Showing that $\csc\left(\frac{1}{z}\right)$ does not have a Laurent series convergent in $B^{\circ}(0,r)$

Question

I am trying to show that the function, $$f(z)=\csc\left(\frac{1}{z}\right),$$ does not have a Laurent series about $0$ that converges to $f$ in $B^{\circ}(0,r)$ for any $r>0$.

I thought I could solve this problem by contradiction. For this, I compute the series for $\csc(z)$, as $$\csc(z)=\frac{1}{z}+\frac{z}{6}+\frac{7}{360}z^3+..$$ This means that $$\csc\left(\frac{1}{z}\right)=z+\frac{1}{6 \ z}+\frac{7}{360 \ z^3}+..$$ It was at this point that I could not see a contradiction, as the series appears to converge in $0<|z|<r$.

Any advice would be greatly appreciated.

Answer 1

A Laurent series must always converge on some annulus. In your case, you want it to converge on $B(0,r)\setminus\{0\}$, for some $r>0$. But that's no possible, since $\csc$ has a pole at every number of the form $\dfrac1{n\pi}$ ($n\in\mathbb N$), and $\lim_{n\to\infty}\dfrac1{n\pi}=0.$

Answer 2

Let $\sigma$ be the set of singularities of $f$ on $\mathbb C$.

Clearly, $$\sigma=\left\{\frac1{n\pi}\right\}\qquad{n\in\mathbb Z\setminus \{0\}}$$

You can easily prove $0$ is a limit point of $\sigma$.

If some Laurent series converges to $f$ in $B^\circ(0,r)$ where $r>0$, then $$B^\circ(0,r)\cap \sigma=\emptyset$$ which is contradictory to the definition of limit point.