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So, I'm trying to classify the singular points of the following function:

$$ f(z)=e^{\cot(\frac {1}{z})} $$

Obviously, when z is zero, the function tends to approach infinity, so that must be a singularity. To characterize the singularity, I think that I must expand it out into a Laurent series and see what it looks like. Here's what I did:

$$\cot(z)=\frac{1}{z}-\frac{z}{3}-\frac{z^3}{45}+\cdots$$

$$\cot\left(\frac{1}{z}\right)=z-\frac{1}{3z}-\frac{1}{45z^3}+\cdots$$

Let's call this function above A for simplicity!

$$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+\cdots$$

$$e^A=1+A+\frac{A^2}{2!}+\frac{A^3}{3!}+\cdots$$

Since A consists of an infinite number of z's with negative powers increasing each term, we notice two things. First, $z=0$ is indeed a singularity. Second, does this allow us to conclude that the singularity $z=0$ is an isolated, essential singularity? Is there any other singularity, such as infinity?

EDIT: So, to consider $z=infinity$, I considered $f(\frac{1}{z})$ at the point $z=0$. Expanding out this, we end up with:

$$\cot(z)=\frac{1}{z}-\frac{z}{3}-\frac{z^3}{45}+\cdots$$

Let's call this expansion B!

$$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+\cdots$$

$$f(\frac{1}{z})=e^{\cot(z)}=1+B+\frac{B^2}{2!}+\frac{B^3}{3!}+\frac{B^4}{4!}+\cdots$$

Thus, that $\frac{1}{z}$ term in B will continually increase in power as the terms go further out. For this reason, I think that it's also an isolated,essential singularity. What are your thoughts; am I thinking of this correctly? Are there any other singularities?

Incognito
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1 Answers1

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Some comments on the part of your work studying $\cot$.

The series for $\cot z$ that you write down is not defined everywhere: that Laurent series only converges for the annulus $0 < |z| < \pi$. Correspondingly, the Laurent series for $\cot(1/z)$ you wrote is only valid for $\frac{1}{\pi} < |z| < \infty$. In particular, it's not defined near $0$, and thus can't tell you if $\cot(1/z)$ has any singularities at $z=0$ or if so, what type they are.

In fact, we can see that $\cot(1/z)$ doesn't even have a Laurent series in any neighborhood of $0$: any annulus $0 < |z| < R$ must include a pole of $\cot(1/z)$, and thus cannot be the annulus of convergence of a Laurent series for $\cot(1/z)$. This fact let us conclude $0$ is an essential singularity of $\cot(1/z)$.

Note that any neighborhood of $z=0$ also contains infinitely many zeroes of $\cot(1/z)$ as well: you can't say that this function converges to $+\infty$ as $z \to 0$. In fact, if it did converge to $+\infty$, then $z=0$ would have to be a pole!

  • Very, very interesting response, Hurkyl. So, you're stating that the $\cot(z)$ series is convergent up to a radius of $\pi$ since that is where the function in undefined, correct? Further, if a Laurent series is not valid at a singularity, do we always conclude that the singularity is essential? I think that's correct, right?

    Should I instead expand the function about $z=\pi$?

     – Incognito Mar 12 '14 at 19:43
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    I'm not quite sure I understand precisely what you're asking in your second question. Note there are other series for $\cot(z)$ defined on other intervals $n \pi < |z| < (n+1) \pi$: only the Laurent series defined on $0 < |z| < \pi$ can tell us about the precise details of the singularity that $\cot(z)$ has at $z=0$. The possible annuli of convergence for Laurent series is like the radius of convergence of a Taylor series: for any annulus that doesn't contain a singularity of $f(z)$, there is a Laurent series for $f(z)$ that converges in the annulus. –  Mar 12 '14 at 19:47
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    $\sin(1/z)$ has no singularities on $0 < |z| < +\infty$, for example, so it has a Laurent series that converges everywhere but zero; and since this is defined on a punctured neighborhood of $0$, it can tell us about the singularity of $0$. The poles of $\cot(1/z)$, alas, prevent us from using this approach to learn about the singularity of $\cot(1/z)$ at $0$... but the fact that the poles prevent us from doing so does tell us it's essential: if the singularity at $0$ was just a pole, there would be some neighborhood of $0$ without any other singularities. –  Mar 12 '14 at 19:50
  • That definitely makes sense. So, using this approach, I can expand cot(1/z) about the point z=1/$\pi$ (http://www.wolframalpha.com/input/?i=series%28cot%281%2Fx%29%29+at+x%3D1%2Fpi) to evaluate its behavior? Since this has a term $\frac{1}{z-\frac{1}{pi}}$, we conclude that it would have a simple pole at that point, correct? If we then plug that into the exponential series, we would find an infinite number of them and also conclude that those points are essential singularities? – Incognito Mar 12 '14 at 19:56
  • @user108149: Yes, something like that. It should be possible to solve the problem in a similar way, but without appealing to series -- e.g. just from knowing the singularities of $\exp$ and $\cot$ and which values of $\cot(1/z)$ equal a singularity of $\exp$ -- but the details to justify doing so escape me at the moment. –  Mar 12 '14 at 20:00
  • This is actually wrong. z=0 is actually a NON-isolated singularity, but the other singularities at $\frac{1}{npi}$ are isolated, essential singularities. – Incognito Mar 14 '14 at 02:45