Is this the correct way to find the Laurent Expansion of the complex function $f(z) = \frac{z}{z^2 - 1}$? The function
$$
f(z) = \frac{z}{z^2 - 1}
$$
has singularities at $z = \pm 1$. I will expand about the point $z = 1$. Then, with the substitution $w = z+1$,
\begin{align*}
f(x) =& \frac{z}{z^2 - 1} \\[3mm]
=& \frac{z}{(z+1)(z-1)} \\[3mm]
=& \frac{w-1}{w(w-2)} \\[3mm]
=& \frac{1-w}{2w} \cdot \frac{1}{1 - \frac{w}{2}} \\[3mm]
=& \frac{1-w}{2w} \sum_{n=0}^\infty \frac{w^n}{2^n} \\[3mm]
=& \sum_{n=0}^\infty \frac{w^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{w^n}{2^{n+1}} \\[3mm]
=& \sum_{n=0}^\infty \frac{(z+1)^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{(z+1)^n}{2^{n+1}}
\end{align*}
This is almost in the form of a Laurent expansion, but not quite. How can I continue this to get the correct final result?
 A: At the penultimate line you have
$$\sum_{n=0}^\infty \frac{w^{n-1}}{2^{n+1}}
-\sum_{n=0}^\infty \frac{w^n}{2^{n+1}}.$$
This is
$$\frac1{2w}+\frac14+\frac w8+\frac{w^2}{16}+\cdots-\frac12-\frac w4-\frac{w^2} 8-\cdots=\frac1{2w}-\sum_{n=0}^\infty\frac{w^n}{2^{n+2}}.$$
Does this look more like a Laurent series?
A: From where you left off:
\begin{align}
&= \sum_{n=0}^\infty \frac{(z+1)^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{(z+1)^n}{2^{n+1}}\\[4px]
&=\frac{(z+1)^{-1}}{2}+
\sum_{n=1}^\infty \frac{(z+1)^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{(z+1)^n}{2^{n+1}}\\[4px]
&=\frac{(z+1)^{-1}}{2}+
\sum_{n=0}^\infty \frac{(z+1)^{n}}{2^{n+2}} - \sum_{n=0}^\infty \frac{(z+1)^n}{2^{n+1}}\\[4px]
&=\frac{(z+1)^{-1}}{2}+\sum_{n=0}^{\infty}
\left(\frac{1}{2^{n+2}}-\frac{1}{2^{n+1}}\right)(z+1)^n\\[4px]
&=\frac{(z+1)^{-1}}{2}+\sum_{n=0}^{\infty}
-\frac{(z+1)^n}{2^{n+2}}
\end{align}
A: You did the right thing, and can conclude as the others suggested, but notice that this way you are expanding around $z=-1$ and not around $z=1$.
