To approach this problem, let’s go through each part carefully:

Part a: Show that the fields $E(z,t) = A \sin(kz - \omega t) \mathbf{i}$ and $B(z,t) = A \sin(kz - \omega t) \mathbf{j}$ satisfy the equation $\nabla \times \mathbf{B} = C \frac{\partial \mathbf{E}}{\partial t}$, given $\omega = \frac{k}{C}$.

1. Calculate $\frac{\partial \mathbf{E}}{\partial t}$:

   • Given $\mathbf{E}(z,t) = A \sin(kz - \omega t) \mathbf{i}$,
   • Taking the time derivative: $\frac{\partial \mathbf{E}}{\partial t} = -A \omega \cos(kz - \omega t) \mathbf{i}$

2. Calculate $\nabla \times \mathbf{B}$:

   • Given $\mathbf{B}(z,t) = A \sin(kz - \omega t) \mathbf{j}$,
   • We need to compute $\nabla \times \mathbf{B}$.
   • In Cartesian coordinates, the curl of $\mathbf{B}$ for a function only dependent on $z$ is: $\nabla \times \mathbf{B} = \left( \frac{\partial B_y}{\partial z} \right) \mathbf{i} - \left( \frac{\partial B_x}{\partial z} \right) \mathbf{j}$
   • With ( \mathbf{B} = A