Deriving relation between amplitude of $\vec{B}$ and $\vec{E}$ in EM waves From the complex form of wave function of $\vec{B}$ and $\vec{E}$ (propagation along $z$)
ie. $\vec{E}(z,t) = \vec{E}_0 \, e^{i(kz- \omega t)}$
and using the relation ${\vec\nabla \times} \,\vec{E} = -\partial \vec{B}/\partial t$
How do I get the relation between electric and magnetic amplitudes mathematically?
Since $\vec{E}$ and $\vec{B}$ are both functions of $z$ so $\rm curl$ w.r.t. to any other coordinate axis will be $0$?
 A: If $\mathbf{E}$ is propagating along $\hat{\mathbf{z}}$, it does not mean its polarization is along the same axis. For instance, if $\mathbf{E_{0}} = E_{0}\hat{\mathbf{x}}$, then $\mathbf{B_{0}} = B_{0}\hat{\mathbf{y}}$. The Faraday law gives you an immediate relation between the derivatives of the electric components and the time derivatives of the magnetic field components from which you can get the equation that relates the amplitudes:
$$ \mathbf{B_{0}} = \frac{1}{c}(\hat{\mathbf{z}} \times \mathbf{E_{0}})$$
A: The magnetic filed has also the solution
$$\vec{B}(z,t) = \vec{B}_0 \, e^{i(kz- \omega t)}$$
with
$$\vec\nabla\cdot \vec E=0\quad ,
\vec\nabla\cdot \vec B=0\quad\Rightarrow\\
\vec k\cdot\vec E_0=0\quad,\vec k\cdot\vec B_0=0$$
thus $~\vec E_0\perp\vec k~,\vec B_0\perp\vec k$
with
$$\frac{1}{c^2}\vec E^T\,\left(\vec\nabla\times \vec E+\frac{\partial \vec B}{\partial t}\right)-
\vec B^T\,\left(\vec\nabla\times \vec B-\frac{1}{c^2}\frac{\partial \vec E}{\partial t}\right)=0\tag 1$$
you obtain
$$~\vec E_0\cdot\vec B_0=0$$

Maxwell equations
$$ \vec\nabla\times\vec E+\frac{\partial\vec B}{\partial t}=0\\
\vec\nabla\times\vec B-\frac{1}{c2}\frac{\partial\vec E}{\partial t}=0$$
Equation (1)
$$\frac{1}{c^2}\vec E^T\,\left(\vec\nabla\times \vec E+\frac{\partial \vec B}{\partial t}\right)-
\vec B^T\,\left(\vec\nabla\times \vec B-\frac{1}{c^2}\frac{\partial \vec E}{\partial t}\right)=0$$
$$-\frac{2\,i\,\rm e^{2\,i\,(k\,z-\omega\,t)}}{c^2}\vec E_0\cdot\vec B_0=0$$
