Finding Taylor Series of $ f(z) = \frac{z^2-2z+5}{(z-2)(z^2+1)}$ in $1<|z|<2$. I have to find Taylor series of
$f(z)=\frac{z^2-2z+5}{(z-2)(z^2+1)}$, for $1<|z|<2$.
I have started with
\begin{eqnarray*}
f(z)&=&\textstyle\frac{z^2 -2z +5}{(z-2)(z^2+1)}\\[4pt]
&=&\textstyle\frac{z^2+1-2(z-2)}{(z-2)(z^2+1)}\\[4pt]
&=&\textstyle\frac{1}{z-2}-\frac{2}{z^2+1}\\[4pt]
&=&\textstyle\frac{1}{-2\left(1-\frac{z}{2}\right)}-\frac{2}{z^2\left(1+\frac{1}{z^2}\right)}\\[4pt]
&=&-\sum\limits_{n=0}^{\infty}\frac{z^n}{2^{n+1}}-2\sum\limits_{n=0}^{\infty}(-1)^nz^{-2n-2}.
\end{eqnarray*}
But, this looks like Laurent series. Am I right about that?

First decompose your expression into partial fractions. $$\frac{z^2-2z+5}{(z-2)(z^2+1)}=\frac{A}{z-2}+\frac{Bz+C}{z^2+1}.$$ You will find that $A=1,B=0,C=-2.$ Then $$\frac{z^2-2z+5}{(z-2)(z^2+1)}=\frac{1}{(-2)(1-\frac{z}{2})}+\frac{-2}{1+z^2}$$ $$=(-1/2)(1-\frac{z}{2}+\frac{z^2}{4}-...)-2(1-z^2+z^4-...)$$ $$=\sum_{n=0}^{\infty}a_nz^n$$ where $a_n=\frac{-1}{2^{n+1}}$ if $n$ is odd and $$a_n=\frac{-1}{2^{n+1}}+2(-1)^{1+n/2}\text { if $n$ is even.}$$
