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Show that the set of all poles of $f(z) = \tan(\frac{\pi z}{2})$ is $P = \{2k+1: k\in \mathbb Z \}$. Also, show that every $p\in P$ is a simple pole of $f$, i.e. a pole of order $1$.

My effort:
I will state some definitions I am using. $f$ has a pole at $p$ if there exists $R > 0$ such that $f$ does not vanish on $D(p,R)\setminus \{p\}$, and the function $\frac1f$ defined as $0$ at $p$ is holomorphic on $D(p,R)$. As usual, $D(p,R) = \{z\in \mathbb C: |z-p| < R\}$. If $f$ has a pole at $p$, there exists $R > 0$, a holomorphic function $h: D(p,R) \to \mathbb C$ with $h(z)\ne 0$ for all $z\in D(p,R)$ and a unique positive integer $n$ such that $f(z) = (z-p)^{-1}h(z)$ for all $z\in D(p,R)\setminus \{p\}$. $n$ is called the order or multiplicity of the pole.

  1. Every $p\in P$ is a pole of $f$.

Take $p = 2k+1$ for some $k\in \mathbb Z$. Then, $f$ is not defined at $p$. In fact, $|f|\to\infty$ as $z\to p$. How do I find $R > 0$ such that $\frac1f$ is holomorphic on $D(p,R)$?

  1. Every $p\in P$ is a simple pole of $f$.

To show this, we should be able to write $f(z) = (z-p)^{-1}h(z)$ for some holomorphic function $f$, in some punctured disc around $p$.

  1. The set $P$ consists of all poles of $f$, i.e. if $p$ is a pole of $f$, then $p\in P$.

This is probably the hardest to show? I am confused with the definition of a pole. Is $|f|\to \infty$ at $p$ a requirement for $p$ to be a pole?


Reference:

Order of a zero: Suppose that $f$ is holomorphic in a connected open set $\Omega$, has a zero at a point $z_0\in \Omega$, and does not vanish identically on $\Omega$. Then there is a neighborhood $U\subset\Omega$ of $z_0$, a non-vanishing holomorphic function $g$ on $U$, and a unique positive integer $n$ such that $f(z) = (z-z_0)^n g(z)$ for all $z\in U$. We say that $f$ has a zero of order or multiplicity $n$. If $n = 1$, this is a simple zero.

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    $\begingroup$ An isolated singularity $p$ is a pole of $f$ if and ony if $|f(z)| \to \infty$ as $z \to p$. $\endgroup$ Sep 21, 2021 at 9:36
  • $\begingroup$ I see. That shows that every $p\in P$ is a pole of $f$, right? What about the other aspects of the problem? Thank you. @KaviRamaMurthy $\endgroup$ Sep 21, 2021 at 9:39
  • $\begingroup$ $p$ is a pole of order $1$ if $(z-p)f(z)$ has a finite non-zero limit as $z \to p$. $\endgroup$ Sep 21, 2021 at 9:42

1 Answer 1

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  1. If $p=2k+1$ for some $k\in\Bbb Z$, then\begin{align}\tan\left(\frac{\pi z}2\right)&=\tan\left(\frac\pi2\bigl((z-2k-1)+2k+1\bigr)\right)\\&=\tan\left(\frac\pi2(z-2k-1)+k\pi+\frac\pi2\right)\\&=\tan\left(\frac\pi2(z-2k-1)+\frac\pi2\right)\\&=-\cot\left(\frac\pi2(z-2k-1)\right)\end{align}and therefore$$\frac1{\tan\left(\frac{\pi z}2\right)}=-\tan\left(\frac\pi2(z-2k-1)\right).$$And $-\tan\left(\frac\pi2(z-2k-1)\right)$ is a holomorphic function which maps $2k+1$ into $0$.
  2. Since$$\left(-\tan\left(\frac\pi2(z-2k-1)\right)\right)'=-1-\tan^2\left(\frac\pi2(z-2k-1)\right),$$which is equal to $-1$ when $z=2k+1$, $2k+1$ is a simple zero of $1/\tan\left(\frac{\pi z}2\right)$. So, near $2k+1$, you can write $1/\tan\left(\frac{\pi z}2\right)$ as $(z-2k-1)\varphi(z)$, where $\varphi$ is holomorphic and $\varphi(2k+1)\ne0$. So, near $2k+1$,\begin{align}\tan\left(\frac{\pi z}2\right)&=\frac1{(z-2k-1)\varphi(z)}\\&=(z-2k-1)^{-1}h(z),\end{align}where $h(z)=\frac1{\varphi(z)}$.
  3. There are no other poles since$$\tan\left(\frac{\pi z}2\right)=\frac{\sin\left(\frac{\pi z}2\right)}{\cos\left(\frac{\pi z}2\right)}$$and $\cos$ is defined everywhere and its set of zeros is the set $P$.
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  • $\begingroup$ Thank you for your answer. Could you please revisit my definitions of poles and their order in the edited post once? I would also like to know if I am missing anything in the definition of a pole, namely the fact that $p$ is a pole $\implies |f(z)| \to \infty$ as $z\to p$. $\endgroup$ Sep 21, 2021 at 9:55
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    $\begingroup$ There is nothing missing in your definition of pole. And it turns out that $f$ has a pole at $p$ if and only if $\lim_{z\to p}|f(z)|=\infty$ (this is property; it's not part of the definition). $\endgroup$ Sep 21, 2021 at 9:58
  • $\begingroup$ Is that a consequence of the fact that $\frac{1}{f}$ on $D(p,R)$ is a holomorphic extension of $1/f$ defined on $D(p,R)\setminus\{p\}$? $\endgroup$ Sep 21, 2021 at 10:01
  • $\begingroup$ That and the fact that that extension maps $p$ into $0$. $\endgroup$ Sep 21, 2021 at 10:04
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    $\begingroup$ If $f(p)=0$ and $f'(p)\ne0$, then, near $p$,$$f(z)=a_1(z-p)+a_2(z-p)^2+a_3(z-p)^3+\cdots=(z-p)\bigl(a_1+a_2(z-p)+a_3(z-p)^2+\cdots\bigr),$$and therefore $p$ is a simple zero of $f$. $\endgroup$ Sep 22, 2021 at 9:36

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