= Solution
A <simple zero> of an <analytic function> has a local factorization $f(z)=(z-z_0)u(z)$, where $u$ is an <analytic function> and $u(z_0)=f'(z_0)\ne0$. Since $1/u$ is also an <analytic function> near $z_0$, its <Taylor series> gives
$$
\frac1{f(z)}=\frac1{z-z_0}\left[\frac1{f'(z_0)}+O(z-z_0)\right].
$$
The nonzero coefficient of $(z-z_0)^{-1}$ proves that this is a <simple pole>, with
$$
\boxed{\operatorname{Res}_{z=z_0}\frac1{f(z)}=\frac1{f'(z_0)}.}
$$
Multiplying by the <Taylor series> $g(z)=g(z_0)+O(z-z_0)$ similarly gives
$$
\boxed{\operatorname{Res}_{z=z_0}\frac{g(z)}{f(z)}=\frac{g(z_0)}{f'(z_0)}.}
$$
The assumption $g(z_0)\ne0$ ensures that the <pole> is not canceled. In either formula the <residue> is precisely the coefficient of $(z-z_0)^{-1}$ in the <Laurent series>.
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