Finding the set of all poles of $\tan(\frac{\pi z}{2})$

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.

• An isolated singularity $p$ is a pole of $f$ if and ony if $|f(z)| \to \infty$ as $z \to p$. Sep 21, 2021 at 9:36
• 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 Sep 21, 2021 at 9:39
• $p$ is a pole of order $1$ if $(z-p)f(z)$ has a finite non-zero limit as $z \to p$. Sep 21, 2021 at 9:42

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 $$P$$.
• 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 $z\to p$. Sep 21, 2021 at 9:55
• 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). Sep 21, 2021 at 9:58
• Is that a consequence of the fact that $\frac{1}{f}$ on $D(p,R)$ is a holomorphic extension of $1/f$ defined on $D(p,R)\setminus\{p\}$? Sep 21, 2021 at 10:01
• That and the fact that that extension maps $p$ into $0$. Sep 21, 2021 at 10:04
• 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 $f$. Sep 22, 2021 at 9:36