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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.

xyz
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    Adding some more details would help in addressing your concerns. Why do you bring up the example you give? Did it give you any difficulties with this definition, and where? How did you come up with this? – Ninad Munshi Jun 29 '21 at 20:45
  • Hello @NinadMunshi . This example is an exercise from an worksheet I have been given. I was just wondering because I was discussing this matter with some colleagues of mine and we aren't sure about using the definition I stated for this particular example, or for any other example that in which $f$ doesn't represent the quocient of two polynomials. Thanks for your comment. – xyz Jun 29 '21 at 20:47
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    You have, of course, identified the poles at $w=0$ and $w=z$. The other poles are where $\tan \pi w = 0.$ – mjw Jun 29 '21 at 21:07
  • Hey @mjw . I have identified all the poles. My only doubt is in calculating its orders. That's why I presented the initial result and asked if it was valid in this specific case. Thanks for your help. – xyz Jun 29 '21 at 21:08
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    The textbook $\textbf{Complex Analysis}$ by Stein and Shakarchi also requires the function $1/f$ to be holomorphic in a full neighborhood of $z_0$ for $z_0$ to be a pole of $f$. – mjw Jun 29 '21 at 21:09
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    The zeros of $\tan \pi w$ are the zeros of $\sin \pi w.$ These are simple zeros. The pole of $f(w)$ at $w=0$ is a double pole, and if $z$ is an integer, then also a double pole there. – mjw Jun 29 '21 at 21:10
  • So if I prove that $1/f$ is holomorphic in $w=0,w=z,w=\pm 1,\pm 2, \dots$ for the specific case I presented and calculate the multiplicity of these values as zeros of $1/f$ I obtain the order of these poles (of $f$) ? – xyz Jun 29 '21 at 21:11
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    Right, holomorphic in a neighborhood of these points. – mjw Jun 29 '21 at 21:13
  • Thanks! If you want to elaborate a formal answer so I can give you the checkmark be my guest! – xyz Jun 29 '21 at 21:13
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    Okay, I'd be happy to do that. But better yet, I think you contributed as much as me. Why don't you summarize everything as an answer and accept that? – mjw Jun 29 '21 at 21:15
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    Decided to post an answer. Had a couple extra comments to make involving $\cos \pi w.$ – mjw Jun 29 '21 at 21:34

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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 $z_0$.

Alex Nolte
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To summarize what we discussed in the "chat":

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 $w_0$.

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 $\cos \pi(k+1/2) =0.$

mjw
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  • I forgor to mention that $z \in \mathbb{C\backslash Z}$. Hope you can edit something out of this! – xyz Jun 29 '21 at 21:41
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    Okay, "grayed" out the part about $z\in \mathbb{Z}$, since you say $z \notin \mathbb{Z}$ is given. – mjw Jun 30 '21 at 04:03