Classify the singularity of the function $f(z)=\frac{1}{z-\sin z}$ at the origin.
Honnestly, I don't know what I have to do on this problem. I know we could develop $z-\sin z = \sum_{k \geq 1} \frac{(-1)^n z^{2n+1}}{(2n+1)!} = 0$.
Is anyone could give me more informations how to solve it?
Using the Taylor expansion of $\sin(z)$, we find that near $z=0$, $$z-\sin(z) \approx \frac{x^3}{6}.$$ So it is a third order pole at $z=0$. Indeed, multiply by $z^3$ and you will see that there is no long a pole there since $$\lim_{z\to 0} \frac{z^3}{z-\sin(z)} = 6.$$