Consider the function $f:z\mapsto 1/\sqrt{z}$, defined, say on the right half-plane $Re(z)>0$. (We can resolve ambiguity by taking the branch that is positive for real $z$).

Let $U$ be the right half-plane, and $a=0$.

Then the following conditions, lifted from Wikipedia's page on essential singularities, appear to hold:

1) $f(z)$ is not defined at $a$ but is analytic in the region $U$. Moreover, every open neighborhood of $a$ has non-empty intersection with $U$.

2) $\lim_{z\rightarrow a}f(z)$ does not exist.

3) $\lim_{z\rightarrow a}{1\over f(z)}$ exists (and is equal to zero).

According to that Wikipedia page, it follows that $a=0$ is a pole of $f$.

But it seems to me that $f$ has no Laurent series at $a=0$, which makes me skeptical that this really is a pole.

This seems to leave three possibilities: Either Wikipedia is wrong, or I am wrong, or I have misunderstood Wikipedia. Which of these is correct?

  • $\begingroup$ $a$ should be an element of $U$. which is not in your case. $\endgroup$ – Julian Mejia May 22 at 21:43
  • $\begingroup$ @JulianMejia: no, $a$ does not have to be an element of $U$. But we need a neighbourhood $V$ of $a$ such that $V\setminus\{a\}\subset U$, which we don't have. $\endgroup$ – TonyK May 22 at 21:50
  • $\begingroup$ @TonyK, yeah I think understand what you mean, but I was just trying to be consistent with the wikipedia page. It considers $a\in U$, so we can say $f$ is holomorphic at $U\setminus\{a\}$ and has a singularity in $a$. $\endgroup$ – Julian Mejia May 22 at 22:38

The conditions in the "Alternate descriptions" section of the Wikipedia article are wrong. The correct definition of a pole (as the term is normally used in complex analysis) requires that $f$ be defined on an entire deleted neighborhood of $a$, not just in an open set that intersects every neighborhood of $a$ (that is, a pole is by definition an isolated singularity). Since the right half-plane does not contain any deleted neighborhood of $0$, your function $f$ does not have have a pole at $0$.

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  • $\begingroup$ Perfect; thank you. Is it correct terminology then to say that $0$ is not a singularity of $f$? $\endgroup$ – user167949 May 22 at 22:47
  • $\begingroup$ I would say $0$ is a singularity of $f$, but not an isolated singularity. I don't have a precise definition of "singularity" in mind when I say this though, and in basic complex analysis it is common to restrict attention to isolated singularities. In particular, the usual "classification of singularities" (removable, pole, essential) you hear about is only for isolated singularities. $\endgroup$ – Eric Wofsey May 22 at 22:48
  • $\begingroup$ Thank you again. $\endgroup$ – user167949 May 23 at 0:29
  • $\begingroup$ 0 is a branch point of f(z) $\endgroup$ – Lawrence Mano May 23 at 21:30

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