# Residues/Classifying singularities

#### nacho

##### Active member
Hi,
so first of all
I am not entirely confident with the terminology when it comes to classifying singularities.
Could someone give me an example of the different types, or explain what they mean? My confusion stems from the terms:
essential singularity, isolated singularity, removable singularity and pole of order m
I would prefer a non-mathematical jargonised definition, as these are readily available in my textbook, however I find them difficult to comprehend.

I know that $e^{\frac{1}{Z}}$ has an essential singularity at $z=0$
But I don't see what really distinguishes this and any other type of singularity. Except that you can't let z approach 0 and 'remove' it, i guess.

Whereas $\frac{1}{z-1}$ has a removable singularity at z=1. Is this also known as an isolated singularity?

$\frac{1}{z^2}$ does this have a pole of order 2 at z=0
WHat about $\sin{\frac{1}{z^2}}$ ?

Secondly, I have attached a residue question.
so far, I want to know if i am on the correct track.

$\cosh z = \frac{e^Z+e^{-Z}}{2}$

The poles are:

$z = 2$ is a removable pole? (i dont know if this is correct terminology)
$z=1$ is a pole of order 2
$z=0$ is a removable pole.

Am I on the right track so far?

I feel i might have missed something with the treatment $\cosh z$ part

Any help is very much appreciated, thanks!

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#### chisigma

##### Well-known member
Hi,
so first of all
I am not entirely confident with the terminology when it comes to classifying singularities.
Could someone give me an example of the different types, or explain what they mean? My confusion stems from the terms:
essential singularity, isolated singularity, removable singularity and pole of order m

I would prefer a non-mathematical jargonised definition, as these are readily available in my textbook, however I find them difficult to comprehend.

I know that $e^{\frac{1}{Z}}$ has an essential singularity at $z=0$

But I don't see what really distinguishes this and any other type of singularity. Except that you can't let z approach 0 and 'remove' it, i guess.

Whereas $\frac{1}{z-1}$ has a removable singularity at z=1. Is this also known as an isolated singularity?

$\frac{1}{z^2}$ does this have a pole of order 2 at z=0

WHat about $\sin{\frac{1}{z^2}}$?...
I agree with You about the fact that in most textbooks the argument 'singularities' is trated with great superficiality. Generally specking a function f(z) has a singularity in $z=z_{0}$ if in that point it is not analytical. A singularity is said to be isolated if in all the points where is $|z - z_{0}| < r, r > 0$ there aren't other singularities. A singularity is said to be removable if with a different definition of f(*) in $z=z_{0}$ the singularity disappears. In the other cases the singularity is [or better should be...] classified according to its Laurent expansion around $z_{0}$...

$\displaystyle f(z) = \sum_{n = - \infty}^{ + \infty} c_{n}\ (z - z_{0})^{n}\ (1)$

If $\displaystyle \lim_{z \rightarrow z_{0}} f(z)\ (z-z_{0})^{k} = c_{- k} \ne 0$ then $z_{0}$ is a pole of order k. If n isn't bounded in its negative values then $z_{0}$ is an essential singularity. This definition however don't take into account that an f(z) can be non analytic in $z= z_{0}$ and the Laurent expansion of f(z) around $z = z_{0}$ may not exists...

Kind regards

$\chi$ $\sigma$

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