I need to find the principal part of the Laurent expansion of the function in the annulus described in the title.
In my attempt, I have replaced $z$ by $w:=z - \sqrt[3]{2 \pi}$. I know the Taylor expansion of the $\sin$ function. I have tried to use a approach similar to the one used in this answer, where we know $g(z)$ and try to compute coefficients one by one of the function $f(z):=1/g(z)$ using the fact that $f(z)g(z)=1$.
One of the doubts I am having with this method is: why can we write such an $f$ without negative powers of $z$? If I can do that in our example then I guess all of the negative coefficients are equal to $0$... Can you please help me clarify and give some hints to the needed computation? Thanks a lot.