Let $f(z)=\frac{e^z}{1+z^2}$ be given. It has a two isolated singularities $\{i,-i\}$.
I know from the Laurent expansion that it has poles of order 1 at these points. However, I want to proof this not using the Laurent expansion.
This means I want to get to the following conclusion:
$\exists\ a_1,..,a_m\in \mathbb{C},m\in\mathbb{N},a_m\ne0$ so that
$f(z)-\sum_{k=1}^m \frac{a_k}{(z-a)^k}$
has a removable singularity at $a$. (In our case these would be $\{i,-i\}$)
Somehow all my approaches led to nothing and I could not show it this way.
Hint
Find $A, B \in \mathbb C$ such that $$g(z) = f(z) - \frac{A}{z-i} -\frac{B}{z+i} \tag0$$ is holomorphic on $\mathbb C \setminus\{i,-i\}$ and with existing limits $\lim\limits_{z \to \pm i} g(z)$.