Solution
= Solution
The <Taylor series> of the <entire function> $g$ gives
$$
\frac{g(z)}{z-z_0}=\frac{g(z_0)}{z-z_0}+g'(z_0)+O(z-z_0).
$$
The coefficient of $(z-z_0)^{-1}$, and therefore the <residue>, is
$$
\boxed{\operatorname{Res}_{z=z_0}\frac{g(z)}{z-z_0}=g(z_0)}.
$$
If $g(z_0)=0$, the apparent <pole> is a <removable singularity> and the <residue> is zero.