Say I have ended up with the expression: $\dfrac{ke^{ix}-ke^{ix}}{e^{ix}-e^{ix}}$.
Let's rewrite it where $A = k$ and $B = e^{ix}$: $\dfrac{AB-AB}{B-B} = \dfrac{(1-1)AB}{(1-1)B}$.
Again let's rewrite it where $C = (1-1)$: $\dfrac{(1-1)AB}{(1-1)B} = \dfrac{CAB}{CB} = A$.
It seems contrived and wrong; I'm trying to get rid of the infinity, but the equation I am trying to derive from equation is probably just wrong.
EDIT:
I got here by trying to solve a differential equation for a piece-wise function $V(x)$, where: $$V(x) = \begin{cases} x \lt 0 & 0 \\ 0 \leq x \leq L & V_0 \\ x \gt L & 0 \end{cases} $$
The differential equation is the Time Independent Schrödinger Equation
$\psi( x ) = \frac{-\hbar^2\mathrm{d}^2\psi(x)}{2m\mathrm{d}x^2} + V(x)\psi(x) = E\psi(x) \rightarrow \frac{\mathrm{d}^2\psi(x)}{\mathrm{d}x^2} - \frac{2m(E-V(x))}{\hbar^2}\psi(x) = 0.$
To make things easier: $\dfrac{\mathrm{d}^2\psi(x)}{\mathrm{d}x^2} - k^2\psi(x) = 0$ where: $k^2 = \dfrac{2m(E-V(x))}{\hbar^2}.$
So when $L < x < 0$ in regions 0 and 2 $k_2^2 = k_0^2 = \dfrac{2m(E-0)}{\hbar^2} = \dfrac{2mE}{\hbar^2}$, and $k_1 = \frac{2m(E-V_0)}{\hbar^2}.$
For this scenario we assume $V_0 < E$ (physical assumption).
We end up with 3 equations
$
\psi_0(x) = Ae^{ik_0x} + Be^{-ik_0x},$
$\psi_1(x) = Ce^{ik_1x} + De^{-ik_1x},$
$\psi_2(x) = Fe^{ik_2x} + Ge^{-ik_2x}.$
Also, due to our physical constraints (no back propagation) $G = 0$, thus: $\psi_2(x) = Fe^{ik_2x}$.
We also know: $\psi_2(L) = Fe^{ik_2L} = \psi_1(L) = Ce^{ik_1L} + De^{-ik_1L} $
and: $ psi_0(0) = psi_1(0) = A + B = C + D.$
As the function $\psi(x)$ must be continuous (and exists in a Hilbert Space)
I put everything into an automatic solver to solve for the system, it results in:
$$ A = \frac{F\left(k_{0} \left(k_{1} + i k_{2} + \left(k_{1} - i k_{2}\right) e^{2 L k_{1}}\right) + i k_{1} \left(k_{1} + i k_{2} + \left(- k_{1} + i k_{2}\right) e^{2 L k_{1}}\right)\right) e^{- L \left(k_{1} - i k_{2}\right)}}{4 k_{0} k_{1}} $$ $$ B = \frac{F \left(k_{0} \left(k_{1} + i k_{2} + \left(k_{1} - i k_{2}\right) e^{2 L k_{1}}\right) + i k_{1} \left(- k_{1} - i k_{2} + \left(k_{1} - i k_{2}\right) e^{2 L k_{1}}\right)\right) e^{- L \left(k_{1} - i k_{2}\right)}}{4 k_{0} k_{1}} $$ $$ C = \frac{F\left(k_{1} - i k_{2}\right) e^{L \left(k_{1} + i k_{2}\right)}}{2 k_{1}} $$ $$ D \frac{F\left(k_{1} + i k_{2}\right) e^{- L \left(k_{1} - i k_{2}\right)}}{2 k_{1}} $$
I attempted to normalize F, but this is wrong because its a piecewise function and not equal too the whole probability distribution:
$$ \int\limits_{0}^{L} F \overline{F}\, dx = 1 $$
That's how I got:
$$ \frac{i G e^{i L k_{2}} \overline{G}}{- k_{2} e^{i L \overline{k_{2}}} + e^{i L \overline{k_{2}}} \overline{k_{2}}} - \frac{i G \overline{G}}{- k_{2} + \overline{k_{2}}} = 1 $$
$$ \left|{G}\right| = \sqrt{- \frac{i \left(- k_{2} + \overline{k_{2}}\right) e^{i L \overline{k_{2}}}}{e^{i L k_{2}} - e^{i L \overline{k_{2}}}}} $$
But its wrong, thats how I got here