Computing with Cauchy Residue theorem

Question

how do I calculate $$\operatorname{Res}\left(\frac{1}{z^2 \cdot \sin(z))}, 0\right)$$ What is the order of the pole? $3$?

Answer 1

You are right, $z=0$ is a pole of order 3 because $$ \lim_{z\to 0}\frac{z^3}{z^2\sin z}=\lim_{z\to 0}\frac{z}{\sin z}=1, $$ and since $$ \frac{1}{z^2\sin z}=\frac{1}{z^3(1-z^2/3!+o(z^3))}=\frac{1}{z^3}(1+\frac{z^2}{6}+o(z^3))=\frac{1}{z^3}+\frac{1}{6z}+o(1) $$ it follows that $$ \text{Res}(\frac{1}{z^2\sin z},0)=\frac16. $$

Answer 2

Yes, the pole is of order 3. Thus

$$\begin{align}\mathrm{Res}_{z=0} \frac{1}{z^2 \sin{z}} &= \frac{1}{2!} \lim_{z \rightarrow 0} \frac{d^2}{d z^2} \left [ z^3 \frac{1}{z^2 \sin{z}} \right ] \\ &= \frac{1}{2}\lim_{z \rightarrow 0} \frac{d^2}{d z^2} \left ( \frac{z}{\sin{z}} \right )\\ &= \frac{1}{2} \lim_{z \rightarrow 0} [z (\csc^3{z} + \cot^2{z} \csc{z}) - 2 \cot{z} \csc{z}]\\ &= \lim_{z \rightarrow 0}\frac{1}{2} \left (\frac{2}{z^2}+\frac{1}{3}-\frac{2}{z^2} \right )\\ &= \frac{1}{6} \end{align}$$