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

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 $$\frac{1}{5!}=\frac{1}{120}$$.

• The Laurent series is correct and the function has in $z=0$ an essential singularity. Jan 13, 2022 at 22:13

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.