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?

Since the poles of $f$ are $\pm i$ and $2$, the radius of convergence of the Taylor series of $f$ around $0$ is $1$ (the smallest distance from $0$ to the poles).
It is then divergent for $1<|z|<2$.
For $|z|<1$ we have $f(z)=\sum\limits_{k\geqslant0}a_kz^k$,
where $a_k=\begin{cases}
\displaystyle -\frac1{2^{k+1}}+2(-1)^{\frac k2+1}&\textrm{if $k$ is even,}\\[8pt]
\displaystyle -\frac1{2^{k+1}}&\textrm{if $k$ is odd.}\end{cases}$
