Confusion about Essential Singularity in 1/(1-z)

[CantorSet]

Hi everyone,

I'm brushing up on some complex analysis and I came across the following example, which is driving me crazy:

I learned that [tex]z_0[/tex] is a pole of order k of [tex]f(z)[/tex] if

[tex]f(z)=\frac{g(z)}{(z-z_0)^k}[/tex]

where [tex]g(z)[/tex] is analytic on some open disk around [tex]z_0[/tex].

And an essential singularity is an isolated singularity that is not a pole of any order.

Then I learned that if [tex]z_0[/tex] is a pole of order k , the singular part of the Laurent series of [tex]f(z)[/tex] has finitely many terms. If its an essential singularity, the singular part of the Laurent series has infinitely many terms.

Then the book gives this example:

[tex]f(z)=\frac{1}{1-z}[/tex]

And right off the bat, I'm thinking 1 is a pole of order 1, since it's already in the form of the definition of a pole of order k. But when they expanded the Laurent series for this in the annulus [tex]1 < \left|z\right| < \infty [/tex] they get

[tex] \frac{1}{(1-z)} = -\frac{1}{z}-\frac{1}{z^2}-\frac{1}{z^3} - ... [/tex]

So that [tex]z = 1 [/tex] is actually an essential singularity. But doesn't this contradict with the fact that it's already in the form [tex] \frac{1}{(1-z)}[/tex] which means its a pole?

 

[Dickfore]

Hi everyone,
[tex] \frac{1}{(1-z)} = -\frac{1}{z}-\frac{1}{z^2}-\frac{1}{z^3} - ... [/tex]

So that [tex]z = 1 [/tex] is actually an essential singularity. But doesn't this contradict with the fact that it's already in the form [tex] \frac{1}{(1-z)}[/tex] which means its a pole?

The expansion you had entered is around [itex]z = \infty[/itex], not [itex]z = 1[/itex] and it is a Taylor series with respect to [itex]1/z[/itex], so the point at infinity is actually regular.

 

[CantorSet]

The expansion you had entered is around [itex]z = \infty[/itex], not [itex]z = 1[/itex] and it is a Taylor series with respect to [itex]1/z[/itex], so the point at infinity is actually regular.

I see that the series is centered around zero, but how is it centered around infinity? Also, what do you mean by "regular"?

I guess I'll summarize my question to this:

Why is [itex] z = 1 [/itex] an essential singularity of

[tex]\frac{1}{1-z}[/tex]

when it can be expressed in the form of a pole:

[tex] f(z)=\frac{g(z)}{(z-z_0)^k} [/tex]

 

[Hurkyl]

Why is [itex] z = 1 [/itex] an essential singularity of

[tex]\frac{1}{1-z}[/tex]

It's not.

Then I learned that if [tex]z_0[/tex] is a pole of order k , the singular part of the Laurent series of [tex]f(z)[/tex] has finitely many terms.

That's not true. The theorem you're trying to think of requires something about the domain of convergence of the series...

 

[CantorSet]

It's not.

That's not true. The theorem you're trying to think of requires something about the domain of convergence of the series...

Thanks, Hurkyl.

I'm citing all of this from Princeton Review's "Cracking the Math Subject GRE" where they try to summarize all of complex analysis in 5 pages, so they're definitely sweeping a lot of stuff under the rug. They were also wrong in stating that [itex] z = 1 [/itex] is an essential singularity of

[tex]\frac{1}{1-z}[/tex]

But you also say they are wrong in stating the if a singularity is a pole of order k, then the singular part of its Laurent Series has only finitely many terms, namely it stops after [tex]a_{-k}[/tex]?

But is it wrong because I did not specify the exact conditions for when the statement is true? Or is it totally wrong?

 

[Hurkyl]

But is it wrong because I did not specify the exact conditions for when the statement is true? Or is it totally wrong?

Because you did not state the conditions.

 

[Dickfore]

I see that the series is centered around zero, but how is it centered around infinity?

Does [itex]z = 0[/itex] belong to the domain [itex]|z| > 1[/itex] where the stated expansion holds? How about [itex]z = \infty[/itex]?

Also, what do you mean by "regular"?

How do you examine the behavior of a function in the neighborhood around [itex]z = \infty[/itex]? A regular point is a point at which the funcion is analytic.

 

[Ryker]

Thanks, Hurkyl.

I'm citing all of this from Princeton Review's "Cracking the Math Subject GRE" where they try to summarize all of complex analysis in 5 pages, so they're definitely sweeping a lot of stuff under the rug. They were also wrong in stating that [itex] z = 1 [/itex] is an essential singularity of

[tex]\frac{1}{1-z}[/tex]

But you also say they are wrong in stating the if a singularity is a pole of order k, then the singular part of its Laurent Series has only finitely many terms, namely it stops after [tex]a_{-k}[/tex]?

But is it wrong because I did not specify the exact conditions for when the statement is true? Or is it totally wrong?

I'm reviving this thread from the dead, because I'm just going through the same book and was confused for a couple of minutes myself, as well. But the book doesn't actually state that this is an essential singularity. In fact, what they do is show that the function [itex]\frac{1}{1-z}[/itex] can be written as a Laurent series [itex] \frac{1}{(1-z)} = -\frac{1}{z}-\frac{1}{z^2}-\frac{1}{z^3} - ... [/itex], for - and this is the important part! - [itex]|z| > 1[/itex]. This expansion was obtained by them explicitly noting in the first place that if [itex]|z| > 1[/itex], then [itex]\frac{1}{|z|} < 1[/itex], implying that this series has nothing to do with the function for [itex]|z| \leq 1[/itex], which is where the singularity in question lies. The latter is absolutely a simple pole, however, and the Laurent series is just a tool that allows you to "Taylor serialize" the series where you otherwise couldn't.

Maybe someone else reading this post will find this addition useful, but if not, that's fine, as well Just thought I'd try to clear up the confusion and note that the authors of the book never stated this was an essential singularity (they actually never stated it's a pole, either, they just say this function has a singularity at z = 1).

 

[bfisch]

I'm reviving this thread from the dead, because I'm just going through the same book and was confused for a couple of minutes myself, as well. But the book doesn't actually state that this is an essential singularity. In fact, what they do is show that the function [itex]\frac{1}{1-z}[/itex] can be written as a Laurent series [itex] \frac{1}{(1-z)} = -\frac{1}{z}-\frac{1}{z^2}-\frac{1}{z^3} - ... [/itex], for - and this is the important part! - [itex]|z| > 1[/itex]. This expansion was obtained by them explicitly noting in the first place that if [itex]|z| > 1[/itex], then [itex]\frac{1}{|z|} < 1[/itex], implying that this series has nothing to do with the function for [itex]|z| \leq 1[/itex], which is where the singularity in question lies. The latter is absolutely a simple pole, however, and the Laurent series is just a tool that allows you to "Taylor serialize" the series where you otherwise couldn't.

Maybe someone else reading this post will find this addition useful, but if not, that's fine, as well Just thought I'd try to clear up the confusion and note that the authors of the book never stated this was an essential singularity (they actually never stated it's a pole, either, they just say this function has a singularity at z = 1).

The book says explicitly on page 313: "Since there are infinitely many terms in the Laurent series, z=1 is an essential singularity." The book is wrong.

 

[Ryker]

The book says explicitly on page 313: "Since there are infinitely many terms in the Laurent series, z=1 is an essential singularity." The book is wrong.

Well, I don't have the book anymore, so I can't check, but I don't recall them saying that explicitly in regards to this particular problem. So yeah, I can't really argue much further.