Weak theorem about an almost identical function: PROOF

[Kouznetsov]

Hi Dmitrii, though I didnt completely follow your proof it triggered coming up with a much more simplified proof. As a prerequisite we only need Picard's big theorem:

Picard's big theorem applied to entire functions. Each entire non-polynomial function takes on every complex number with at most one exception infinitely often.

First it can be shown that \( J \) (as an entire function) does not even omit one value from \( \mathbb{C} \). (I give the proof in a next post, though this assertion is not really necessary for the following conclusions.)
We know that \( J(k)=k \) for each \( k\in\mathbb{Z} \), but by the above theorem for each \( k\in\mathbb{Z} \) there have to be infinitely other \( z\in \mathbb{C}\setminus \mathbb{Z} \) with \( J(z)=k \).
Hence \( J(\mathbb{C}\setminus\mathbb{Z})=\mathbb{C} \).

That means if we have a function \( g(z)=f(J(z)) \), where \( f \) is a superexponential with singularities only at \( \{z\in\mathbb{Z}:z\le -2\} \), then \( g \) has singularities outside \( \mathbb{C}\setminus\{z\in\mathbb{Z}:z\le -2\} \).

Henryk, I like your proof. Indeed, it is shorter.

Now, please, prove that some of singularities are in the right hand side of the complex plane.

Small hint:
From the asymptotic behavior, at 1<b<exp(1/e), tetration is periodic and the period is imaginary. This means, that J(z)=F^(-1)(G(z)), is also periodic.
Can it be, that h(z) is periodic (with period unity) and j(z)=h(z)+z is also periodic (with imaginary period)?

Is it possible to unwrap the periodic function with some arcsin in such a way, that the resulting function is not periodic but still entire?