Find the Laurent series for $\sin πz/(4z^2-1)$ about the point $z=1/2$ I really don't know where I should start. I thought first expand $\sin(\pi z - 1/2)$ using Maclaurin Series and break down the denominator by factoring, but I have no idea from here. 
 A: Let $z=1/2+w$ then
$$\frac{\sin (\pi z)}{4z^2-1}=\frac{\sin (\pi/2 +\pi w)}{4(1/2+w)^2-1}
=\frac{\cos(\pi w)}{4w(1+w)}.$$
Now recall that the expansion of $\cos(\pi w)$ and $(1+w)^{-1}$ at $0$:
for $w\in\mathbb{C}$,
$$\cos(\pi w)=\sum_{k=0}^{\infty}\frac{(-1)^k(\pi w)^{2k}}{(2k)!}=1-\frac{\pi^2 w^2}{2}+O(w^4)$$
and  for $|w|<1$,
$$\frac{1}{1+w}=\sum_{k=0}^{\infty}(-1)^k w^k=1-w+w^2+O(w^3).$$
Hence the first few terms of the Laurent expansion at $w=0$ can be found by expanding 
$$\frac{1}{4w}\left(1-\frac{\pi^2 w^2}{2}+O(w^4)\right)\left(1-w+w^2+O(w^3)\right).$$
Can you take it from here?
A: Your expansion must be in negative powers of $\frac{1}{z-\frac{1}{2}}$ or any factor of that.
$$\frac{1}{4z^2-1}= \frac{1}{4z^2}\frac{1}{1-(\frac{1}{2z})^2}$$
Taylor expansion of, $$sin(\pi z)=\pi z - \frac{(\pi z)^3}{3!} +\frac{(\pi z)^5}{5!}-\frac{(\pi z)^7}{7!}...........$$
$$sin(\pi z)\frac{1}{4z^2-1}=sin(\pi z) \frac{1}{4z^2}\frac{1}{1-(\frac{1}{2z})^2} $$
$$=\frac{1}{4}\frac{1}{1-(\frac{1}{2z})^2}\Bigg(\frac{\pi}{z}- \frac{\pi ^3 z}{3!}+\frac{\pi^5 z^3}{5!}-\frac{\pi^7 z^5}{7!}....................\Bigg)$$
This is probably the simplest laurent expansion i can think,the other methods might involve you finding cauchy product.I hope this helps
