Problem 5.6.14: Instantaneous to phasor

$\omega =4\times 10^{8}rad/s$, $\beta =4rad/m$, and $\omega t-\beta x$ means travel towards $+x$.

and with $e^{-j\pi /2}=-j$:

$\mathrm{Re}\{-0.8j{e}^{-j4x}\}=-0.8\sin 4x$, the snapshot at $t=0$. At $x=0$, $\mathrm{Re}\{-0.8j{e}^{j\omega t}\}=0.8\sin \omega t$: the field peaks a quarter period ($T/4=3.93\mathrm{ns}$) after the cosine reference. The factor $-j$ means a lag of $90^{\circ }$ (Figure 5.6.26).