Talk:Essential singularity

I think there is a slight inconsistency in the definitions. The existence of an infinite principal part is a slightly different criterion than what comes out of the Casorati-Weierstrass theorem. Take, for instance f(z)=sec(1/z), The Laurent expansion for sec(w) changes on each annulus (n-1/2)pi<|w|<(n+1/2)n. In consequence there is no Laurent expansion of f(z) in any punctured neighbourhood of z=0, and hence no way to discuss the existence or otherwise of an infinite principal part. This is, of course due to the fact that the singularity is not isolated.

131.111.16.20 (talk) 13:34, 28 March 2014 (UTC)

Shouldn't it be that a singularity is not essential iff lim z->a |f(z)| is finite or infinity not lim z->a f(z)? The problem is that in the case of a pole, the infinity reached in the limit will depend on "direction" which a is approached by z. The way this theorem is currently stated seems to make sense only if we compactify the complex plane, setting all infinite points on the compact plane to the same point. Newlyformed 06:14, 14 August 2007 (UTC)

Well, as far as I know the only way one defines the complex infinity is by compactification. So, for example, it is true that

$\lim _{|z|\to \infty }z^{3}=\infty .$

Oleg Alexandrov (talk) 15:17, 14 August 2007 (UTC)

'The point a is an essential singularity...only if...:

   * The function f has poles in every neighbourhood of a, meaning that the singularity is not isolated.'

Not true? —Preceding unsigned comment added by 80.41.125.149 (talk) 16:01, 18 March 2009 (UTC)

I don't think this is in fact true. Consider the function exp(1/z) - it has an essential singularity at z=0 (which you can see by writing down its Laurent series, $1+{\frac {1}{z}}+{\frac {1}{2z^{2}}}+{\frac {1}{6z^{3}}}+\ldots$) but doesn't have any other poles near zero, since if z is nonzero, 1/z is a perfectly well-behaved complex number, and we know how to define exp(w) for any (non-infinite) complex number.--Druiffic (talk) 18:08, 23 July 2009 (UTC)

This looks totally wrong:

$\lim _{z\to a}f(z)$ exists but $\lim _{z\to a}{\frac {1}{f(z)}}$ does not exist, then a is a zero of f and a pole of 1/f.

"a" is a zero of f if the limit exists? That's not right. I think the author meant "if the limit exists and is zero". But maybe I'm missing something. Ceresly (talk) 14:14, 27 March 2010 (UTC)

What you're missing is that if the limit of f exists and is nonzero then the limit of 1/f will also exist. Jowa fan (talk) 05:33, 1 April 2010 (UTC)

How can a be a zero of f if f is not defined at a, as previously stipulated? A zero is just a point z where f(z) = 0. Akwdb (talk) 06:17, 14 August 2019 (UTC)

The caption of exp(1/z) is wrong, IMO. It says that "a pole [...] would be uniformly white". Even 1/|z| would have radially distributed grey levels. And 1/z would have star-like hue and radially distributed luminance. richy (talk) 20:16, 5 May 2010 (UTC)

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Am I mistaken in the notion that essential singularities can occur without the complex plane, take for instance y = sin(1/x), it has an essential singularity at 0. 74.142.201.171 (talk) 04:34, 8 October 2010 (UTC)

No, you are correct. But in the real case, there exist functions that are differentiable but not analytic, so the difference between a pole and an essential singularity isn't so clear. In the real case it's common to talk about discontinuities rather than singularities; there's a separate page for classification of discontinuities. Jowa fan (talk) 00:23, 11 October 2010 (UTC)

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Name of the file with a photo of the model has a typo in the function expression, according to the catalog of models [1] the function is

$6w=e^{1 \over 6z}$

So, I think that I'm correct, and this should not be changed in the article to correspond with the file name. 79.132.172.210 (talk) 21:48, 15 November 2012 (UTC)