Residue of $\frac{1}{f}$ by $f$'s Taylor series, integrating $\frac{1}{z^2\sin(z)}$ over the unit circle.

Question

I wish to calculate the integral $\int_{\partial B(0,1)} \frac{1}{z^2 \sin(z)}$, the integrand has a pole of order 3 at $0$, so I tried to calculate the residue of the integrand at $0$. I found very similar post which do so for $\frac{1}{z\sin(z)}$ : Compute $\int_{\gamma}\frac{dz}{z\sin(z)}$ . I don't understand the step: $\frac{1}{z\cdot(z-z^3/6 + O(z^5) )} = \frac{1}{z^2} - \frac{1}{6} + O(z^2)$, and in general can one generalize the method such that by knowing $\frac{1}{\sum _0^\infty a_n(z-z_0)^n}$ to obtain the residue of $\frac{1}{f}$?

Answer 1

Here are some more details: \begin{align} \frac1{z\sin z}&=\frac 1{z\Bigl(z-\cfrac{z^3}6+O(z^5)\Bigr)}=\frac 1{z^2\Bigl(1-\cfrac{z^2}6+O(z^4)\Bigr)}=\frac 1{z^2}\cdot\Bigl(1+\cfrac{z^2}6+O(z^4)\Bigr)\\ &=\frac 1{z^2}+\frac16+\frac 1{z^2}O(z^4)=\frac 1{z^2}+\frac16+O(z^2) \end{align}