# Solve $i^{i^{i^\ldots}}$ [duplicate]

How to find $$i^{i^{i^\ldots}} \quad :\quad i=\sqrt{-1}$$ I'm able to find the solution for the finite powers using $$i=e^{i(2k\pi+\frac{\pi}{2})}\quad:\quad k\in\mathbb{Z}$$ $$i^{i}=e^{-(2k\pi+\frac{\pi}{2})}$$ $$i^{i^{i}}=e^{-i(2\pi k+\frac{\pi}{2})}=-i$$ $$i^{i^{i^i}}=e^{(2\pi k+\frac{\pi}{2}) }$$ $$\text{and so on}$$ but what should be the approach to solve for infitie powers$\space$?

## marked as duplicate by Math Lover, Simply Beautiful Art, lab bhattacharjee, Did, StubbornAtomAug 12 '18 at 4:33

• Note that one should define exponentiation to be single-valued by choosing a branch, which would make this question clearer... – Simply Beautiful Art Aug 12 '18 at 3:03
• Your method of computing the iterates is not correct. Note that $(i^i)^i \ne i^{(i^i)}$. – Shalop Aug 12 '18 at 3:18
• @Shalop What should be the correct method of computing the iterates? – Arpit Yadav Aug 12 '18 at 3:21
• Going by your rules, we say $i^z = e^{\big(2k+\frac12\big)i\pi z}$. Let us call this function $f(z)$. Now compute $f(i)$. Then compute $f(f(i))$. Then compute $f(f(f(i)))$. You will see that it is not the same as what you are doing. – Shalop Aug 12 '18 at 3:44
• – Simply Beautiful Art Aug 12 '18 at 4:04

$X = i^{i^{i^{\cdot}}}$

So $X = i^X$.

So $\log (X) = X \log (i)$.

So ${\log (X) \over X} = \log (i)$.

So $X = i W(-i)$, where $W$ is the Lambert $W$ or PolyLog function.

Mathematica evaluates this to $X = 0.44 + 0.36 i$.

Here's a graph on the imaginary plane of 200 successive exponentiations with the solution as a red dot:

• This is a fixed point but proving that it converges is more work. – parsiad Aug 12 '18 at 1:18
• @parsiad. Fair enough. Good thing the OP didn't ask for a proof of convergence! – David G. Stork Aug 12 '18 at 1:19
• @DavidG.Stork Does series converge to any particular value? – Arpit Yadav Aug 12 '18 at 1:21
• This is purely formal, obviously. – Did Aug 12 '18 at 3:51