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Michael Hardy
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$\operatorname{Coth}(\alpha_n a) \to i$ when $n \to \infty $

I have a question about the following statement from an article here. He states on page 490 that:

https://sci-hub.ru/10.1002/mana.201200319

Let $\alpha = \dfrac{\rho_1\lambda^2}{\kappa_1 + i\kappa_2\lambda}$ where $\rho_1, \kappa_1, \kappa_2 > 0$. then $$ \operatorname{Coth}(i\alpha L) \to 1, \ \ \text{ when } \ \ n \to \infty $$ where $L > 0$.

Well, firstly, I think he's changing $\alpha$ to $\alpha_{n}$ and leaving $n$ hidden. And doing some mathematical manipulations, I'm getting to

Let $x_n = \Re(\alpha_n L)$ and $y_n = \Im(\alpha_n L)$. Using some properties, I can get $$ \operatorname{Coth}(i\alpha_n L) = \dfrac{\cos(x_n)\sin(x_n)}{\cosh^2 (y_n) - \cos^2(x_n)} - \dfrac{\tanh(y_n)} {\frac{\cos^2(x_n)}{\cosh^2 (y_n)} - 1} $$

From the equality above can I conclude what I want? Why?

user253963
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