Laurent series of $1/(z^2-z)$ I'm re-learning this stuff. It is a little confusing to me how to obtain the Laurent expansion of rational functions.
I have to determine the Laurent series of
$$
\frac{1}{z^2-z},
$$
centered at $z=-1$ that converges at $z=1/2$.
The "centered" part is clear to me, but what does it mean that converges at $z=1/2$?
The partial fractions is 
$$
-\frac{1}{z} + \frac{1}{z-1},
$$
and I know that I have to find the series of each term centered at $z=-1$. The annulus where I can make a Laurent expansion is for example $0<|z|<1$, right? (I am taking into account that $1/z$ is analytical everywhere except at zero and that $1/(1-z)$ has radius of convergence $1$. It's the part that it converges at $1/2$ that I don't know how to fit with these arguments.
Second question. Can I make another Laurent expansion for 1<|z| ? How should it be?
 A: The poles of the function are at $0$ and $1$, and you want an expression for your function that will be convergent at $1/2$. The request is for an expansion about $-1$, that is, in powers of $t=z+1$. Let’s make the substitution $z=t-1$: The expression $-1/z+1/(z-1)$ then becomes
$$
\frac{-1}{-1+t}+\frac1{-2+t}=\frac1{1-t}+\frac{-1/2}{1-t/2}\,,
$$
which of course we want to expand in series so as to give a result convergent for $t=3/2$. In my display above, we can easily expand the second part of the RHS, namely $(-1/2)\big/(1-t/2)$, because $t=3/2$ will give a geometric series in powers of $3/4$, perfect. But we can’t do that for $1/(1-t)$, that would give a series in powers of $3/2$, not convergent. So let’s rewrite that first part:
$$
\frac1{1-t}=\frac{-1/t}{1-1/t}=\frac{-1}t\sum_{n=0}^\infty\left(\frac1t\right)^n=-\sum_{n=1}^\infty\left(\frac1t\right)^n\,.
$$
Thus, the complete expansion is
$$
-\sum_{n=-\infty}^{-1}t^n-\sum_{n=0}^\infty\frac{t^n}{2^{n+1}}\,,
$$
if I haven’t made any careless computational errors.
A: Your $f(z)=\frac{1}{z^2-z}$ is analytic at the point of investigation (center of series), so we hope for a series in non-negative powers of $(z+1)$ which is called Taylor's series.
$f(z)=f(-1)+\frac{f'(-1)}{1!}.(z+1)+\frac{f''(-1)}{2!}.(z+1)^2+\frac{f'''(-1)}{3!}.(z+1)^3+...$
For $0<|z|<1$ you can expect negative powers of $z$ in the series expansion because $z=0$ is a singularity (a pole) of $f(z)$.
$f(z)=\frac{-1}{z}-\frac{1}{1-z}$
$=\frac{-1}{z}+\frac{1}{z}(1+1/z +1/z^2+...)$
$=1/z^2 +1/z^3+...$
