Let $X=\Bbb C_\infty$ be the Riemann sphere $C\cup \{\infty\}$. Given finitely many points $p_1,\dots,p_n\in X$ and a corresponding complex numbers $r_1,\dots,r_n$, can we construct a meromorphic $1$-form $\omega$ on $X$ with simple poles at $p_i$'s with residue $r_i$, and with no other poles? Of course we assume $\sum_i r_i=0$ (residue theorem).
Let $z$ be the coordinate on $\Bbb C$ and $w=1/z$ be the coordinate on $X-\{0\}$. If $p_i\neq \infty$ for all $i$, then we can consider $\omega=\sum_{i=1}^n \dfrac{r_i}{z-p_i}dz$. At $\infty$ we have $$\omega=\sum_{i=1}^n \dfrac{r_i }{w(p_iw-1)}dw=\sum_{i=1}^n\left(\frac{-r_i}{w}+\frac{r_ip_i}{p_iw-1} \right)dw$$ and the condition $\sum r_i=0$ implies that $\omega$ is holomorphic at $\infty$, so we are done. But how can we handle the case when some $p_i=\infty$?
Note that each $\frac{r_i}{z-p_i}$ term has two poles: one at $p_i$ and one at $\infty$, with residues $r_i$ and $-r_i$ respectively. With this in mind, consider what happens when the $p_i=\infty$ term is simply dropped from the sum.