Does it make sense to think of $e^{ikx}\equiv $cos$(kx)+i$sin$(kx)$ as a right propagating wave? I am rather confused by the imaginary term here.
Context:
\begin{equation} \phi(x)=\begin{cases} Ae^{ik_1x}+Be^{-ik_1x}, & \text{if $x<0$}.\\ Ce^{ik_2x}+De^{-ik_2x}, & \text{if $0 \leq x \leq a$}.\\ Ee^{ik_3x}+Fe^{-ik_3x}, & \text{$x>a$}. \end{cases} \end{equation}
Here we assume the particle is travelling to the right from $x=-\infty$ and $\phi(x)$ are the eigenstates
Propagation is only well-defined if you have a time-dependent component. For the Schrodinger equation, you would always have to multiply by something like $e^{-i \omega t}$ to find the time-dependent part. You would get a wave such as $e^{i(kx-\omega t)}$, which shows that for increasing $t$, we need to have increasing $x$ to keep the same phase and thus the wave is right-propagating. The imaginary term does not matter for this definition.