The residue theorem for a null-homologous cycle

By [L1], the set $A:=\{a\in S:n(\mathrm{\Gamma },a)\mathrm{\ne }0\}$ is finite. For each $a\in A$ choose $r_a>0$ so small that the closed discs $\overline{D(a,r_a)}$ are pairwise disjoint, lie in $\mathrm{\Omega }$, meet no pole other than $a$, and are disjoint from ${\mathrm{\Gamma }}^{\ast }$. Let $C_a$ be the positively oriented circle $\mathrm{\mid }\zeta -a\mathrm{\mid }=r_a$.

Put [step 1.1, L4] $\mathrm{\Delta }:={\sum }_{a\in A}n(\mathrm{\Gamma },a)C_a\mathrm{.}$ For every $p\in \mathbb{C}\setminus (\mathrm{\Omega }\setminus A)$ one has $n(\mathrm{\Gamma },p)=n(\mathrm{\Delta },p)$. Indeed, if $p\mathrm{\notin }\mathrm{\Omega }$ then admissibility makes $n(\mathrm{\Gamma },p)=0$, and every $C_a$ lies in $\mathrm{\Omega }$, so gives $n(\mathrm{\Delta },p)=0$ as well. If $p=a\in A$, then [L4] gives $n(C_a,a)=1$ and $n(C_b,a)=0$ for $b\mathrm{\ne }a$, so $n(\mathrm{\Delta },a)=n(\mathrm{\Gamma },a)$. Therefore $\mathrm{\Gamma }$ and $\mathrm{\Delta }$ are homologous in $\mathrm{\Omega }\setminus A$.

The function $f$ is holomorphic on $\mathrm{\Omega }\setminus A$, so [L3] applied to step 2.1 yields ${\int }_{\mathrm{\Gamma }}f(z)dz={\int }_{\mathrm{\Delta }}f(z)dz={\sum }_{a\in A}n(\mathrm{\Gamma },a){\int }_{C_a}f(z)dz\mathrm{.}$

Each $C_a$ encloses only the pole $a$, so [L2] gives ${\int }_{C_a}f(z)dz=2\pi i\mathrm{Res}(f,a)\mathrm{.}$ Substituting this into step 3.1 proves the displayed formula. Since the set $A$ is finite, the residue sum has only finitely many nonzero terms.