Home exercise 5.6.1: Phasors in both directions

The $x$-term is already a cosine, with the phasor $5e^{j30^{\circ }}{e}^{-j\beta z}$. For the $y$-term, $-2\sin \theta =2\cos (\theta +90^{\circ })$, so its phasor is $2e^{j90^{\circ }}{e}^{-j\beta z}=j2e^{-j\beta z}$:

$3-j4=5e^{-j53.13^{\circ }}$, so

With $u=\mathrm{Re}\{U{e}^{j\omega t}\}$ and $U$ independent of $t$, the derivative passes through the real part:

Choosing $C=0$, the zero-mean antiderivative and the only one that is itself time-harmonic, the phasors are $-{\omega }^{2}U$ and $U/(j\omega )$.

At $t=0$, $z=0$: (a) gives $E_x=\mathrm{Re}\{4.33+j2.5\}=4.33=5\cos 30^{\circ }$ and $E_y=\mathrm{Re}\{j2\}=0=-2\sin 0$, and (b) gives $H_y=\mathrm{Re}\{3-j4\}=3=5\cos 53.13^{\circ }$.