why Residue of 1/Z^2 at Z=0 different by Laurent series and Residue theorem?

Question

If you Find Residue of $\frac{1}{ Z^2}$ at Z=0 by Laurent series its 1 while by using formula it is 0 . Also for $\frac{1}{ Z}$ at Z =0 We get residue by Laurent series and by formula same which is 1

Here by formula I mean $\frac{1}{(n-1)!}$ * $\lim_{z\to 0}$ { $\frac {d^{n-1} }{dz^{n-1}}$ $(Z-a)^n $ $\frac{f(Z)}{(z-a)^n}$ }

Coefficient of $Z^{-2}$ is 1 in laurent series

Answer 1

Since$$\frac1{z^2}=\frac1{z^2}+\frac{\color{red}0}{z}+0+0\times z+0\times z^2+\cdots,$$the residue of $\frac1{z^2}$ at $0$ is $0$, not $1$.