Laurent series for $f(z) = (z^2-4z+5)^{-1}$ centered at $z_{0} = 1$ I know there's plenty of questions on Laurent series, but I haven't found any suitable for my case.
I have to find the series for $f(z) = \dfrac{1}{z^2-4z+5}$ for a disk centered at $z_{0} = 1$ with a radius $R = \sqrt{2}$ and later for an annulus centered also at $z_ {0} = 1$ and from radius $\sqrt{2}$ to $\infty$.
Figuring out how to find the series for the disk, I suppouse I will be able to find the one for the annulus.
I have rewritten the expression as:
\begin{equation}
f(z) = \frac{A}{z-2+i} + \frac{B}{z-2-i} = \frac{i/2}{z-2+i} + \frac{-i/2}{z-2-i}
\end{equation}
So I am able to find the series for each expression,
\begin{equation}
\frac{i/2}{z-1-1+i} = \frac{i/2}{(-1+i)(1+\frac{z-1}{-1+i})} = \frac{i}{2(-1+i)}\sum_{n=0}^{\infty} (-1)^{n} \left( \frac{z-1}{-1+i} \right) ^{n}
\end{equation}
The same for the other one,
\begin{equation}
\frac{-i/2}{z-1-1-i} = \frac{i}{2(1+i)}\sum_{n=0}^{\infty} (-1)^{n} \left( \frac{z-1}{-1-i} \right) ^{n}
\end{equation}
In both cases, the radius of convergence is $\sqrt{2}$
\begin{equation}
\left| \left( \frac{z-1}{-1-i} \right)  \right| < 1 \rightarrow |z-1| < \left| -1-i \right| = \sqrt{(-1)^{2} + (-1)^{2}} = \sqrt{2} 
\end{equation}
I think I haven't made any mistake until now. My problem is that I haven't found out a way to add the sums, if that is the operation that I should do, to obtain the result. To be more precisely, I don't even know where to start to get the result:
\begin{equation}
f(z) = \sum_{n=0}^{\infty} \frac{(z-1)^{n}}{(\sqrt{2})^{n+1}} \sin (n+1)\frac{\pi}{4} = \frac{1}{2}+\frac{1}{2}(z-1)+\frac{1}{4}(z-1)^{2}-\frac{1}{8}(z-1)^{4} + \mathcal{O}(z-1)^{5}
\end{equation}
Thanks for your time

Hint: These and de Moivre should help:  $$(-1+i)^{-1} = -\sqrt{2}^{-1}(\cos \pi/4 +i \sin \pi/4) $$
$$(-1-i)^{-1} = -\sqrt{2}^{-1}(\cos \pi/4 -i \sin \pi/4) $$
