Find the singular points, the pole type and compute the residue of: $\frac{\sinh(z)}{z^4(1-z^2)}$

Question

I am trying to find the singular points, toe pole type and compute the residue of: $$\frac{\sinh(z)}{z^4(1-z^2)}$$ With $z_0=0$ I am ok since it is easy to find the residue by Laurent expansion (it is a principal pole. Am i right? infinite times in the denominator).

But with $z=1$and $z=-1$ I have a problem. How can I compute the Laurent expansion around 1 and -1? Also are $z=1$, $z=-1$ simple poles?

Thanks!

Answer 1

In order to determine the Laurent expansion at $1$ and $-1$ see Claude Leibovici's answer.

To evaluate the residues, you may also note that $-1$ and $+1$ are simple poles. Therefore $$\mbox{Res}\left(\frac{\sinh(z)/z^4}{1-z^2},\pm 1\right)=\left.\frac{\sinh(z)/z^4}{(1-z^2)'}\right|_{z=\pm 1} =\left.\frac{\sinh(z)/z^4}{-2z}\right|_{z=\pm 1}=-\frac{\sinh(1)}{2}.$$ See the detailed explanation here.

Answer 2

Hint

Around $z=1$, the Taylor expansion of $\sinh(z)$ would be useful $$\sinh(z)=\sinh (1)+(z-1) \cosh (1)+\frac{1}{2} (z-1)^2 \sinh (1)+O\left((z-1)^3\right)$$

Similarly, around $z=-1$ $$\sinh(z)=-\sinh (1)+(z+1) \cosh (1)-\frac{1}{2} (z+1)^2 \sinh (1)+O\left((z+1)^3\right)$$

I am sure that you can take it from here.