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.
- 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 $D(p,R)$?
- 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 $p$.
- 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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