Understanding relation between Laurent Series and Singularities

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!

1 Answer

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 $$0$$.

• Ohhh, it works! But could we see this using Laurent series? Nov 6 '18 at 0:09
• $\frac z {\sin \, z}$ is analytic near $0$. Its power series expansion (in $|z|<\pi$) is of the form $1+a_1z+\cdots$. Dividing by $z$ you get the Laurent series for $\frac 1 {\sin\, z}$ which is of the form $\frac 1 z+g(z)$ with $g$ analytic. Nov 6 '18 at 0:35