Let $f$ be a holomorphic function in an open $D \subset \mathbb C$ except for a sequence {$p_n$} of poles. Show that ...

Question

Let $f$ be a holomorphic function in an open $D \subset \mathbb C$ except for a sequence {$p_n$} of poles. Show that {$p_n$} does not converge to any point {$p_0$} $\in D.$

Knowing that a holomorphic function is a complex-valued function of one or more complex variables that is complex differentiable in a neighborhood of every point in its domain. So a function is holomorphic at $z_0$ if $lim_{z \to z_0} \frac{f(z)-f(z_0)}{z-z_0}$. But $f$ is not holomorphic for the sequence, so then the elements of the sequence cannot converge to elements of $D$ so that $lim_{p_n \to p_0} \frac{f(p_n)-f(p_0)}{p_n-p_0}$ does not exist for any point. Is this what i need to do? If so,The fact that $f$ is not holomorphic in the sequence does not guarantee this.

Grateful for the attention.

Answer 1

When the power series $\sum_{n\ge0}c_n(z-z_0)^n$, converges to $f(z)$, it always does so on the open disk $\lvert z-z_0\rvert \lt r$, where $r=\limsup _{n\to\infty}\dfrac 1{\lvert c_n\rvert ^{1/n}}$. This implies that the singular points are isolated.

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In the case of several complex variables, $f(z_1,\dots, z_n)=\sum_{k_1,\dots, k_n=0}^\infty c_{k_1,\dots, k_n}(z_1-a_1)^{k_1}\dots (z_n-a_n)^{k_n}$, and an analogous statement can be made.

Answer 2

If $(p_n)$ is a constant sequence then it does converge. If we assume that $(p_n)$ is a sequence of distinct poles of $f$ then the result is true. In this case $p_n \to p$ implies that $p=p_n$ for at most one n value of $n$, say $n=n_0$ and $p\notin \{p_k: k >n_0\}$. But then $f$ is analytic at $p$ which implies that there is a disk around $p$ in which $f$ is bounded. But then this disk cannot contain any pole $p_k$ so $p_n$ does not converge to $p$.