Let $f=\frac{2}{z}-\frac{3}{z-2}+\frac{1}{z+4}$ . Find the Laurent series in all possible regions of convergence about z=0 and read the residuum.
I am not sure if I have to consider all 3 regions (inside circle pf radius 2,
inside annulus $2<r<4$ and ouside circle of radius 4). Can someone guide me on how to continue? Thank you!
Best Answer
Note that for $z\neq -4,0,$ we can write $$\frac1{z+4}=\cfrac1{4-(-z)}=\frac1{4}\cdot\cfrac1{1-\left(-\frac{z}{4}\right)}$$ and $$\frac1{z+4}=\cfrac1{z-(-4)}=\frac1{z}\cdot\cfrac1{1-\left(-\frac{4}{z}\right)}.$$
Now, one of these can be expanded as a multiple of a geometric series in the disk $|z|<4,$ and the other can be expanded as a multiple of a geometric series in the annulus $|z|>4$. 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 region works for which rewritten version, and find the respective expansions in both cases.
Likewise, we can rewrite $\frac1{z-2}$ in two similar forms, one of which is expandable in $|z|<2$ and one of which is expandable in $|z|>2.$
Using these expansions will give you three different Laurent expansions of $f(z),$ one for each of the regions $0<|z|<2,$ $2<|z|<4,$ and $|z|>4$.