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?

  • $\begingroup$ Ummm… that's not a rational function… How could you do partial fraction decomposition? $\endgroup$
    – xbh
    Jan 23, 2019 at 6:16

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

  • $\begingroup$ So there's no way to avoid the tedious multiplication of series terms? Also why is $e^z$ not a rational function? $\endgroup$ Jan 23, 2019 at 15:54
  • 1
    $\begingroup$ No. But how many terms of the expansion do you need? A rational function is by definition a ratio of polynomials. $\endgroup$
    – Robert Z
    Jan 23, 2019 at 16:00
  • $\begingroup$ not alot , but would prefer to avoid it if possible. Thank you. $\endgroup$ Jan 23, 2019 at 16:37
  • 1
    $\begingroup$ In order to have the Laurent expansion up to $w^2$ you need the expansion of $e^w$ up to $w^3$. $\endgroup$
    – Robert Z
    Jan 23, 2019 at 16:40

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