I have tried to determine residues of the below function via Mathematica and Matlab, but they lead me nowhere. For small enough $\alpha$, I figured out what are the poles, but nothing about residues. My method was somehow straight-forward, using Taylor series. Does anybody have an idea to calculate its residues?
$\frac{e^{-\sqrt{z(z+r))}}}{1+\alpha\sqrt{z(z+r)} + (1-\alpha \sqrt{z(z+r)})e^{-2\sqrt{z(z+r))}}}$
Thanks in advance.
$$\lim_{\alpha\to 0}\frac{e^{-\sqrt{z(z+r))}}}{1+\alpha\sqrt{z(z+r)}+(1-\alpha \sqrt{z(z+r)})e^{-2\sqrt{z(z+r))}}}=$$ $$\lim_{\alpha\to 0}\frac{1}{e^{\sqrt{z(z+r))}}\left(1+\alpha\sqrt{z(z+r)}+\frac{1-\alpha\sqrt{z(z+r)}}{e^{2\sqrt{z(z+r)}}}\right)}=$$ $$\frac{1}{\lim_{\alpha\to 0}\left(e^{\sqrt{z(z+r))}}\left(1+\alpha\sqrt{z(z+r)}+\frac{1-\alpha\sqrt{z(z+r)}}{e^{2\sqrt{z(z+r)}}}\right)\right)}=$$ $$\frac{1}{\lim_{\alpha\to 0}\left(e^{\sqrt{z(z+r))}}\left(1+\alpha\sqrt{z(z+r)}+e^{-2\sqrt{z(z+r)}}\left(1-\alpha\sqrt{z(z+r)}\right)\right)\right)}=$$ $$\frac{1}{e^{\sqrt{z(z+r))}}\left(1+0\cdot\sqrt{z(z+r)}+e^{-2\sqrt{z(z+r)}}\left(1-0\cdot\sqrt{z(z+r)}\right)\right)}=$$ $$\frac{1}{e^{\sqrt{z(z+r))}}\left(1+0+e^{-2\sqrt{z(z+r)}}\left(1-0\right)\right)}=$$ $$\frac{1}{e^{\sqrt{z(z+r))}}\left(1+e^{-2\sqrt{z(z+r)}}\right)}=$$ $$\frac{1}{e^{-\sqrt{z(z+r)}}+e^{\sqrt{z(z+r)}}}=$$ $$\frac{e^{\sqrt{z(z+r)}}}{1+e^{2\sqrt{z(z+r)}}}=\frac{\text{sech}\left(\sqrt{z(z+r)}\right)}{2}$$