Solution
= Solution
A <holomorphic function> has a zero of order $n$ at $a$ when it is <holomorphic> on a neighbourhood of $a$ and factors as
$$
\boxed{f(z)=(z-a)^n h(z),\qquad h\text{ holomorphic},\quad h(a)\ne0}.
$$
Equivalently, $f(a)=f'(a)=\cdots=f^{(n-1)}(a)=0$ and $f^{(n)}(a)\ne0$. The <order of a zero of a holomorphic function> is the degree of the first nonzero term of its <Taylor series>.