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 $0$?

Solution 1:

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.

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