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If I want to use the principal branch of $\log (\log(\log z))$, am I correct in thinking that I need branch cuts where $\log (\log z)$ is negative and real, where $\log z$ is negative and real, and where $z$ is negative and real?

Would there then be a single cut on $(-\infty, e]$?

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There is a branch of $\log(z)$ that maps ${\mathbb C} - [0,\infty)$ onto ${\mathbb R} \times (0,2\pi)$, which is a subset of ${\mathbb C} - [0,\infty)$. So one can iterate $\log(z)$ as many times as you'd like, the image will only get smaller with each iteration and will always be within the set ${\mathbb C} - [0,\infty)$ where this branch of $\log(z)$ is defined.

Note this is not the commonly used principal branch of $\log(z)$, but a different one that allows ${\mathbb C} - [0,\infty)$ to be the domain for all the iterations.

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The Maple code $$ expression := ln(ln(ln(x))): FunctionAdvisor(branch_cuts, expression, plot = `2D`); $$ produces $$[\ln(\ln(\ln(x)))\& (x = \exp(\alpha) \in (\alpha, RealRange(-infinity, Open(0)))), \&(x = \exp(\exp(\alpha)) \in (\alpha, RealRange(-infinity, Open(0)))), x < 0] $$ enter image description here

PS. Indeed, $z\to \ln(z)$ maps the open upper halfplane onto the strip $S:=\{z:\Re(z) \in (-\infty ,\infty) , \Im(z) \in (0,\pi)\}.$ Then $\ln(\ln(z))$ maps the upper open halfplane in the subset of $S$ and so on with $\ln(\ln(\ln(z))).$ The similar thing with the lower open halfplane. PPS. Also $\ln(z)$ maps the ray $(e,\infty)$ on the ray $(1,\infty)$, then $\ln(\ln(z))$ maps it on $(0,\infty)$ and $\ln(\ln(\ln(z)))$ maps on $(-\infty,\infty)$.

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