As a practice problem we were asked to find the Laurent expansion of $f(z)=\frac{ze^z}{z-1}$ about $z=1$. Upon later reviewing the solution I see that they let $z=(z-1)+1$ and $e^z = e*e^{z-1}$. This resulted in

$f(z) = e(1+\frac{1}{z-1})e^{z-1} = e(1+\frac{1}{z-1})\sum_{n=0}^{\infty}\frac{(z-1)^n}{n!} = \frac{e}{z-1}+e\sum_{n=0}^{\infty}(\frac{n+2}{n+1})\frac{(z-1)^n}{n!}$

Here they used the Taylor expansion of $e^{z-1}$. Why is this still considered the Laurent Expansion even though they only expansion was used was the Taylor Expansion? Is it because $\frac{e}{z-1}$ still captures the singularity of the original function?

  • $\begingroup$ $e^z = \sum_{k=0}^\infty \frac{1}{k!} z^k \implies \frac{e^z}{z} = \sum_{k=-1}^\infty \frac{1}{(k+1)!} z^k$.. Every Laurent expansions with finitely many negative terms are obtained this way. Things like $e^{1/z} +e^z - 1 = \sum_{k=-\infty}^\infty \frac{1}{|k|!} z^k$ are a little different. $\endgroup$
    – reuns
    Sep 26, 2017 at 18:50

1 Answer 1


A Taylor expansion is a Laurant expansion.

i.e. a Taylor expansion $= \sum_\limits{n=0}^{\infty} c_n(z-a)^n$

Laurant expansion $= \sum_\limits{n=-\infty}^{\infty} c_n(z-a)^n$

If all of the $c_n$ terms are zero when $n<0$ then the Laurant expansion is a Taylor expansion.

However in the example you provide there is one term of $e(z-1)^{-1}$

Depending on your function, its poles, and the radius of convergence, there may be many or no terms with negative exponents.

  • $\begingroup$ Would $e$ be considered the residue of $f(z)$ in this case as well? $\endgroup$ Sep 26, 2017 at 18:29
  • $\begingroup$ it would........ $\endgroup$
    – Doug M
    Sep 26, 2017 at 18:32

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