# Finding isolated singularities

I am having trouble categorizing the singularities of the following complex valued function:

$$f(z) = \frac{z^2}{\sin(z)}$$

It seems like the isolated singularities are $2n\pi$ where $n\in\{0,\pm1,\pm2,\cdots\}$ but I am having difficulty determining whether we have removable or poles or essential singularities? In the book it says we have a removable singularities if given an isolated singularities:

$$|f(z)| \mbox{ is bounded as } z\to z_0$$

Which I think means that the limit as $z\to z_0$ exists, which means that we ought to be able to calculate $f(z_0)$ to determine its value but for instance $f(0)$ does not exist but the limit does for its modulus. Does that mean it is not a removable singularity and possibly a pole or even an essential singularity (truth be told I am not fully sure what that means). Any help would be really great!

• I think I may have solved this question... I am pretty sure at $z=0$ we a removable singularity, and at $2n\pi,n\in\{\pm1,\pm2\cdots\}$ we have poles all with a zero order 1... I still wouldn't mind a more thorough solution though! Apr 11, 2014 at 20:08
• You mean $n \pi$, not $2 n \pi$. Apr 11, 2014 at 20:12
• Oh shoot, yeah $n\pi$. I will fix my comment. Okay I will fix this later... but yeah $n\pi$ Apr 11, 2014 at 20:13

## 1 Answer

As $z \to n \pi$, $\sin(z) \to 0$, while $z^2 \to (n \pi)^2$. If $n \ne 0$, this implies $|f(z)| \to \infty$, which says you have a pole. For $n = 0$, the limit is an "indeterminate form". Now use the fact that $\sin(z)/z \to 1$ as $z \to 0$.