= Solution
The <simple zero> assumption gives the local factorization
$$
h(z)=(z-a)k(z),\qquad k(a)=h'(a)\ne0,
$$
where $k$ is <holomorphic>. In a small disc $k$ stays nonzero, so $q=g/k$ is <holomorphic> there. Expanding $q$ to first order,
$$
\frac{g(z)}{h(z)}=\frac{q(z)}{z-a}=\frac{q(a)}{z-a}+\text{a function holomorphic near }a.
$$
The <residue> is therefore
$$
\boxed{\operatorname{Res}_a\left(\frac gh\right)=q(a)=\frac{g(a)}{h'(a)}}.
$$
If $g(a)=0$, the quotient has a <removable singularity> rather than a <simple pole>, and the same formula correctly gives zero. This is the <residue formula for a quotient at a simple denominator zero>.
Back to article page
