# Laurent expansion for sin(1/z)

I know the Taylor expansion of $$\sin(z)$$, but I still don’t understand how to expand it into the Laurent series. If I use the Taylor expansion $$\sin(z)=\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n+1}}{(2n+1)!},$$ it still is centered at $$0$$. Maybe I need an example of it. Does $$\sin\left(\frac{1}{z}\right)$$ have a Laurent expansion centered at $$0$$?

• Hello, there. What exactly do you know about Laurent expansions? Nov 22, 2021 at 13:37
• The singularities of $\sin\left(\frac{1}{z}\right)$ occur at $z_n=\frac{1}{n\pi}$ for $n\in\mathbb{Z}$. $0$ is the accumulation point of $z_n$ because $\lim_{n\to\infty}z_n=\lim_{n\to-\infty}z_n=0$. So it is unclear to me that the Laurent series expansion centered at $0$ converges anywhere at all. However, this also seems off to me, so I must be missing something. It would be helpful if someone could tell me if I am correct or not. Nov 22, 2021 at 13:50
• Wait, never mind. Obviously I am wrong. I was confusing the singularities with the zeroes. That was quite the brain fart. Nov 22, 2021 at 14:01

The series you have tried to write (look it up as it is not written correctly) converges for every complex value of $$z$$. So, if you replace $$z$$ with $$\frac{1}{z}$$ in that formula, you get a Laurent series, the one you are looking for, obviously converging for every $$z$$ except $$z=0$$. $$z=0$$ is called an essential singularity of $$\sin{\left(\frac{1}{z}\right)}$$. This type of series contains an infinite number of negative power terms.
• If by "remnant" you mean the residue, the series does have a residue because the residue is the $a_{-1}$ coefficient of the series. So what is the value of the residue here?