[proofplan] We prove the two implications by comparing a pole with the reciprocal function. In the forward direction, the local normal form of a pole writes $f$ as a nonvanishing holomorphic factor divided by a positive power of $z-a$, and this forces the modulus to diverge. Conversely, if $|f(z)| \to \infty$, then $f$ is nonzero on a punctured neighborhood of $a$, so the reciprocal is holomorphic there and tends to $0$. The removable singularity theorem extends the reciprocal across $a$, and the finite-order factorization of a nonzero [holomorphic function](/page/Holomorphic%20Function) with a zero then gives the local pole form for $f$. [/proofplan] [step:Use the local pole form to force unbounded growth] Assume that $a$ is a pole of $f$. Throughout the proof, for $c \in \mathbb{C}$ and $s > 0$, write $B(c,s) := \{z \in \mathbb{C} : |z-c| < s\}$, and write $\mathbb{N} := \{1,2,3,\dots\}$. Since $U$ is open and $a \in U$, choose $r_U > 0$ such that $B(a,r_U) \subset U$. By the local form in the definition of a pole, after replacing the local radius by its minimum with $r_U$, there exist $r \in (0,r_U]$, an integer $m \in \mathbb{N}$, and a holomorphic map \begin{align*} g: B(a,r) \to \mathbb{C} \end{align*} such that $B(a,r) \subset U$, $g(a) \neq 0$, and \begin{align*} f(z) = \frac{g(z)}{(z-a)^m} \end{align*} for every $z \in B(a,r) \setminus \{a\}$. Since $g$ is continuous at $a$ and $g(a) \neq 0$, there exists $\rho \in (0,r]$ such that \begin{align*} |g(z)| \geq \frac{|g(a)|}{2} \end{align*} for every $z \in B(a,\rho)$. Hence, for every $z \in B(a,\rho) \setminus \{a\}$, \begin{align*} |f(z)| = \frac{|g(z)|}{|z-a|^m} \geq \frac{|g(a)|}{2|z-a|^m}. \end{align*} Let $M > 0$ be given. Define \begin{align*} \delta := \min\left\{\rho,\left(\frac{|g(a)|}{2M}\right)^{1/m}\right\}. \end{align*} If $z \in U$ and $0 < |z-a| < \delta$, then $z \in B(a,\rho) \setminus \{a\}$ and \begin{align*} |f(z)| \geq \frac{|g(a)|}{2|z-a|^m} > M. \end{align*} This proves $\lim_{z \to a}|f(z)| = \infty$. [guided] Assume that $a$ is a pole of $f$. For $c \in \mathbb{C}$ and $s > 0$, the notation $B(c,s)$ means the open disc $\{z \in \mathbb{C} : |z-c| < s\}$, and $\mathbb{N}$ means $\{1,2,3,\dots\}$. Since $U$ is open and $a \in U$, choose $r_U > 0$ such that $B(a,r_U) \subset U$. The definition of a pole gives the precise local structure we need: after decreasing the local radius if necessary, there are $r \in (0,r_U]$, an integer $m \in \mathbb{N}$, and a holomorphic map \begin{align*} g: B(a,r) \to \mathbb{C} \end{align*} with $B(a,r) \subset U$ and $g(a) \neq 0$ such that \begin{align*} f(z) = \frac{g(z)}{(z-a)^m} \end{align*} for every $z \in B(a,r) \setminus \{a\}$. The only possible obstruction to $|f(z)|$ becoming large would be cancellation from the numerator. But $g(a) \neq 0$, and holomorphic functions are continuous, so the numerator stays bounded away from $0$ near $a$. More precisely, continuity of $g$ at $a$ gives a number $\rho \in (0,r]$ such that \begin{align*} |g(z)-g(a)| < \frac{|g(a)|}{2} \end{align*} whenever $z \in B(a,\rho)$. By the [reverse triangle inequality](/theorems/2300), this implies \begin{align*} |g(z)| \geq |g(a)| - |g(z)-g(a)| > \frac{|g(a)|}{2}. \end{align*} Therefore, for every $z \in B(a,\rho) \setminus \{a\}$, \begin{align*} |f(z)| = \frac{|g(z)|}{|z-a|^m} \geq \frac{|g(a)|}{2|z-a|^m}. \end{align*} Now fix $M > 0$. To make the right-hand side larger than $M$, it is enough to require \begin{align*} |z-a|^m < \frac{|g(a)|}{2M}. \end{align*} Define \begin{align*} \delta := \min\left\{\rho,\left(\frac{|g(a)|}{2M}\right)^{1/m}\right\}. \end{align*} Then $0 < |z-a| < \delta$ implies $z \in B(a,\rho) \setminus \{a\}$ and \begin{align*} |f(z)| \geq \frac{|g(a)|}{2|z-a|^m} > M. \end{align*} This is exactly the punctured-neighborhood definition of $\lim_{z \to a}|f(z)| = \infty$ inside $U$. [/guided] [/step] [step:Construct a removable reciprocal from the infinite limit] Assume conversely that $\lim_{z \to a}|f(z)| = \infty$ in the punctured-neighborhood sense inside $U$. Since $U$ is open and $a \in U$, choose $r > 0$ such that $B(a,r) \subset U$. Applying the infinite-limit condition with $M = 1$, choose $\rho \in (0,r]$ such that \begin{align*} 0 < |z-a| < \rho \implies |f(z)| > 1. \end{align*} In particular, $f(z) \neq 0$ on $B(a,\rho) \setminus \{a\}$. Define \begin{align*} F: B(a,\rho) \setminus \{a\} \to \mathbb{C} \end{align*} by \begin{align*} F(z) = \frac{1}{f(z)}. \end{align*} Since $f$ is holomorphic and nonzero on $B(a,\rho) \setminus \{a\}$, the reciprocal rule for holomorphic functions shows that $F$ is holomorphic on $B(a,\rho) \setminus \{a\}$. The infinite-limit condition also gives $F(z) \to 0$ as $z \to a$: given $\varepsilon > 0$, apply the condition to $M = 1/\varepsilon$ to obtain $\eta > 0$ such that $0 < |z-a| < \eta$ and $z \in U$ imply $|f(z)| > 1/\varepsilon$, hence $|F(z)| < \varepsilon$. Therefore $F$ is bounded in a punctured neighborhood of $a$. By the Removable Singularity Theorem (citing a result not yet verified in the wiki: Removable Singularity Theorem), the map \begin{align*} \widetilde{F}: B(a,\rho) \to \mathbb{C} \end{align*} defined by $\widetilde{F}(a)=0$ and $\widetilde{F}(z)=F(z)$ for $z \neq a$ is holomorphic on $B(a,\rho)$. [/step] [step:Factor the zero of the reciprocal and recover the pole form] The holomorphic function $\widetilde{F}$ satisfies $\widetilde{F}(a)=0$. Also, $\widetilde{F}$ is not identically zero on $B(a,\rho)$, because $\widetilde{F}(z)=1/f(z) \neq 0$ for every $z \in B(a,\rho) \setminus \{a\}$. By the finite-order factorization theorem for zeros of holomorphic functions (citing a result not yet verified in the wiki: Zeros of holomorphic functions have finite order), there exist an integer $m \in \mathbb{N}$, a radius $\sigma \in (0,\rho]$, and a holomorphic map \begin{align*} h: B(a,\sigma) \to \mathbb{C} \end{align*} such that $h(a) \neq 0$ and \begin{align*} \widetilde{F}(z) = (z-a)^m h(z) \end{align*} for every $z \in B(a,\sigma)$. Since $h(a) \neq 0$ and $h$ is continuous, after decreasing $\sigma$ if necessary we may assume $h(z) \neq 0$ for every $z \in B(a,\sigma)$. Define \begin{align*} G: B(a,\sigma) \to \mathbb{C} \end{align*} by \begin{align*} G(z) = \frac{1}{h(z)}. \end{align*} The reciprocal rule for holomorphic functions shows that $G$ is holomorphic on $B(a,\sigma)$, and $G(a)=1/h(a) \neq 0$. For $z \in B(a,\sigma) \setminus \{a\}$, we have \begin{align*} \frac{1}{f(z)} = \widetilde{F}(z) = (z-a)^m h(z). \end{align*} Taking reciprocals gives \begin{align*} f(z) = \frac{G(z)}{(z-a)^m}. \end{align*} This is the local form of a pole of order $m$ at $a$. Hence $a$ is a pole of $f$. [/step]