= Solution
A <simple zero> means $h(a)=0$ and $h'(a)\ne0$. The local factorization is $h(z)=(z-a)h_1(z)$, where $h_1$ is <holomorphic> and $h_1(a)=h'(a)\ne0$. Thus
$$
\frac{g(z)}{h(z)}=\frac{g(z)/h_1(z)}{z-a}.
$$
The numerator is <holomorphic> near $a$, so its constant Taylor coefficient is the coefficient of $(z-a)^{-1}$ in the quotient. Therefore
$$
\boxed{\operatorname{res}\left(\frac gh;a\right)=\frac{g(a)}{h'(a)}.}
$$
If $g(a)=0$, the possible <pole> cancels and the <residue> is zero; the formula does not require $g/h$ actually to have a <pole>. This is the <residue formula for a quotient at a simple denominator zero>.
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