# How can one calculate $i^{i^{i^{i^{.^{.^.}}}}}$?

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 $i$.

#### Solutions Collecting From Web of "How can one calculate $i^{i^{i^{i^{.^{.^.}}}}}$?"

Suppose $z=i^{i^{i^{i^{.^{.^.}}}}}$. Then, $i^z=z$. And now…

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.

If you iterate over a few towers, you’ll see that it does converge towards a point in the complex plane. This point is approximately $0.438283+0.360592i$

edited for significance