Laurent series of $f(z) = \frac{5}{z^2 + (2i - 1)z - 2i}$ Let $f: \mathbb{C} \setminus \left\{ 1, -2i \right\} \rightarrow \mathbb{C}$ with $$f(z) = \frac{5}{z^2 + (2i - 1)z - 2i}$$
Determine the Laurent series of $f$ in the annuli $\left\{ z \in \mathbb{C} : 0 < |z + 2i| < \sqrt{5} \right\}$.
I think I didn't understand properly how the Laurent series works. Can someone explain me how to approach problems like this?

By a standard fraction decomposition, one has
$$
f(z) = \frac{5}{z^2 + (2i - 1)z - 2i}=\frac{1-2i}{z+2i}+\frac{2i-1}{1-z}
$$wich rewrites
$$
\frac{1-2i}{z+2i}+\frac{2i-1}{1-z}=\frac{1-2i}{z+2i}+\frac{2i-1}{1+2i}\cdot\frac{1}{1-\frac{z+2i}{1+2i}}=\frac{1-2i}{1+2i}\cdot \frac1Z+\frac{2i-1}{1+2i}\cdot\frac{1}{1-Z}
$$with
$$
Z=\frac{z+2i}{1+2i},\qquad |Z|<1,
$$ giving
$$
f(z)=\tilde{f}(Z)= \frac{1-2i}{1+2i}\cdot \frac1Z+\frac{2i-1}{1+2i}\cdot\sum_{n=0}^\infty Z^n.
$$ Can you take it from here?
