= Solution
On a positively oriented contour encircling $z_0$ once inside an annulus of analyticity, the <Laurent series> coefficients are
$$
\boxed{a_n=\frac1{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{n+1}}dz,\qquad n\in\mathbb Z.}
$$
This follows by inserting the convergent Laurent expansion and integrating term by term: only the term with exponent $-1$ has nonzero contour integral. Uniform convergence on compact circles in the annulus justifies that step.
For the function at hand,
$$
\frac1{z(z-1)}=-\frac1z\frac1{1-z}=-\frac1z\sum_{j=0}^\infty z^j.
$$
The <geometric series> converges for $|z|<1$, and the pole at zero excludes the centre. Thus
$$
\boxed{\frac1{z(z-1)}=-\sum_{n=-1}^\infty z^n,\qquad 0<|z|<1.}
$$
The coefficients are $a_n=-1$ for $n\geq-1$ and zero for $n<-1$. The coefficient formula gives the same result on $|z|=\rho<1$: $a_n=(2\pi i)^{-1}\oint z^{-n-2}/(z-1)\,dz$, and expansion of $(z-1)^{-1}$ selects exactly those coefficients.
The annulus is essential. For $|z|>1$, expand instead in inverse powers:
$$
\frac1{z(z-1)}=\frac1{z^2}\frac1{1-z^{-1}}=\sum_{j=0}^\infty z^{-j-2},\qquad |z|>1.
$$
These are different Laurent expansions about the same centre on the two annuli separated by the singularity at $z=1$. The series requested in the PDF is the inner one, not a global expansion.
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