Show that the box scheme
$$\frac{1}{2k}\Big[(U_j^{n+1}+U_{j+1}^{n+1})-(U_j^n + U_{j+1}^n)\Big]+\frac{a}{2h}\Big[(U_{j+1}^{n+1}-U_{j}^{n+1})+(U_{j+1}^n - U_{j}^n)\Big]$$
is convergent for the one-way wave equation
$$u_t+au_x=0, ~~~-\infty<x<\infty, ~t>0,~ x\in\mathbb R$$
Using von Neumann stability analysis and letting
$$U_j^n=g(\xi)^ne^{ijh\xi}$$
where
$$h\xi=\theta$$
one finds that
$$(g+g e^{i\theta})-(1+e^{i\theta})+\frac{ak}{h}((g e^{i\theta}-g)+(e^{i\theta}-1)=0$$
which reduces to
$$g=\frac{(1+e^{i\theta})-\frac{ak}{h}(e^{i\theta}-1)}{(1+e^{i\theta})+\frac{ak}{h}(e^{i\theta}-1)}$$
The goal is to show that
$$|g|\leq 1$$
or
$$\Bigg|\frac{(1+e^{i\theta})-\frac{ak}{h}(e^{i\theta}-1)}{(1+e^{i\theta})+\frac{ak}{h}(e^{i\theta}-1)}\Bigg|\leq 1$$
However, I cannot seen to get this argument to work. If I try
$$g=\frac{(1+e^{i\theta})-z}{(1+e^{i\theta})+z}$$
where
$$z=\frac{ak}{h}(e^{i\theta}-1)$$
then I see that
$$z\leq 0 ~~\Rightarrow ~~ |g|\leq 1$$
However, this doesn't make any sense as $z$ is defined as a complex number. Is there a different approach of attack for the final step of this problem?
I think that the scheme is stable for all choices of $a,k,$ and $h$ (and therefore convergent by the Lax Equivalence Theorem). However, I cannot find a way of reducing the final inequality to find the valid range of values for
$$\frac{ak}{h}$$