# Complex towers: $i^{i^{i^{…}}}$

If $w = z^{z^{z^{...}}}$ converges, we can determine its value by solving $w = z^{w}$, which leads to $w = -W(-\log z))/\log z$. To be specific here, let's use $u^v = \exp(v \log u)$ for complex $u$ and $v$.

Two questions:

• How do we determine analytically if the tower converges? (I have seen the interval of convergence for real towers.)
• Both the logarithm and Lambert W functions are multivalued. How do we know which branch to use?

In particular $i^{i^{i^{...}}}$ numerically seems to converge to one value of $i2W(-i\pi/2)/\pi$. How do we establish this convergence analytically?

(Yes, I have searched the 'net, including the tetration forum. I haven't been able to locate the answer to this readily.)

• You may look for "Shell-Thron-region", which is the complex extension for the "Euler-interval" for real numbers (where the infinite power tower of a real basis converges), see for instance math.eretrandre.org/hyperops_wiki/… . i is inside that region, so the infinite power-tower converges (I think, only the principal branch is always selected). As far as I recall the original article of D. Shell is online available for more details. – Gottfried Helms Aug 10 '12 at 6:06
• Thron's article and Shell's article. In any event, remember that the power function itself is multivalued... – J. M. is a poor mathematician Aug 10 '12 at 6:26
• "How do we determine analytically if the tower converges" (jstor.org/discover/10.2307/2300357) had some analysis for determining the convergence of infinite exponentials. "Both the logarithm and Lambert W functions are multivalued. How do we know which branch to use?" IIRC, the branches of h, are split-level with-respect-to W, i.e. inside the Shell-Thron region you have to use W branch n, and outside the Shell-Thron region you have to use W branch n+-1 (I don't remember), if you want the real part/imaginary part of the result to fall within some specific range. – Andrew Robbins Aug 17 '12 at 14:18

(Edit-version 2)

If you have a base $b$ such that you look at $b^{b^{b^{...}}}$ then find a solution for $t$ such that $b = t^{1/t}$ If you have such a $t$, then look at its logarithm $u = \ln(t)$. If $|u| \le 1$ then the infinite tower is a convergent expression. Note that the function h(x), such that $t = h(b) \to t^{1/t} = b$ is multivalued and you take the principal value) I've made a picture about this "Shell-Thron-region" in the tetration-forum (the picture reflects only the upper halfplane, the full picture is symmetric around the x-axis).

The blue curve indicates the complex bases $b$ on the boundary between convergence and divergence of the resp. infinite powertower. Outside of this curve the powertower diverges. To each point on this curve there is another point $t$ in the complex plane associated (which is on the magenta curve). I connected some example points $b=t^{1/t}$ with a grey line. The yellow curve indicates the points $u$, the logs of the points $t$.

Inside (and on) the yellow circle (with radius 1 ) are all points $u$ whose exponentials $t$ are inside (and on) the magenta curve and whose associated bases $b$ are inside (and on) the blue curve -and thus whose infinite powertower with base b is convergent.
For values $b$ outside the blue curve, the corresponding $t$ are outside the magenta and the corresponding $u$ are outside the yellow curve the infinite powertower diverges.
(Note, that due to the multivaluedness we can have values $u$ and $t$ outside their curves which have corresponding $b$ inside the blue curve, but that doesn't matter for the question since there must then another value $u$ and $t$ inside their curved regions)

Your example is $b=i$, and that value is inside the blue curve, so the infinite powertower is convergent.

You might try using one of the following series expansions:

${a_1}^{{a_2}^{{.^{.^{a_n}}}}} = {\large\rm T}_{k=1}^{n} a_k = \sum_{k_j \ge 0 \atop 1 \le j \le n} \prod_{i=1}^{n}\frac{{(k_{i-1} \ln(a_i))}^{k_i}}{(k_i)!}$

which Barrow (link above) gives a variant of without logarithms, or

$a^{a^{a^{.^{.^{.}}}}} = \exp_a^{\infty}(1) = \sum_{k=0}^{\infty} \frac{\ln(a)^{k}}{k!} (k+1)^{(k-1)} = \sum_{k=0}^{\infty} \frac{(a - 1)^{k}}{k!} \sum_{j=0}^{k}\left[{k \atop j}\right] (j + 1)^{(j - 1)}$

which is just a substitution of variables in the Lambert-W series, the second series is just the Stirling transform of the first.