Solving equation system of complex funtions

Question

Does there exist two complex functions $f$ and $g$ satisfy below equation system? $$ \begin{cases} f=e^g\\ g=e^f \end{cases} $$ What about analytic funtions?

Answer 1

Let $z$ be a solution to the equation $z=\exp(z)$, for example $z = -W(-1)$, where $W$ is a branch of the Lambert function. Numerically, one such value is $z \approx 0.3181315052 - 1.337235701 i$. Put $f = g = z$ for a solution to your problem.

Answer 2

Here is a partial answer: if there exist analytic functions $f$ and $g$ satisfying the equations, they must be constants. Suppose the equations are satisfied by analytic functions. Then $f'=e^g g',\ g' = e^f f' = e^{f+g}g'$, so $g'(e^{f+g}-1)=0$. If $g'=0$, then $g$ is constant and so is $f$. If $e^{f+g}-1=0$, then since $f+g$ is analytic (and continuous), $f+g$ must be constant, say, $f+g=c$. Then $f'=-g'$, so $f'=e^g g' = -g'$, $g'(e^g + 1)=0$, and either $g$ and $f$ are constants, or $e^g = -1$. If $e^g =-1$, then, since $g$ is continuous, $g$ must be constant, and so is $f$.

Now it remains to prove that there exists a constant pair of solutions $f, g$.

Answer 3

Merging the equation system, we get $f=e^{e^f}$. Mutiplying at both sides of this equation by $e^f$, i.e., $fe^f=e^fe^{e^f}$, let $fe^f=\zeta$, then $f=W(\zeta)$ and $g=e^f=W(\zeta)$, $f$ and $g$ maybe belong to different branch of $W(\zeta)$.