How to find the Laurent series expansion of $\frac{2}{z^2-4z+8}$ by long division? I'm trying to find the Laurent series expansion for
$$ \frac{2}{z^2-4z+8} $$
using polynomial long division.
However, I noticed that if I divide leading with the $8$ term, then I will only get positive power terms, namely
$$ \frac{1}{4}+\frac{1}{8}z+\frac{1}{32}z^2 + ... $$
Whereas if I divide leading with the $z^2$ term, then I will get only negative power terms, as in
$$ \frac{2}{z^2} + \frac{8}{z^3}+\frac{16}{z^4} + ... $$
What is going on?
(I'm doing this to find the residues, i.e. the coefficient of the $\frac{1}{z}$ term.)
 A: Well you wan't to find a function
$$
f(z):= \sum_{n=-\infty}^\infty a_n z^n
$$
defined around $0$ such that $f(z)(z^2-4z+8)=2$. If
$$
f(z)=\frac{2}{z^2} + \frac{8}{z^3}+\frac{16}{z^4} + \cdots 
$$
then $f$ is not defined at $0$. However, if
$$
f(z)= \frac{1}{4}+\frac{1}{8}z+\frac{1}{32}z^2 + \cdots
$$
then $f$ is defined at $0$ and doing some algebra you can check that $f(z)(z^2-4z+8)=2$. Thus, your first approach is the correct one. Of course this means that the coefficient for $\frac{1}{z}$ is $a_{-1}=0$.
A: The first expansion is a correct power series expansion in the disk centered at $0$ with the two bad points at its edge. The second is a correct Laurent series expansion outside that disk.
The first one is valid in the disk, and it shows (although circuitously) that the residue at $0$ is $0$, because the function is in fact holomorphic there. This was really already visible from the algebraic expression.
The second expansion is only valid outside that disk centered at $0$, so does not give info in a straightforward way about what happens inside...
A: You could use the fact that $res(z_{0},f)=\lim_{z \to z_{0}} (z-z_{0})f(z)$ once $f(0)=\frac{1}{4}$$-f$ is bounded in a neighbourhood of $0$ and so $res(0,\frac{2}{z^2-4z+8})=0$. If you prefer you can also notice that $f$ is analytic in a neighbourhood of $0$ then through its Taylor series $res(0,\frac{2}{z^2-4z+8})=0$.
