Question about finding Laurent Series over closed region and classifying singularity Represent $\sin(\pi x/(x+1))$ Laurent Series about the region $0<|x+1|<2$:
Its true that $$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$
So the $$\sin(\pi x/(1+x))=\sum (-1)^{n-1} \frac{(\pi x/(x+1))^{2n-1}}{(2n-1)!}$$
I'm not so sure what to do after this. Would its singularity be at z=-1 and how would you determine what kind of singularity it is and whether or not its a pole?  
What's puzzling me a bit here is understanding how that disk is playing a role here.  Can someone also explain that a bit more?  

What you have written is not a Laurent series: a Laurent series about a point $a$ should be of the form
$$ \sum_{k=-\infty}^{\infty} a_k (z-a)^k. $$
In this case, the function has an essential singularity at $x=-1$, because (as the power series you have for the $\sin{x}$ shows) $\sin$ has an essential singularity at $x=\infty$.
Now, to involve the disc. Write $z=1+x$, so the set is $\{ z: 0 < \lvert z \rvert < 2 \}$. The function is then
$$ \sin{\left( \frac{\pi(x+1-1)}{x+1} \right)} = \sin{\left( \frac{\pi(z-1)}{z} \right)} = \sin{\left( \pi - \frac{\pi}{z} \right)} = \sin{\left(\frac{\pi}{z} \right)} $$
using the symmetry of the sine. Therefore, because the Maclaurin series for $\sin{x}$ converges for any $x$, the expansion
$$ \sin{\left(\frac{\pi}{z} \right)} = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}\left(\frac{\pi}{z} \right)^{2k+1} = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}\left(\frac{\pi}{x+1} \right)^{2k+1} $$
converges for any $z \neq 0$, so this series is a valid Laurent expansion for the function in the region requested (and, in fact, beyond).
