0
$\begingroup$

Let $f(z)$ be a complex-valued meromorphic function.

If we say that $k$ is "the order of zero or pole" of $f(z)$ the the point $z=z_0$. What does this exactly mean?

As far as I understand it, if the function has no zeros or poles at $z_0$, we should say that $k=0$, and if it has a zero or order $r$, that $k=r$. Is this correct?

And if $f$ has a pole of order $s$, should it be $k=s$ or $k=-r$?0

Thanks in advance for any information

$\endgroup$
  • $\begingroup$ Can you indicate where you see that phrase used? I think it may depend a bit on the context. $\endgroup$ – Willie Wong Sep 14 '16 at 6:00
  • $\begingroup$ @WillieWong see the complex-analysis tag, and I don't think it has a different meaning in other context $\endgroup$ – reuns Sep 14 '16 at 7:57
  • $\begingroup$ Well, the phrases "order of zero" and "order of pole" are both entirely clear. But your question is whether the number $k$ being "order of zero or pole" should be set equal to the "order of pole" or its negative when you have a pole. For example, if I were to write something about the Laurent expansion of a function, I may say that the summation starts at "$n$" being the order of zero or pole of the function, and in this case obviously I would mean that negative of the usual "order of pole". But there may also be contexts where the text is written in such a way that the usual sense is used. $\endgroup$ – Willie Wong Sep 14 '16 at 13:35
4
$\begingroup$

Let $f\colon \mathbb{C}\to \mathbb{C}$ be a meromorphic function. Suppose $f$ has a pole at $z = a$. Then there exists a postive integer $m$ and an analytic function $g$ such that $g(a)\neq 0$ and $$ f(z) = \frac{g(z)}{(z - a)^m} $$ We say that $f$ has a pole of order $m$ at $a$.

The definition for the order of a zero is analagous. The reference is Conway's Functions of One Complex Variable I

$\endgroup$
  • $\begingroup$ I think you might need to add that $g$ is non zero at $a$. $\endgroup$ – Joel Cohen Sep 14 '16 at 6:36
  • $\begingroup$ @JoelCohen Thanks $\endgroup$ – user259242 Sep 14 '16 at 6:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.