Taylor Expansion about a singular point instead of the Laurent Expansion

Question

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?

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.