Show that for every polynomial \(p(z)\) there exists a meromorphic function \(f(z)\) such that \[ f(z+1)=p(z)f(z). \]
Solution
For a nonzero polynomial, factor \[ p(z)=c\prod_{j=1}^d(z-a_j), \] with the roots repeated according to multiplicity. Choose \(\lambda\in\mathbb C\) such that \(e^\lambda=c\). Then \[ \boxed{ f(z)=e^{\lambda z}\prod_{j=1}^d\Gamma(z-a_j)} \] is a meromorphic solution. If \(p\equiv0\), take \(f\equiv0\).
1. Every nonzero polynomial can be put in the displayed factored form.
By the fundamental theorem of algebra, if \(p\) has degree \(d\), then \[ p(z)=c\prod_{j=1}^d(z-a_j) \] for some \(c\in\mathbb C^*\) and roots \(a_j\in\mathbb C\), counted with multiplicity. Since the exponential map is surjective onto \(\mathbb C^*\), there is \(\lambda\in\mathbb C\) with \(e^\lambda=c\).
2. The function\[ f(z)=e^{\lambda z}\prod_{j=1}^d\Gamma(z-a_j) \]is meromorphic on \(\mathbb C\).
The exponential factor is entire. Each function \(\Gamma(z-a_j)\) is meromorphic on \(\mathbb C\), with only isolated poles. A finite product of meromorphic functions is meromorphic, so \(f\) is meromorphic.
3. The function in 2 satisfies\[ f(z+1)=p(z)f(z). \]
The functional equation of \(\Gamma\) is \[ \Gamma(w+1)=w\Gamma(w). \] Therefore \[ \begin{aligned} f(z+1) &=e^{\lambda(z+1)} \prod_{j=1}^d\Gamma(z+1-a_j)\\ &=e^\lambda e^{\lambda z} \prod_{j=1}^d (z-a_j)\Gamma(z-a_j)\\ &=c\prod_{j=1}^d(z-a_j) \left(e^{\lambda z}\prod_{j=1}^d\Gamma(z-a_j)\right)\\ &=p(z)f(z). \end{aligned} \]
4. The zero polynomial also admits a meromorphic solution.
If \(p\equiv0\), take \(f\equiv0\). Then \[ f(z+1)=0=p(z)f(z) \] for every \(z\).