I have a question regarding an exercise that is concerned with a Laurent series expansion.
This is my solution for a):
Consider the annuli
$$
A_1:=\{z \in \mathbb{C} \mid 0<|z-1|<2 \} \\
A_2:=\{z \in \mathbb{C} \mid 0<|z+2i|<2 \}.
$$
We have, $$ |-2i-1| =|2i+1| =\sqrt{5} >2. $$ So $1 \notin A_2, \ -2i \notin A_1$.
Consider $$ f(z) =\frac{5}{z^2+(2i-1)z-2i} =\frac{5}{(z+2i)(z-1)} $$ and $$ f_1: A_1 \to \mathbb{C}, \ g(z)=\frac{5}{z+2i}=(z-1) f(z) \\ f_2: A_2 \to \mathbb{C}, \ h(z)=\frac{5}{z-1}=(z+2i) f(z). $$
By the above discussion, $g$ and $h$ are holomorphic on their respective domains as they have no singularities. By the Laurent series theorem we obtain the representations $$ f_1(z)=\sum_{n=-\infty}^{\infty} a^{(1)}_n (z-1)^n $$ $$ f_2(z)=\sum_{n=-\infty}^{\infty} a^{(2)}_n (z+2i)^n, $$ where $$ a^{(1)}_n =\frac{1}{2 \pi i} \int_{|z-1|=\rho_1} \frac{f_1(z)}{(z-1)^{n+1}} $$ $$ a^{(2)}_n =\frac{1}{2 \pi i} \int_{|z+2i|=\rho_2} \frac{f_2(z)}{(z+2i)^{n+1}}, $$ with $0<\rho_i<2, \ i=1,2$. From this we obtain the representations $$ f(z) =\frac{f_1(z)}{z-1} =\sum_{n=-\infty}^{\infty} a^{(1)}_n (z-1)^{n-1} $$
$$ f(z) =\frac{f_2(z)}{z+2i} =\sum_{n=-\infty}^{\infty} a^{(2)}_n (z+2i)^{n-1}. $$
on $A_1,A_2$ respectively. Using the substitution $k=n-1$ we obtain
$$ f(z)=\sum_{k=-\infty}^{\infty} a^{(1)}_{k+1} (z-1)^k $$
$$ f(z) =\sum_{k=-\infty}^{\infty} a^{(2)}_{k+1} (z+2i)^k. $$
By defining $a_k:=a^{(1)}_{k+1}, b_k:=a^{(2)}_{k+1}$, we obtain the Laurent Series for $f$ around $1$ and $-2i$. q.e.d.
However, I am not entirely satisfied with this. Considering the wording of b) I think I am supposed to find a more explicit formula for the Laurent Series. But how does one derive a more explicit formula in this case? The only thing I was able to observe is that $f$ has poles at the singularities, therefore the Laurent Series expansions should have finite principal part. But I do not see how this helps.