Simplification of a complicated fraction

Question

I am going over a physics text and I have difficulty to see how one can go from $$2A = (1+ \frac{\alpha}{ik})(1+\frac{ik}{\alpha})\frac{Fe^{ika}e^{-\alpha a}}{2} + (1- \frac{\alpha}{ik})(1-\frac{ik}{\alpha})\frac{Fe^{ika}e^{\alpha a}}{2}$$ to $$\frac{Ae^{-ika}}{F} = \cosh(\alpha a) + i(\frac{\alpha^2 - k^2}{2k \alpha})\sinh(\alpha a)$$ Could you give me some hints on how to proceed? I, obviously, know $\cosh(x) = \frac{e^x+e^{-x}}{2}$ and $\sinh(x) = \frac{e^x-e^{-x}}{2}$ but the connection between the two steps is not obvious to me.

Answer 1

Dividing by $2e^{ik\alpha}F$ gives $$\frac{Ae^{-ik\alpha}}F=\left(1+\frac\alpha{ik}\right) \left(1+\frac{ik}\alpha\right)\frac{e^{-a\alpha}}4 +\left(1-\frac\alpha{ik}\right) \left(1-\frac{ik}\alpha\right)\frac{e^{a\alpha}}4.$$ Next, $$\left(1+\frac\alpha{ik}\right) \left(1+\frac{ik}\alpha\right)=\frac{(ik+\alpha)^2}{ik\alpha} =\frac{2k\alpha+i(k^2-\alpha^2)}{k\alpha}$$ and similarly $$\left(1-\frac\alpha{ik}\right) \left(1-\frac{ik}\alpha\right)=\frac{2k\alpha-i(k^2-\alpha^2)}{k\alpha}.$$ Then, $$\frac{Ae^{-ik\alpha}}F=\frac{e^{-a\alpha}+e^{a\alpha}}2 +i\frac{k^2-\alpha^2}{2k\alpha}\frac{e^{-a\alpha}-e^{a\alpha}}2 $$ which equals $$\cosh a\alpha+i\frac{\alpha^2-k^2}{2k\alpha}\sinh a\alpha.$$

Answer 2

Hint: Multiply your first equation by $$\frac{e^{-ika}}{2F}$$ and expanding your paranthesis we get $$\left(2+\frac{\alpha}{ik}+\frac{ik}{\alpha}\right)e^{-\alpha a}+\left(2-\frac{ \alpha}{ik}-\frac{ik}{\alpha}\right)e^{\alpha a}$$