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
Assume that the heuristic statement "$A+iB=i^{i\,\cdots \text{infinity times}}$" as written in the OP is rigorously described by the limit, if it exists, of the equation
$$\begin{align} z_{n+1}&=i^{z_n}\\\\ &=e^{z_n\log(i)}\\\\ &=e^{i\pi z_n/2}\\\\ \end{align}$$
subject to the initial condition $z_0=i$.
If $\lim_{n\to \infty}z_n=A+iB$ exists, then
$$\begin{align} A+iB&=e^{i\pi (A+iB)/2}\\\\ &=e^{-\pi B/2}\cos(\pi A/2)+ie^{-\pi B/2}\sin(\pi A/2)\tag1 \end{align}$$
Taking the modulus on both sides of $(1)$, we obtain
$$A^2+B^2=e^{-\pi B}$$
Taking the ratio of the imaginary and real parts of both sides of $(1)$, we obtain
$$\frac{B}{A}=\tan(\pi A/2)$$
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*}
