1
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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 $0$.

$\endgroup$
2
  • $\begingroup$ Ohhh, it works! But could we see this using Laurent series? $\endgroup$ Nov 6 '18 at 0:09
  • 1
    $\begingroup$ $\frac z {\sin \, z}$ 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. $\endgroup$ Nov 6 '18 at 0:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.