Find the residue of $\frac1{z-\sin z} $at its pole.

Question

$f(z)=\frac{1}{z-\sin z}$

Now $z-\sin z=z-(z-\frac {z^3}{3!}+\frac{z^5}{5!}\cdots)$

After solving $z=0$ is a pole of order $3$ for $f(z)$ and $$f(z)=\frac1{z^3(\frac1{3!}-\frac{z^2}{5!}+ \cdots )}$$ What is next?

Answer 1

We have $$\left({z^3\over3!}-{z^5\over5!}+\cdots\right)\sum_{n=-3}^\infty a_nz^n=1$$ and the residue is $a_{-1}$. So for the constant term, we must have$${a_{-3}\over3!}=1$$ and for the quadratic term, we must have $$-{a_{-3}\over5!}+{a_{-1}\over3!}=0$$ which give $$a_{-1}={3\over10}$$