andydude Wrote:See Analytic solution of F(z+1)=exp(F(z)) in complex z-plane for more information.
About the uniqueness: It is well known that if we have a solution \( \alpha \) of the Abel equation \( \alpha(z+1)=f(\alpha(z)) \) then for any 1-periodic function \( \phi \) also \( \beta(z)=\alpha(z+\phi(z)) \) is a solution to the Abel equation. (Because \( \beta(z+1)=\alpha(z+1+\phi(z+1))=\alpha(z+\phi(z)+1)=f(\alpha(z+\phi(z))=f( \beta( z)) \)).
So let \( F \) be one solution of
(*) \( F(z+1)=\exp(F(z)) \) with
(**) \( \lim_{y\to\infty} F(x+iy) = L \) and \( \lim_{y\to -\infty} F(x+iy)=L^\ast \)
then \( G(z)=G(z+\sin(2\pi z)) \) is another solution of (*). Let us now consider (**). We know that \( \sin(z)=-i\frac{e^{iz}-e^{-iz}}{2} \) and \( \sin(x+iy)=i\frac{-e^{ix}e^{-y}+e^{-ix}e^y}{2} \)
\( G(x+iy)=F\left(x+iy+i\frac{-e^{2\pi ix}e^{-2\pi y}+e^{-2\pi ix}e^{2\pi y}}{2}\right) \). As \( e^{-2\pi y}\to 0 \)
at least for x=0 also
\( \lim_{y\to\infty} G(iy)=\lim_{y\to\infty} F(i(y+e^{2\pi y}))=L \)
[edit] fixed some negligences. [/edit]