Laurent Series and Residue for $f(z) = z^4 \sin(1/z)$ at $z=0$

Question

I had to find the Laurent series and residue of the function. We know that the $res_{z_0} = a_{-1}$, the coefficient of $\frac{1}{z-z_0}$ of the Laurent series.

For $f(z) = z^4 \sin(1/z)$ on the annulus $0<|z|< \infty$ I got the following Laurent Series about $z_0=0$ \begin{align} f(z) &= z^4 \sin(1/z)\\ &= z^4 \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} (1/z)^{2n+1}\\ &= z^4 \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} z^{-2n-1}\\ &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} z^{-2n+3}. \end{align}

I am confused if this is correct since it has infinitely many terms that have a quotient of $z$, this means that it should be an essential singularity at $z=0$, but I thought that it had a pole at $z=0$. Is this correct or did I do something wrong in my calculations? Maybe my idea of an essential singularity is wrong.

And if this Laurent series is correct, then is it true that (for $n=2$) we get $\frac{1}{5!} z^{-1}$ so that the residue is $\frac{1}{5!}=\frac{1}{120}$.

Answer 1

You are right about the value of the residue. And $0$ is an essential singularity indeed. Being an essential singularity means that, if your Laurent series is $\sum_{n=-\infty}^\infty a_nz^n$, then there are infinitely many negative integers $n$ with $a_n\ne0$. That's the case here; it's not zero for every negative odd integer.