Here is a related problem. First notice that $z=1$ is a simple pole of the function. The residue of the function at $z=1$, which is the coefficient of $(z-1)^{-1}$ in the Laurent series, is $1$ and you will see this from the series once it is derived. Here is how you compute the series
$$ \frac{1}{z^2-z}=\frac{1}{(z-1)z}=\frac{ 1 }{(z-1)(1+(z-1))}$$
$$=\frac{1}{z-1}(1-(z-1)+(z-1)^2-(z-1)^3-\dots) $$
$$ \frac{1}{z-1}-1+(z-1)-(z-1)^2+\dots $$
$$ = \frac{1}{z-1} - \sum_{k=0}^{\infty}(-1)^k(z-1)^k, \quad 0<|z-1|<1 $$
You can see from the above series that the coefficient of $(z-1)^{-1}$ is one as we said in the beginning. For the other series, $|z-1|>1$, we have
$$ \frac{1}{z^2-z}=\frac{1}{z(z-1)} = \frac{1}{(z-1)((z-1)+1)} = \frac{1}{(z-1)^2(1+\frac{1}{(z-1)})} $$
$$ =\frac{1}{(z-1)^2}( 1-(z-1)^{-1}+(z-1)^{-2}+\dots )$$
$$ =\frac{1}{(z-1)^2}+\sum_{k=0}^{\infty}(z-1)^{-k}, \quad |z-1|>1. $$
Note that, the series
$$ \frac{1}{1+x}=1-x+x^2-x^3+\dots\,, $$
was used in the above derivation.