[proofplan] The proof is a direct translation between the local factorisation that defines a pole and the local factorisation that defines a zero. If $f$ has a pole of order $m$, then locally $f(z)=(z-a)^{-m}g(z)$ with $g$ holomorphic and nonzero at $a$; after shrinking the ball, $g$ is nowhere zero, so $1/f=(z-a)^m/g(z)$ extends holomorphically and vanishes to order $m$. Conversely, if $1/f$ extends as a [holomorphic function](/page/Holomorphic%20Function) with a zero of order $m$, factor that extension as $(z-a)^m h(z)$ with $h(a)\ne 0$, shrink so $h$ is nonzero, and invert the factorisation to recover a pole of order $m$. [/proofplan]

[step:Invert the local factorisation of a pole]Assume first that $a$ is a pole of order $m$ of $f$. By the definition of a pole of order $m$, there exist $\rho > 0$ and a holomorphic function $g: B(a,\rho) \to \mathbb{C}$ such that $B(a,\rho) \subset U$, $g(a)\ne 0$, and \begin{align*} f(z)=\frac{g(z)}{(z-a)^m} \end{align*} for every $z \in B(a,\rho)\setminus\{a\}$. Since $g$ is continuous at $a$ and $g(a)\ne 0$, there exists $r$ with $0<r\le \rho$ such that $g(z)\ne 0$ for every $z \in B(a,r)$. Therefore $f(z)\ne 0$ for every $z \in B(a,r)\setminus\{a\}$. Define $F: B(a,r)\to \mathbb{C}$ by \begin{align*} F(z)=\frac{(z-a)^m}{g(z)}. \end{align*} The function $F$ is holomorphic on $B(a,r)$ because $g$ has no zeros there. For $z \in B(a,r)\setminus\{a\}$, the defining factorisation of $f$ gives \begin{align*} F(z)=\frac{1}{f(z)}. \end{align*} Moreover $F(z)=(z-a)^m h(z)$ with $h: B(a,r)\to\mathbb{C}$ defined by $h(z)=1/g(z)$, and $h(a)=1/g(a)\ne 0$. Hence $F$ has a zero of order $m$ at $a$.[/step]

[guided]Assume that $a$ is a pole of order $m$ of $f$. The definition of a pole of order $m$ says precisely that the singular part of $f$ is one factor of $(z-a)^{-m}$ and that the remaining factor is holomorphic and nonzero at $a$. Thus there are $\rho>0$ and a holomorphic function $g: B(a,\rho)\to\mathbb{C}$ such that $B(a,\rho)\subset U$, $g(a)\ne 0$, and \begin{align*} f(z)=\frac{g(z)}{(z-a)^m} \end{align*} for every $z\in B(a,\rho)\setminus\{a\}$. The reciprocal of $f$ will involve $1/g(z)$, so we need $g$ to be nonzero on a whole neighbourhood of $a$, not only at $a$. Since $g$ is holomorphic, it is continuous. Because $g(a)\ne 0$, continuity gives a radius $r$ with $0<r\le \rho$ such that $g(z)\ne 0$ for every $z\in B(a,r)$. On the punctured ball $B(a,r)\setminus\{a\}$, we then also have $f(z)\ne 0$, because $(z-a)^m\ne 0$ and $g(z)\ne 0$ there. Now define the proposed holomorphic extension of the reciprocal as the map $F: B(a,r)\to\mathbb{C}$ given by \begin{align*} F(z)=\frac{(z-a)^m}{g(z)}. \end{align*} This function is holomorphic on all of $B(a,r)$: the numerator $z\mapsto (z-a)^m$ is a polynomial, and the denominator $g$ is holomorphic and nowhere zero on $B(a,r)$, so $1/g$ is holomorphic there. For every $z\in B(a,r)\setminus\{a\}$, the pole factorisation gives \begin{align*} \frac{1}{f(z)}=\frac{(z-a)^m}{g(z)}=F(z). \end{align*} Thus $F$ is exactly the holomorphic extension of the reciprocal. Finally, $F$ has the required order of vanishing at $a$. Indeed, define $h: B(a,r)\to\mathbb{C}$ by $h(z)=1/g(z)$. Then $h$ is holomorphic, $h(a)=1/g(a)\ne 0$, and \begin{align*} F(z)=(z-a)^m h(z). \end{align*} By the definition of a zero of order $m$, this says that $F$ has a zero of order $m$ at $a$.[/guided]

[step:Invert the local factorisation of a zero] Conversely, suppose there exists $r>0$ such that $B(a,r)\subset U$, $f(z)\ne 0$ for every $z\in B(a,r)\setminus\{a\}$, and the reciprocal map on $B(a,r)\setminus\{a\}$ extends to a holomorphic function $F: B(a,r)\to\mathbb{C}$ with a zero of order $m$ at $a$. By the definition of a zero of order $m$, there exists a holomorphic function $h: B(a,r)\to\mathbb{C}$ such that $h(a)\ne 0$ and \begin{align*} F(z)=(z-a)^m h(z) \end{align*} for every $z\in B(a,r)$. Since $h$ is continuous at $a$ and $h(a)\ne 0$, choose $s$ with $0<s\le r$ such that $h(z)\ne 0$ for every $z\in B(a,s)$. For $z\in B(a,s)\setminus\{a\}$, the extension property gives $F(z)=1/f(z)$, hence \begin{align*} f(z)=\frac{1}{F(z)}=\frac{1}{(z-a)^m h(z)}=\frac{1/h(z)}{(z-a)^m}. \end{align*} Define $g: B(a,s)\to\mathbb{C}$ by $g(z)=1/h(z)$. Then $g$ is holomorphic on $B(a,s)$ and $g(a)=1/h(a)\ne 0$. Therefore \begin{align*} f(z)=\frac{g(z)}{(z-a)^m} \end{align*} for every $z\in B(a,s)\setminus\{a\}$. By the definition of a pole of order $m$, $a$ is a pole of order $m$ of $f$. [/step]