3
$\begingroup$

I have been reviewing Dennis Zill's Complex Analysis text and he defines a singular point as a point $z$ at which a function $f$ fails to be analytic. Now he goes on to talk about isolated singularities and calls everything else a nonisolated singularity. My confusion however, stems from how other people define it.

I've seen descriptions of singular points which seem to imply that they only include branch points, branch cuts and isolated singularities. There's also a definition from a Fullerton professor's page which goes like this: a point $a$ is a singular point if $f$ is not analytic at $a$ but every neighborhood of $a$ contains a point $b$ such that $f$ is analytic at $b$.

But consider the complex function $f(z)=|z|^2$. This function is nowhere analytic, since it is only differentiable at $z=0$. If I'm understanding this correctly, by Zill's definition every point in the complex plane is a singular point. But using the other two definitions above, no point is a singular point.

So what's going on here? What's the best way to define a singular point? Many thanks in advance!

$\endgroup$
  • $\begingroup$ Actually, I think I may have found the answer to my own question here. Churchill and Brown explicitly give $f(z)=|z|^2$ as an example of a function which has no singular points because for any given point, there are no other points in any neighborhood of it which are analytic. It makes sense intuitively, as every point being singular would render the concept moot. $\endgroup$ – Steve Mar 30 '14 at 19:20
2
$\begingroup$

There are 3 types of isolated singularities. There are removable singularities, poles, and essential singularities.

A point z is a removable singularity for f if f is defined in a neighborhood of the point z, but not at z, but f can be defined at z so that f is a continuous function which includes z. Here is an example of this: if f(z) = z is defined in the punctured disk, the disk minus 0, then f is not defined at z=0, but it can certainly be extended continuously to 0 by defining f(0) = 0. This means at z=0 is a removable singularity.

A point z is a pole for f if f blows up at z (f goes to infinity as you approach z). An example of a pole is z=0 for f(z) = 1/z.

A point z is an essential singularity if the limit as f approaches z takes on different values as you approach z from different directions. An example of this is exp(1/z) at z=0. As z approaches 0 from the right, exp(1/z) blows up and as z approaches 0 from the left, exp(1/z) goes to 0.

$\endgroup$
  • $\begingroup$ Well these are all true statements, but there is no point for the function $|z|^2$ which is an isolated singularity. $\endgroup$ – Steve Mar 30 '14 at 19:16

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.