I have to found the Laurent Series of $f(z)=\frac{e^z}{z^2-1}$ on the annulus $0<|z|<1$. I'm not sure what to do here. I know I can expand the $e^z$: $f(z)=\frac{1+z+z/2!..}{z^2-1}$, but I am not sure if this is the right approach. I am also a little confused about the "on the annulus" part.
So my question is how to find the right series :)
Thanks in advance!
An idea:
$$\frac{e^z}{z^2-1}=-e^z\frac1{1-z^2}=-\left(1+z+\frac{z^2}2+\frac{z^3}6+\ldots+\frac{z^k}{k!}+\ldots\right)\left(1+z^2+z^4+\ldots+z^{2k}+\ldots\right)=$$
$$=-\left(1+z+\frac32z^2+z^3+\frac32z^4+\ldots\right)$$
The above is valid for $\;|z|<1\;$, and observe we get a power series, as expected from an analytic function on $\;0<|z|<1\;$ . We cannot expand this annulus anymore as the function has two singular points on $\;|z|=1\;$ ...
In fact, the "annulus thing" is just for the purpose of considering a Laurent series. After the work above is done, we can see that taking off the point $\;z=0\;$ is moot and we can add it, working simply with $\;|z|<1\;$ .