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 $0$.


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

Thanks in advance!


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 $3$.

  • $\begingroup$ I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Apr 21 at 18:54
  • $\begingroup$ 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? $\endgroup$ – Pedro Gomes Apr 22 at 10:26
  • $\begingroup$ 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 $0$. $\endgroup$ – José Carlos Santos Apr 22 at 10:29
  • $\begingroup$ 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? $\endgroup$ – Pedro Gomes Apr 22 at 10:36
  • $\begingroup$ 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? $\endgroup$ – 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 $3$.


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