2 \pi k + \pi / 2) A\right). \tag{3}
$$
Your desired results follow from (2) and (3) if we assume $k = 0$.
This corresponds to taking the principal logarithm in defining $i^x$.
For $k = 0$:
\begin{align*}
A^2 + B^2 &= e^{- \pi B} \\
\frac{B}{A} &= \tan \left(\pi A / 2\right)
\end{align*}
</s>