An infinite exponentiation is something which is raised to a power which is something raised to a power ad infinitum> Suppose

G = a^aa…
which can be represented as
G = a^G

However a little manipulation turns it into a seemingly trivial problem. The manipulation is to take the G-th root of both sides giving

G^1/G = a

Now if we want a value of a that gives G as a solution we need only take the G-th root of G and we have the answer. For example, for G=2, a=2^½=√2. Thus

√2^√2√2… = 2

A previous study worked out the analysis when G is a real number It was found that there is convergence if and only if a<e^1/e where e=2.721828….

When a and G are complex number the equation a=G^1/G is actually two equations. Let G=R·e^iΦ and a=r·^iθ. Then

r·^iθ = (R·e^iΦ1/R·eiΦ
or, expressed in
terms of logarithms
ln(r) + iθ = (e^−iΦ/R)[ln(R) + iΦ)
or, equivalently
ln(r) + iθ = (cos(Φ) −i·sin(Φ)/R)[ln(R) + iΦ]

The RHS may be rearranged in terms real and imaginary components as

(cos(Φ)ln(R) + Φ·sin(Φ))/R

Thus

ln(r) = (cos(Φ)ln(R) + Φ·sin(Φ))/R
and
θ = (cos(Φ)ln(R) + Φ·sin(Φ))/R

The real component is maximized where d(ln(r))/dR = 0; i.e.,

(ln(R)/R² − 1/R² = 0
and that is where
ln(R) − 1 = 0'
and hence
R = e

(To be continued.)

Setting this equal to zero

G^{(1/G -2)} − ln(G)G^1/G/G² = 0
and dividing G^1/G yields
1/G² − ln(G)/G² = 0
multiplying by G²
further reduces it to
ln(G) = 1
which means
G = e

The maximum value of the function G^1/G is then e^1/e.