= Solution
Use the local factorization for a <pole> of order $n$:
$$
f(z)=(z-a)^{-n}h(z),\qquad h(a)\ne0.
$$
Shrink the disc so that the <holomorphic function> $h$ has no zeros there. Differentiating and dividing gives the <logarithmic derivative>
$$
\boxed{\frac{f'(z)}{f(z)}=-\frac n{z-a}+\frac{h'(z)}{h(z)}}.
$$
The second term is <holomorphic> across $a$. Since $n\geq1$, the first term cannot cancel. Hence $f'/f$ has a <simple pole>, and the coefficient of $(z-a)^{-1}$ in its <Laurent series> is
$$
\boxed{\operatorname{Res}_a(f'/f)=-n}.
$$
This is the pole case of the <integer residue of a logarithmic derivative>.
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