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Let $\{z_k\}_{k\in\mathbb{N}}$ be a sequence of poles of $f(z)$. Suppose that $\lim\limits_{k\to\infty}z_k=z_0$ and that $f$ is holomorphic in a neighborhood of $z_0$ except for $z_0$ and $\{z_k\}$. Prove that $f$ behaves near $z_0$ like near an essential singularity in the following sense

$$\forall\ w\in\mathbb{C}\ \exists\ \xi_n\to z_0:f(\xi_n)\to w$$

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closed as off-topic by user223391, Saad, Xander Henderson, JonMark Perry, Gibbs Jun 18 '18 at 7:24

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A slight modification of the standard proof of C-W works here. Say $V$ is an open set such that $f$ is holomorphic in $W=V\setminus(\{z_0\}\cup\{z_k\})$.

If the conclusion fails then there exist $a\in\Bbb C$, $r>0$ and $\delta>0$ such that $|f(z)-a|>\delta$ for every $z\in W$ with $|z-z_0|<r$. So $g=1/(f-a)$ is bounded in $W$, near $z_0$. Hence $g$ has a removable singularity at each $z_k$, $k\ge N$. Hence $g$ has an isolated singularity at $z_0$, which is hence removable.

Since $f$ has a pole at $z_k$ we must have $g(z_k)=0$; now since $z_k\to z_0$ and $g$ is holomorphic in a neighborhood of $z_0$ we must have $g=0$ near $z_0$, which is impossible.

(Exactly the same as the proof of C-W, except that in the actual C-W we know a priori that $g$ has an isolated singularity at $z_0$, while here we need to prove that by first removing the singularities at the $z_k$.)

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  • $\begingroup$ Thanks, I understand it in full. But, can we only say that $\exists m\in \mathbb{N}$ such that $\forall k\geq m$ we have that $z_k$ is a removable singularity ? This of course doesn't change the proof at all. $\endgroup$ – user554578 Jun 17 '18 at 17:01

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