Simple doubt on singularities and poles theory.

Question

I found the following definition online:

A point $z_0$ is called a pole of order $m$ of $f(z)$ if $1/f$ has a zero of order $m$ at $z_0$.

I was wondering if this definition is valid at every case, and more specifically, at the following case: \begin{equation*} f(w) = \frac{\pi z}{w(z-w)}\cot(\pi w) \end{equation*} Thanks for all the help in advance.

EDIT: Basically what I am asking is if what is stated in this question is valid for any function $f$ we take.

Answer 1

To summarize what we discussed in the "chat":

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Quoting from Complex Analysis by Stein and Shakarchi, p. 74.

A deleted neighborhood of $w_0$ is an open disc centered at $w_0$, minus the point $w_0$. It is the set

$$\{w:0<|w-w_0|<r\}$$ for some $r>0$.

A function $f(w)$ defined in a deleted neighborhood of $w_0$ has a pole at $w_0$ if the function $1/f$, defined to be zero at $w_0$, is holomorphic in a full neighborhood of $w_0$.

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The function \begin{equation*} f(w) = \frac{\pi z}{w(z-w)}\cot(\pi w) \end{equation*}

has poles at $w=0, w=z,$ and $w \in \mathbb{Z}$. The pole at $w=0$ is a double pole$\color{blue}{.}$ $\color{silver}{(\text{when } z\ne 0) \text{ and the pole at }w=z\text{ is double if } z \text{ is an integer}.}$

We've used the fact that the zeros of $\tan \pi w$ are the zeros of $\sin \pi w$ which are the integers, and the zeros of $\sin \pi w$ do not coincide with the zeros of $\cos \pi w$. Otherwise those zeros would be removable.

Also, if $z = k+\pi/2$, with $k\in \mathbb{Z}$, then $z$ is not a pole! That's because $\cos \pi(k+1/2) =0.$

Answer 2

The singularity at $z_0$ must have a neighborhood in which there are no other poles or zeros of $f$. One cannot do this with essential singularities, e.g. $\sin(1/z)$ at $0$.

If a singularity is isolated and has a punctured neighborhood where $f$ is nonzero this does always work. For then $1/f$ has an isolated $0$ at $z_0$. And so we can use the classification of zeros of holomorphic functions to find an $m$ and holomorphic $h$ with $h(z_0) \neq 0$ so that $(1/f(z)) = (z-z_0)^m h(z)$ in a neighborhood of $z_0$.