TITLE: How can one calculate $i^{i^{i^{i^{.^{.^.}}}}}$?
QUESTION [6 upvotes]: I was reading an article that said about the value of $i^i$, and I thought to do the following $$\left(i^{i^{i^{i^{.^{.^.}}}}}\right)$$i.e, infinite powers of $ i $.

REPLY [4 votes]: Let's play fast and loose with things.  Assume the answer is $x+iy$.  Then
$$x+iy=i^{x+iy}=e^{i{\pi\over2}(x+iy)}=e^{-\pi y/2}(\cos(\pi x/2)+i\sin(\pi x/2))$$
from which we can conclude
$$x^2+y^2=e^{-\pi y}\quad\text{and}\quad {y\over x}=\tan(\pi x/2)$$
It follows that $x$ is a solution of
$$(x\sec(\pi x/2))^2=e^{-\pi x\tan(\pi x/2)}$$
I get $x\approx 0.438282936727$ for this equation, which agrees with what Matthew Conroy gave in a comment under Foo Barigno's answer.