Residue of $f(z)=\frac{1}{z^2(e^z-1)}$

Question

I want to find the residue of the function $f(z)=\frac{1}{z^2(e^z-1)}$. I have tried something that I am 99 precent sure about, but still I would appreciate some feedback. What I have done is that: $$\frac{1}{z^2(e^z-1)}=\frac{1}{z^2}\cdot\frac{1}{1+z+\frac{z}{2}+o(z^2)-1}=\frac{1}{z^2}\frac{1}{z+\frac{z}{2}+o(z^2)}$$ $$=\frac{1}{z^2}\cdot\frac{1}{1+\frac{z}{2}+o(z)}=\frac{1}{z^2}\cdot\frac{1}{1-(-\frac{z}{2}+o(z))}$$ $$=\frac{1}{z^2}\cdot[1+(-\frac{z}{2}+o(z))+(-\frac{z}{2}+o(z))^2+\cdots]=\frac{1}{z^2}[1-\frac{z}{2}+\text{greater powers of }z]$$ $$=\frac{1}{z^2}-\frac{1}{2z}+\text{greater powers of }z$$ $$\Downarrow$$ $$\operatorname{Res}(f,0)=-\frac{1}{2}$$

Answer 1

The value found is not correct. You need to expand the exponential at order $3$: \begin{align} \frac{1}{z^2(e^z-1)}&=\frac{1}{z^2\bigl(z+\tfrac{z^2}2+\tfrac{z^3}{6}+o(z^3)\bigr)}=\frac{1}{z^3\bigl(1+\tfrac{z}2+\tfrac{z^2}{6}+o(z^2)\bigr)}\\ &=\frac{1}{z^3}\Bigl(1-\Bigl(\frac{z}2+\frac{z^2}6\Bigr)+\Bigl(\frac{z}2+\frac{z^2}6\Bigr)^2+o(z^2)\Bigr)\\ &=\frac{1}{z^3}\Bigl(1-\Bigl(\frac{z}2+\frac{z^2}6\Bigr)+\frac{z^2}4+o(z^2)\Bigr) \\&=\frac{1}{z^3}\Bigl(1-\frac{z}2+\frac{z^2}{12}+o(z^2)\Bigr)=\frac{1}{z^3}-\frac1{2z^2}+\frac{1}{12z}+o\Bigl(\frac1{z}\Bigr) \end{align}