$$ F_\alpha(z)=\frac{2}{\pi \alpha}\sum_{n=0}^\infty \frac{1}{n^2+z/\alpha^2} $$
The answer is:
$$ F_\alpha(z)=z^{-\frac{1}{2}}\coth(\pi z^{\frac{1}{2}}/\alpha)+\alpha/\pi z $$
It is easy to show that these two expression has the same analytic behavior, i.e. they have the same poles at $z=-n^2\alpha^2$.
Additional concern
It is also asked that if $\alpha\to0$, the analytic function has a branch cut at negative real axis. It is intuitive that all the poles will "merge" into a branch cut in this limit. However, how can I start from the analytical form of $F_\alpha(z)=z^{-\frac{1}{2}}\coth(\pi z^{\frac{1}{2}}/\alpha)+\alpha/\pi z$ and give a somehow more rigors proof of this statement, i.e. can I simplify the expression to some form such as $\log z$ so that I can see directly that it has a branch cut at real negative axis...
This formula is essentially the same thing as the partial fractions expansion of $\cot z$, so you'll find a lot of information if you search for this.
Here's a sketch of a direct argument: write $w=\pi z^{1/2}/\alpha$. We then want to show that $$ \coth w = \frac{1}{w} + \sum_{n\ge 1} \frac{2w}{w^2+n^2\pi^2} . \quad\quad\quad\quad (1) $$ The RHS could in fact be further rewritten as $$ \sum_{n\in\mathbb Z} \frac{1}{w+in\pi} , $$ with the understanding that I will take symmetric partial sums now.
As you already observed, the poles and their residues are the same on both sides. So $F(w)$, defined as the difference of the two sides, is entire. It's also easy to see that both sides of (1) are bounded away from the poles. It follows that $F$ is constant.