Is it possible to give the representations of all possibile trancendental meromorphic solutions of the following functional equation:


$$f(z+1)-f(z)=0 \, .$$


Maybe it is well known for the experts. I have searched for some references for this equation, however, I did not find the answers.


Any reference and comments will be appreciated. 



transcendental entire or meromorphic solution for the function equation f(z+1)=f(z)
These functions satisfying $f(z)=f(z+1)$ are precisely the functions of the form $g(e^{2\pi iz})$ for $g$ holomorphic or meromorphic everywhere but possibly at $0$. That these functions satisfy $f(z)=f(z+1)$ is rather clear, so we need to show the converse.


For simplicity I'll look at functions satisfying $f(z+2\pi i)=f(z)$. Let $\operatorname{Log}\_1,\operatorname{Log}\_2$ we two branches of logarithm with domains $U\_1,U\_2$ such that $U\_1\cup U\_2=\mathbb C\setminus\{0\}$. Define $g(w)=f(\operatorname{Log}\_i w)$ for any $w\in U\_i,i=1,2$. If $w=U\_1\cap U\_2$, we note that $\operatorname{Log}\_1 w-\operatorname{Log}\_2w=2\pi i k$ for some $k\in\mathbb Z$, so $f(\operatorname{Log}\_1 w)=f(\operatorname{Log}\_2 w)$.


Therefore $g(w)$ is a well-defined function on $\mathbb C\setminus\{0\}$. It's clear that if $f$ is holomorphic or meromorphic on $\mathbb C$, then so is $g$ on $\mathbb C\setminus\{0\}$. Finally, we have $g(e^z)=f(\operatorname{Log}\_i e^z)=f(z+2\pi i k)=f(z)$ for some $k\in\mathbb Z$.


We can also show that all nonconstant such functions are transcendental: suppose $\sum\_{i=0}^n F\_i(z)f(z)^i=0$. Let this be the algebraic relation in which the maximum degree among all of the $F\_i$ is the smallest possible. Note that we then also have $\sum\_{i=0}^n F\_i(z+1)f(z)^i=\sum\_{i=0}^n F\_i(z+1)f(z+1)^i=0$, so $\sum\_{i=0}^n (F\_i(z+1)-F\_i(z))f(z)^i=0$. However, unless $F\_i$ is constant, $F\_i(z+1)-F\_i(z)$ has smaller degree and is nonzero, so we get a different algebraic relation, contradicting our choice above. It follows that $F\_i$ must all be constant, but then $f(z)$ must be constant as well (as it takes values in a finite set).


