Is the textbook solution wrong by a sign? Laurent Series 
Find the Laurent series of $\frac{e^z}{z^2 -1}$ about $z = 1$.

Here is my solution:
Factor denominator
$\frac{e^z}{(z-1)(z+1)}$
let $w = z - 1$, and so $z = w + 1$, substitute in
$\frac{e^{w+1}}{w(w+2)}$
Do partial fraction decomposition to get rid of the exponential in numerator
$\frac{e^{w+1}}{w(w+2)}$ = $\frac{A}{w} + \frac{B}{w+2}$
$e^{w+1} = A(w+2) + Bw$
If $w=0$, $e=2A$, $A = e/2$. If $w = -2$, $\frac{1}{e} = -2B$, $B = \frac{-1}
{2e}$
Thus our new equation is 
$\frac{e}{2w} + \frac{-1}{2e(w+2)}$
Because the first term $\frac{e}{2w}$ is already in terms of $w = (z-1), we leave it be. The second term we can expand using geometric series expansion
$\frac{-1}{4e} \cdot \frac{1}{1 - (\frac{-w}{2})}$ 
and so the Laurent series we get is 
$$\frac{e}{2w} - \frac{1}{4e} \cdot \{1 - \frac{w}{2} + \frac{w^2}{4} - \frac{w^3}{8} ... \} $$
However, the textbook solution has same terms, but no negative sign. Where did I go wrong?

You considered $e^{w+1}$ as a rational function (which is not!). Instead you should expand it at $w=0$ as
$$e^{w+1}=e\cdot e^w=e \left(1+w+\frac{w^2}{2}+\frac{w^3}{6}+\dots\right).$$ 
Hence, after decomposing the rational function $\frac{1}{w(w+2)}$,
$$\frac{e^{w+1}}{w(w+2)}=e^{w+1}\left(\frac{1}{2w} - \frac{1/4}{1+w/2}\right)\\=
e\left(1+w+\frac{w^2}{2}+\frac{w^3}{6}+\dots\right)\left(\frac{1}{2w} - \frac{1}{4}\left(1-\frac{w}{2}+\frac{w^2}{4}+\dots\right)\right).$$
Can you take it from here?
Finally the Laurent expansion should be
$$\frac{e}{2w} + \frac{e}{4} \left(1 + \frac{w}{2} + \frac{w^2}{12} +\dots\right).$$
