Find the Laurent series for $\sin πz/(4z^2-1)$ about the point $z=1/2$

Question

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.

Answer 1

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?

Answer 2

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