Computing Pole Order

Question

Determine the type of singularity of $z_0$ of $f(z)=\frac{1}{z-\sin(z)}\:z_0=0$.

$\lim_{z\to 0}\frac{1}{z-\sin(z)}=\infty$ so it is clearly pole.

I do not know if there is a way to compute the order without writing down the Laurent series of f(z)=\frac{1}{z-\sin(z)} around $0$.

Question:

How can I compute the order of the pole without Laurent series?

Thanks in advance!

Answer 1

Since$$z-\sin z=\frac{z^3}{3!}-\frac{z^5}{5!}+\cdots=z^3\left(\frac1{3!}-\frac{z^2}{5!}+\cdots\right),$$you know that$$\frac1{z-\sin(z)}=\frac1{z^3}\times\frac1{\frac1{3!}-\frac{z^2}{5!}+\cdots}$$and, since $\dfrac1{\frac1{3!}-\frac{z^2}{5!}+\cdots}$ is analytic and not $0$ near $0$, you can write it as $a_0+a_1z+a_2z^2+\cdots$ (actually, $a_0=3!=6$) and so$$\frac1{z-\sin(z)}=\frac{a_0}{z^3}+\frac{a_1}{z^2}+\frac{a_2}z+\cdots$$Therefore, the order is $3$.

Answer 2

Using Taylor series around $z=0$ gives $$\frac1{z-\sin{(z)}}=\frac1{z^3\left(\frac16-\frac{z^2}{120}+\dots\right)}$$ So the order of the pole is $3$.