<s>[INST] Finding Laurent Series of $ f(z) = \frac{z^2-2z+5}{(z-2)(z^2+1)}$ at $z=2$. I need to find the Laurent Series for $z=2$, $z=i$, and $1<|z|<2$ for the following function:
$$ f(z)= \frac{z^2 -2z +5}{(z-2)(z^2+1)}.$$
I was thinking if I could get something like $$\frac{1}{1 - (z-2)}$$ I could use geometric series.
Like in this example for $z=-1$:
$$f(z)=\frac{1}{1-z} =\frac{1}{2-(z+1)} =\frac{1}{2\left(1-\frac{z+1}{2}\right)}=\frac{1}{2}\sum \left(\frac{z+1}{2}\right)^n.$$
Unfortunately, I'm not getting anywhere with that approach. What am I missing?
edit: can I rewrite it like this:
$$f(z)=\frac{1}{z-2}-\frac{i}{z+i}+\frac{i}{z-i}$$
and then use Taylor?
 [/INST] *

*$f(z)=\frac{z^2 -2z +5}{(z-2)(z^2+1)}=\frac{z^2+1-2(z-2)}{(z-2)(z^2+1)}=\frac{1}{z-2}-\frac{2}{z^2+1}=\frac{1}{z-2}-\frac{2}{(z-2)^2+4z-3}=\frac{1}{z-2}-\frac{2}{(z-2)^2+4(z-2)+5}=\frac{1}{z-2}-\frac{2}{5}[1+\{\frac{4}5(z-2)+\frac{1}5(z-2)^2\}]^{-1}.$ 


Hence etc. 


*

*$f(z)=\frac{z^2 -2z +5}{(z-2)(z^2+1)}=\frac{z^2+1-2(z-2)}{(z-2)(z^2+1)}=\frac{1}{z-2}-\frac{2}{z^2+1}=\frac{1}{z-2}-\frac{2}{(z-i)(z+i)}=\frac{i}{z-i}+(\frac{1}{z-2}-\frac{i}{z+i})=\frac{i}{z-i}+\frac{1}{z-i-2+i}-\frac{i}{z-i+2i}=\frac{i}{z-i}+\frac{1}{i-2}(1+\frac{z-i}{i-2})^{-1}+\frac{1}{2}(1+\frac{z-i}{2})^{-1}.$ 


Hence etc.
</s>