Let \(E=\mathbb C/\Lambda\), let \([z_i]\) be finitely many distinct points of \(E\), and prescribe at each \(z_i\) the principal part \[ F_i(z)=\sum_{j=1}^{N_i} a_{i,j}(z-z_i)^{-j}. \] Prove that there exists a meromorphic function \[ f:E\to\widehat{\mathbb C} \] with poles at the \([z_i]\) and these prescribed principal parts if and only if \[ \sum_i a_{i,1}=0. \]
Hint: use the residue theorem in one direction and translates of the Weierstrass zeta function in the other.
Solution
A meromorphic function with the prescribed principal parts exists exactly when \[ \boxed{\sum_i a_{i,1}=0.} \]
1. The residue condition is necessary.
Suppose \(f\) is meromorphic on the compact torus \[ E=\mathbb C/\Lambda \] and has the prescribed principal parts. The holomorphic differential \(dz\) on \(\mathbb C\) is \(\Lambda\)-invariant, so it descends to \(E\). At \([z_i]\) the residue of the meromorphic differential \(f(z)\,dz\) is exactly the coefficient of \((z-z_i)^{-1}\) in the Laurent expansion of \(f\), namely \[ \operatorname{Res}_{[z_i]}(f\,dz)=a_{i,1}. \] The sum of the residues of a meromorphic differential on a compact Riemann surface is zero. Therefore \[ \sum_i a_{i,1}=0. \]
2. Derivatives of \(\zeta\) provide elliptic functions with a single prescribed higher-order principal term.
Near \(0\), the Weierstrass zeta function has expansion \[ \zeta(w)=\frac1w+h(w), \] where \(h\) is holomorphic. Hence for every \(j\ge2\), \[ \zeta^{(j-1)}(w) =(-1)^{j-1}(j-1)!\,w^{-j}+\text{holomorphic terms}. \] Define \[ \phi_{i,j}(z) =\frac{(-1)^{j-1}}{(j-1)!} \zeta^{(j-1)}(z-z_i) \qquad(j\ge2). \] Then \(\phi_{i,j}\) has principal part \[ (z-z_i)^{-j} \] at \(z_i\).
Moreover, for every \(\lambda\in\Lambda\) the zeta quasi-periodicity has the form \[ \zeta(z+\lambda)=\zeta(z)+\eta(\lambda). \] Differentiating once kills the additive constant, so every derivative \(\zeta^{(k)}\) with \(k\ge1\) is \(\Lambda\)-periodic. Thus each \(\phi_{i,j}\) is elliptic and descends to a meromorphic function on \(E\).
3. If \(\sum_i a_{i,1}=0\), the simple-pole part is also elliptic.
Set \[ F_1(z)=\sum_i a_{i,1}\zeta(z-z_i). \] For \(\lambda\in\Lambda\), \[ \begin{aligned} F_1(z+\lambda)-F_1(z) &=\sum_i a_{i,1} \bigl(\zeta(z-z_i+\lambda)-\zeta(z-z_i)\bigr)\\ &=\eta(\lambda)\sum_i a_{i,1}\\ &=0. \end{aligned} \] Therefore \(F_1\) is \(\Lambda\)-periodic. Since each translate \(\zeta(z-z_i)\) has principal part \((z-z_i)^{-1}\) at \(z_i\), the principal part of \(F_1\) at \(z_i\) is \[ a_{i,1}(z-z_i)^{-1}. \]
4. Summing the simple- and higher-order pieces realizes all prescribed principal parts.
Assume \[ \sum_i a_{i,1}=0 \] and define \[ f(z) =\sum_i a_{i,1}\zeta(z-z_i) +\sum_i\sum_{j=2}^{N_i} a_{i,j}\frac{(-1)^{j-1}}{(j-1)!} \zeta^{(j-1)}(z-z_i). \] By 2 and 3, \(f\) is \(\Lambda\)-periodic, so it descends to a meromorphic function on \(E\).
Fix \(i\). Because the points \([z_i]\) are distinct in \(E\), every summand centered at \(z_k\) with \(k\ne i\) is holomorphic near \(z_i\). The terms centered at \(z_i\) have, by 2 and 3, principal parts \[ a_{i,1}(z-z_i)^{-1} \quad\text{and}\quad a_{i,j}(z-z_i)^{-j} \quad(2\le j\le N_i). \] Hence the complete principal part of \(f\) at \(z_i\) is exactly \[ \sum_{j=1}^{N_i}a_{i,j}(z-z_i)^{-j}=F_i(z). \] This proves sufficiency and completes the equivalence.