Consider an expression of the form

ζ = α^αα…
 

where α is a complex number.

The infinitude may be expressed as the limit of an iteration of the form

ζ_n = α^ζ_n-1

Thus

ζ = lim_n→∞ ζ_n

The equation satisfied by ζ ;is

ζ = α^ζ
and thus α = ζ^1/ζ

The way to find the complex root is to multiply the ratio 1/ζ top and bottom by the complex conjugate of ζ, ζ*. This results in

α = ζ^ζ*/|ζ|²

For example, suppose a value of α is sought such that the infinite exponentiation converges to 1+i. The complex conjugate of 1+i is 1-i and |1+i|²=2. We then want to find

(1+i)^1/(1+i) = (1+i)^(1-i)/2 = 1.71896 + 0.38333i

The values of the iterations are

It does indeed appear that the iteration is converging to 1+i.

Now consider ζ=i. The complex conjugate of i is −i and |ii*|=1. Thus α would be i^-i. Remarkably i^-i is the real number 4.81048…. The exponential iteration of this number will not produce a complex number and will not converge.

A complex number may be expressed in three different forms. One is the rectangular form as x+iy, (where i is the square root of −1). A second is in polar form as Re^iθ. A third is the logarithmic form as e^ρ+iθ.

Arithmetic operations on two complex numbers can be simplified by choosing the proper forms for the two number. The addition α+β can be carried out most simply if both numbers are in the rectangular form; i.e., if α=x+iy and β=w+iz then α+β=(x+w)+i(y+z). Although multiplication of two complex numbers in rectangular forms is not very difficult multiplication is even easier if both are in polar form or both are in logarithmic form. If α=Re^iθ and β=re^iφ then α*β = (Rr)e^i(θ+φ). If α=e^ρ+iθ and β=e^σ+iφ then α*β = e^{(ρ+σ)+i(θ+φ)}.

For some operations, such as exponentiation, the convenient forms may be different for α and β. For exponentiation let α=e^ρ+iθ and β=w+iz. Then

ζ = α^β = (e^ρ+iθ)^w+iz = e^{(ρ+iθ)(w+iz)}
which reduces to
ζ = e^{(ρw−θz)+i(ρz+θw)}
 

The polar form of ζ is Re^i(ρz+θw) where R=e^ρw−θz. The rectangular form is then Rcos(ρz+θw)+iRsin(ρz+θw).

By way of contrast let α=Re^iθ and β=x+iy. Then

ζ = (Re^iθ)^x+iy = R^x+iye^iθ(x+iy)
= R^xR^iye^iθx−θy = R^xe^iyln(R)e^iθxe^−θy
which upon combining the like terms gives
=[R^xe^−θy] e^{i(yln(R)+xθ)}
or, equivalently

=[e^xln(R)−θy] e^{i(yln(R)+xθ)}
 

This is the polar form of ζ. The rectangular form of ζ is

e^ln(R)x−θycos(yln(R)+xθ) + ie^ln(R)x−θysin(yln(R)+xθ)
 

This works but clearly the logarithmic version is simpler.

Below is a calculator which implements complex exponentiation from the rectangular form of the complex numbers.