= Solution
A <order of a zero of a holomorphic function>[zero] of order $n\geq1$ at $a$ means that $f$ is holomorphic near $a$ and
$$
f(a)=f'(a)=\cdots=f^{(n-1)}(a)=0,\qquad f^{(n)}(a)\neq0.
$$
Equivalently its <Taylor series> starts with the nonzero coefficient of $(z-a)^n$, so
$$
\boxed{f(z)=(z-a)^n h(z),\qquad h\text{ holomorphic near }a,\ h(a)\neq0.}
$$
Unlike a <pole>, a zero is a regular point of the function itself; its reciprocal has a pole of the corresponding order.
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