Question: Why does having infinitely many negative degree terms in a Laurent series imply having an essential singularity at $0$?
Why I am Asking: I came across a question in Ahlfor's Complex Analysis book wanting us to show that $e^z$, $\sin z$, and $\cos z$ all have essential singularities at $\infty$. I was going to write $\sin z$ and $\cos z$ in terms of $e^z$ and go about trying to show that, for instance, $\sin(\frac{1}{z})$ has an essential singularity at $0$. However, I came across the above justification, and I couldn't quite reason why it works…. I was hoping that someone could shed some light on it for me.
Best Answer
An isolated singularity is either a removable singularity, a pole or an essential singularity. For a removable singularity, after removing the singularity you have an analytic function, whose Laurent series is just a Taylor series: all negative degree terms are $0$. For a pole of order $m$, all terms of degree $< -m$ are $0$. So if there are infinitely many nonzero terms of negative degree, it can't be removable and can't be a pole; all that's left is an essential singularity.