$\newcommand\la\lambda\newcommand\al\alpha\newcommand\ka\kappa\newcommand\th\theta$First of all, you copied the expression for $\al$ incorrectly. On p. 489 of the paper linked by you (the fourth display from the bottom of the page), we find $$\al=r(\la)e^{i\th/2},$$ where $\mathbb R\ni\la=\la_n\to\infty$, $$r(\la):=\frac{\rho_1^{1/2}\la}{\sqrt[4]{\ka_1^2+\ka_2^2\la^2}}\to\infty, \quad e^{i\th/2}\to\frac{1-i}{\sqrt2}, $$ and $\rho_1,\ka_1,\ka_2$ are fixed positive real numbers. So, $$|e^{-2i\al L}|=|\exp(-2iLr(\la)e^{i\th/2})| =\exp(-(\sqrt2+o(1))\,Lr(\la))\to0.$$ Thus, $$\coth(i\al L)=\frac{1+e^{-2i\al L}}{1-e^{-2i\al L}}\to\frac{1+0}{1-0}=1,$$ as desired.

$\newcommand\la\lambda\newcommand\al\alpha\newcommand\ka\kappa\newcommand\th\theta$First of all, you copied the expression for $\al$ incorrectly. On p. 489 of the paper linked by you (the fourth display from the bottom of the page), we find $$\al=r(\la)e^{i\th/2},$$ where $\mathbb R\ni\la=\la_n\to\infty$, $$r(\la):=\frac{\rho_1^{1/2}\la}{\sqrt[4]{\ka_1^2+\ka_2^2\la^2}}\to\infty, \quad e^{i\th/2}\to\frac{1-i}{\sqrt2}, $$ and $\rho_1,\ka_1,\ka_2$ are fixed positive real numbers. So, $$|e^{-2i\al L}|=|\exp(-2iLr(\la)e^{i\th/2})| =\exp(\Re(-2iLr(\la)e^{i\th/2})) \\ =\exp(-(\sqrt2+o(1))\,Lr(\la))\to0.$$ Thus, $$\coth(i\al L)=\frac{1+e^{-2i\al L}}{1-e^{-2i\al L}}\to\frac{1+0}{1-0}=1,$$ as desired.