I have a question from my Complex Analysis textbook regarding an exercise which deals with Laurent decompositions. Here is the problem:

Consider the Laurent Series for $f(z)=\frac{(z^2-\pi^2)}{\sin z}$ that is centered at $0$ and that converges for $|z|=1$.
What is the largest open set on which the series converges?

Since $\sin z$ has a simple zero at $\pi$, the function $\frac{\sin z}{(z-\pi)}$ extends to be analytic and nonzero at $z=\pi$.
Hence $\frac{(z^2-\pi^2)}{\sin z}$ extends to be analytic at $z=\pi$.
Similarly, it extends to be annalytic at $z=-\pi$.
We say that the singularities at $z=\pm \pi$ are removable.
The function tends to $\infty$ at $z=0$ and at $z=\pm 2\pi$.
Thus the largest annular domain containing the circle $\{|z|=1\}$ to which the funtion extends analytically is the punctured disk $\{0<|z|<2\pi\}$.
This is then the largest open set on which the series converges.

I am wondering why the singularity at $\pi$ is "removable" and how I might be able to determine that when given another problem. Basically, I'm trying to expand my intuition here, which currently would lead me to believe that the function is not analytic at this apparent singularity.

I am also trying to understand why the zero zeros at $\pi$ and $2\pi$ are treated differently. Why exactly are the zeros at $2\pi$ not removable?

Furthermore, why exactly does $f(z)$ fail to converge for $z<0$ and $z>2\pi$?

I'm having a tough time grasping what is going on here. I would greatly appreciate any/all input!

Thank you so much.


Isolated singularities of an analytic function are either poles, removable singularities or essential singularities. If $|f(z)| \to \infty$ as $z$ goes to the isolated singularity $p$, then by definition $p$ is a pole; if $|f(z)|$ stays bounded as $z \to p$ it is removable, and otherwise it is essential. In particular, if $f(z) = g(z)/h(z)$ where $g$ and $h$ are analytic in a neighbourhood of $p$ and $h(p) = 0$, then $p$ is a pole if $g(p) \ne 0$ or if the order of $p$ as a zero of $g$ is less than its order as a zero of $h$, while it is a removable singularity if $p$ is a zero of $g$ with order greater than or equal to its order as a zero of $h$.

In your case, $\sin(z)$ and $z^2 - \pi^2$ both have zeros of the same order $1$ (simple zeros) at $z=\pi$, so $\pi$ is a removable singularity of $(z^2 - \pi^2)/\sin(z)$.

  • $\begingroup$ Thank you! That is helpful. $\endgroup$ – Hawleyluyah Dec 11 '17 at 7:46
  • $\begingroup$ Am I correct in thinking that the function has three poles at z=0, z=pi, and z=-pi? $\endgroup$ – Hawleyluyah Dec 11 '17 at 7:50
  • 1
    $\begingroup$ No. As I said, $\pi$ is a removable singularity, not a pole. Same for $-\pi$. All other integer multiples of $\pi$ are poles. $\endgroup$ – Robert Israel Dec 11 '17 at 18:36
  • $\begingroup$ Thank you for clarifying! I meant to type in $2\pi$ and $-2\pi$. However, I didn't realize that all other integer multiples are also poles, so that is helpful. $\endgroup$ – Hawleyluyah Dec 11 '17 at 20:44

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