Residue of $\frac{1}{(z^2-1)^3}$ at the singularities.

Question

Find the residue of $$\frac{1}{(z^2-1)^3}$$ at the singularities.

To find which order of pole it is I tried to look to the order of zeros of $(z^2-1)^3$ deriving the function seems to big the hard way so I tried to look at the Laurent series $$\frac{1}{(z^2-1)^3}=\frac{1}{z^6(1-\frac{1}{z^6})^3}$$

Can I do such a thing?

Answer 1

Big Hint: $z^2-1=(z+1)(z-1).$

Answer 2

Note that\begin{align}(z+1)^{-3}&=\frac12\bigl((z+1)^{-1}\bigr)''\\&=\frac12\left(\frac12-\frac{z-1}4+\frac{(z-1)^2}8-\frac{(z-1)^3}{16}+\frac{(z-1)^4}{32}-\cdots\right)''\\&=\frac18-\frac3{16}(z-1)+\frac3{16}(z-1)^2-\cdots\end{align}Therefore\begin{align}\frac1{(z^2-1)^3}&=\frac1{(z-1)^3(z+1)^3}\\&=\frac18(z-1)^{-3}-\frac3{16}(z-1)^{-2}+\frac3{16}(z-1)^{-1}+\cdots\end{align}and so$$\operatorname{Res}\left(\frac1{(z^2-1)^3},1\right)=\frac3{16}.$$Can you compute the other residue now?