for the complex function $f_{(2)}(z)=z^z$,where in the complex plane does the inverse $z_{(2)}(f)$ not exist, same for inverse of functions $f_{(3)}(z)=z^{z^z}$ being $z_{(3)}(f)$ and so on

for instance $z_{(2)}(e^{-\pi /2})=i$ since $i^i=e^{-\pi /2}$

what is $z_{(2)}(i)$

what is $z_{(3)}(e^{-\pi /2})$

what is $z_{(2)}(1+i)$

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    $\begingroup$ Don't we have the Lambert-W-function which under the assumption of "principal values" of the logarithm gives an answer for $z^z =a $ to find out $z$ -even if $a$ and/or $z$ is complex? So I seem not really to understand what your problem is here? Also I have done a little essay on "super-roots" (a term that I don't really like) of higher orders, getting powerseries-solutions for them, for instance what you call $z_{(3)}()$ Perhaps you find here in MSE something when searching for "superroots", or in wikipedia with the same term. For instance math.stackexchange.com/q/3314712 $\endgroup$ – Gottfried Helms Oct 12 '19 at 7:34

at "$z_{(2)}(i)$" and "$z_{(2)}(1+i)$" : (Pari/GP)

\\ input to Pari/GP         \\   output from Pari/GP
\\ -------------------------\\-----------------------------
t=exp(LambertW(log(I)))     \\ %666 = 1.36062 + 1.11944*I
t^t                         \\ %667 =           1.00000*I

t=exp(LambertW(log(1+I)))   \\ %668 = 1.39402 + 0.577732*I
t^t                         \\ %669 = 1.00000 + 1.00000*I                

For $t = z_{(3)}(x)$ I used my series-representation. The value $x=\exp(-\pi/2)$ seems too much out of radius of convergence, so I could not yet obtain a result. But for $x=\exp(\pi/2) \approx 4.81048$ I could obtain $t \approx 1.77053$ and $t^{t^t} \approx 4.81048 $ with error of about $-0.0000000045...$. Here I had to use a procedure for summing divergent series (because the found powerseries at this argument diverges strongly), which is an adaption of the Noerlund-summation with a manually adapted "order" for the summation, but likely the better known Borel-summation should have been possible too.

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  • $\begingroup$ thank you ... are there closed form expressions for these $\endgroup$ – phdmba7of12 Oct 13 '19 at 14:38
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    $\begingroup$ @phdmba7of12 - no as far as I know, no more "closed" than the LambertW is "closed" form... $\endgroup$ – Gottfried Helms Oct 13 '19 at 17:56

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