I am thinking about an example, in order to better understand how Laurent Series help us understand the Poles, Zeros and Essential singularities of a complex function.
I am trying to find the singularities of $\frac{1}{sin(z)} - \frac{1}{z}$
Individually, 1/z has a pole of order 1 at 0
For $\frac{1}{z}$ I am quite confused as to what to do. Given for sin(z), it has zeros at $n \pi$ for all integer $n$, so the inverse has poles of order 1 at those points. But what happens at 0, where $\frac{1}{sin(z)} - \frac{1}{z}$ have poles pushing against each other?
Thank you for some hints on how to see this example!
Write it as $\frac {z-\sin\, z}{z\sin \,z}$ and apply L'Hopital's Rule twice to see that the limit as $z \to 0$ is $0$. The function has a removable singularity at $0$.