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?

Solutions Collecting From Web of "Finding Laurent Series of $ f(z) = \frac{z^2-2z+5}{(z-2)(z^2+1)}$ at $z=2$."

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

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.