How do I find the Laurent series for $\frac{1}{z^2 – 4}$ at $z = 2$?
Can anyone help me out with this? I used partial fraction decomposition and got the $\frac{1}{4}$ part of it, just don't understand the rest. I think I may be treating to much like an infinite series
Best Answer
Note that for $z\neq\pm 2$ we can write $$\frac1{z+2}=\cfrac1{z-2+4}=\frac1{4}\cdot\cfrac1{1-\left(-\frac{z-2}{4}\right)}$$ and $$\frac1{z+2}=\cfrac1{z-2+4}=\frac1{z-2}\cdot\cfrac1{1-\left(-\frac{4}{z-2}\right)}.$$
Now, one of these can be expanded as a multiple of a geometric series in the annulus $|z-2|>4$ and the other can be expanded as a multiple of a geometric series in the annulus $0<|z-2|<4$. That is, we will use the fact that $$\frac1{1-w}=\sum_{k=0}^\infty w^k$$ whenever $|w|<1$. You should figure out which annulus works for which rewritten version, and find the respective expansions in both cases. That will give you two different Laurent expansions of your function, valid in two different annuli.