# Computing Pole Order

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?

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 $$3$$.
• Thanks for your answer! How can you go from $\dfrac1{\frac1{3!}-\frac{z^2}{5!}+\cdots}$ to $a_0+a_1z+a_2z^2+\cdots$? I am not understanding that step since the exponent $n$ is negative? How can it become positive? – Pedro Gomes Apr 22 at 10:26
• I am just using the fact that $\frac1{3!}-\frac{z^2}{5!}+\cdots$ defines an analytic function near $0$ whose value at $0$ is different from $0$. Therefore, $\frac1{\frac1{3!}-\frac{z^2}{5!}+\cdots}$ is also an analytic function and so it has a Taylor series centered at $0$. – José Carlos Santos Apr 22 at 10:29
• I do not understand how one expression can be equal to the other. The original function goes to infinity at $0$ but the proposed one $a_0+a_1z+a_2z^2+\cdots$ is $0$. If the neighbourhood of zero is considered excluding zero, then how can one be sure the two expressions are equal? – Pedro Gomes Apr 22 at 10:36
• I don't understand that. Tell me: if $f$ is an anlytic function defined in the neighborhood of $0$ and without zeros there, do you agree or dont that $\frac1f$ is also analytic there? – José Carlos Santos Apr 22 at 10:38
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 $$3$$.