Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 24F (Riemann Surfaces)

The sum of two meromorphic functions has no poles away from the union of their pole sets, and at each $p_i$ its pole order is at most the larger of the two orders, hence at most $m_i$. Scalar multiplication cannot increase a pole order. The zero function belongs to the set, so $V$ is a complex vector space.

Choose a local coordinate $z_i$ at each $p_i$ and map $f\in V$ to all coefficients of the negative-power terms in its Laurent expansions:The kernel consists of globally holomorphic functions on the compact Riemann surface $R$. By the maximum modulus principle, every such function is constant, so the kernel has dimension one. Rank--nullity givesFor $n=0$, this simply says $\dim V=1$.

Lift $f$ to a doubly periodic meromorphic function on $\mathbb{C}$. Integrate it around a fundamental parallelogram whose boundary avoids all poles. Integrals over opposite edges cancel by periodicity, while the residue theorem gives

Thus the principal-part map from part (b) takes values in the codimension-one hyperplane on which the sum of residue coefficients is zero. The Mittag--Leffler existence criterion on a compact Riemann surface says that prescribed principal parts occur precisely when their residues pair trivially with every holomorphic one-form. On a complex torus the holomorphic one-forms are the scalar multiples of $dz$, so the single condition is exactly the displayed residue sum. Hence the image has dimension
${\sum }_i{m}_i-1$, and the kernel of constants has dimension one. For $n\ge 1$,Equivalently, this is the genus-one case of the Riemann-Roch theorem for a positive divisor.