= Solution
By the finite-order <order of a zero of a holomorphic function>[zero] factorizations,
$$
f(z)=(z-a)^mF(z),\qquad g(z)=(z-a)^nG(z),\qquad F(a)G(a)\neq0,
$$
with $F,G$ holomorphic near $a$. Shrinking the disk if necessary, continuity ensures $G$ never vanishes there, so $F/G$ is holomorphic and nonzero at $a$. Consequently
$$
\frac{f(z)}{g(z)}=(z-a)^{m-n}\frac{F(z)}{G(z)}.
$$
If $n>m$, the exponent is $-(n-m)$, with a holomorphic nonvanishing factor. Hence
$$
\boxed{f/g\text{ has a pole of order }n-m.}
$$
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