Laurent for a function is a generalization of the Taylor series. With a Laurent series, however, the powers can be negative. An advantage of the Laurent series over the Taylor series is expanding around singular points for a given function. Looking for a demonstration of this advantage for an unusual function.
Can Laurent's theorem be used on all Taylor theorem expandable functions?
On
A Taylor series has a radius of convergence.
Some familiar functions that radius is infinite.
$e^x = \sum \frac {x^n}{n!}$
But for others is its not
$\ln (1-x) = -\sum \frac {x^n}{n}$
A Laurent series gives you some flexibility in determining where the series will converge and will not converge.
$\frac {1}{1-x} = \sum_\limits{n=0}^\infty x^n$ converges when $|x|< 1$
$\frac {1}{1-x} = \sum_\limits{n=1}^\infty -x^{-n}$ converges when $|x|> 1$
One place it comes in handy:
If you have contour integral in complex analysis. And your function has a Laurent series representation, and the the contour is inside the convergent part of the Laurent series
$\int_\gamma f(z)\ dz = 2\pi i (a_{-1})$ where $a_{-1}$ is the coefficient of the $z^{-1}$ term
I don't know if it's "unusual", but for example, $f(z) = \sin(1/z)$ has an isolated singularity at $z=0$. You can do a Laurent expansion around that point:
$$ \sin(1/z) = z^{-1} - \frac{z^{-3}}{3!} + \frac{z^{-5}}{5!} - \ldots = \sum_{n=-\infty}^\infty a_n z^n$$ where $a_n = (-1)^{(n+1)/2}/(-n)!$ for odd $n < 0$, $0$ otherwise.
I'm not quite sure what you're referring to as "Laurent's theorem". But any Taylor series is a Laurent series where the terms in negative powers happen to be $0$, so wherever Taylor expands the function, so does Laurent.