Is the textbook solution wrong by a sign? Laurent Series

Question

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?

Answer 1

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 $$\frac{e}{2w} + \frac{e}{4} \left(1 + \frac{w}{2} + \frac{w^2}{12} +\dots\right).$$