This question has been bugging me for a while, I have an answer that is working sufficiently for the program I'm using, but it is a tad slow, and let's say imprecise. It is not an overtly difficult question, and for that I assume someone smarter than me has already found a better explanation.


To begin, we take a value $y \in \mathbb{C}$ such that $|\log(y)| < 1$. Then we look at the exponential:


$$
b^z = e^{\log(y) z/y}\\
$$


The value $b = y^{1/y}$ is inside what we call the Shell-Thron region, in the tetration circles. This is the area in which:


$$
\lim\_{n\to\infty}\exp\_b^{\circ n}(0) \,\,\text{converges}
$$


Or, where the iterated exponential converges.


Now, to highlight the question, we can start with $1 < b < e^{1/e}$. The exponential $b^z$ has two fixed points on the real line. There is an attracting fixed point $1 < y < e$ and a repelling fixed point $\mu > e$. Both of which satisfy:


$$
e^{\log(y)/y} = e^{\log(\mu)/\mu}\\
$$


This relationship carries over for all $|\log(y)| < 1$, not just when it is real valued. From here, my question is simple.


Can anyone describe the holomorphic function:


$$
\begin{align}
f(y): \{y \in \mathbb{C} : \, |\log(y)| < 1\} &\to \{\mu \in \mathbb{C} :\,|\log(\mu)| > 1\}\\
\frac{\log f(y)}{f(y)} &= \frac{\log(y)}{y}\\
\end{align}
$$


And when restricted to real values:


$$
f(y): (1,e) \to (e , \infty)\\
$$


The manner I am solving this currently is a tad involved. But if $h(z) = \frac{\log(z) y}{\log(y)} = \log\_b(z)$, and $h^{\circ n}$ are the iterates of $h$:


$$
f(y) = \lim\_{n\to\infty} h^{\circ n}(e)\\
$$


This formula works on the real line, and moderately well in the complex plane. But I'm not certain if it works for all $|\log(y)| < 1$, there may be issues. It essentially runs off the basis that $e$ is in the Julia set of $b^z$, and/or in the attracting basin of the fixed point $\mu$ for the logarithm $\log\_b(z)$. I don't know if this is true universally though, just for a large part of the domain in question.


This has been bugging me, and any help is appreciated. In essence, does anyone know any good ways of computing/describing/constructing $f$, other than the way I mentioned?



Finding the repelling fixed point of an exponential, knowing only its attracting one
Okay, so I found the answer to this problem! It's a tad different than I thought, so I'll give the run down. Special thanks to @RolandBacher for getting me to think about $e$ as a branching point. The answer was simpler than I thought once I dug through it.


Let


$$
g(y) = y^{1/y}\\
$$


Then, there exists two branches to the inverse of this function, call them $h\_1$ and $h\_2$. The first satisfies:


$$
h\_1 : (1,e^{1/e}) \to (1, e)\\
$$


And the second satisfies:


$$
h\_2 : (1,e^{1/e}) \to (e,\infty)\\
$$


This is just standard calculus when looking at the self-root function. Thereby, our function $f$ is just the analytic continuation of the expression:


$$
f(y) = h\_2(y^{1/y})\\
$$


This solves the problem entirely. I apologize for asking a question when the answer was right in front of me, lmao. Thanks for the help, Roland!


