No magnetic field in the case of oscillating electric charges - proof I am reading the book Plasma dynamics by Dendy (2002) and I stumbled at a derivation of a proof. 
I do not quite understand why equation $1.23$ is equal to zero. Also, equation I.$9$ is $\nabla\times\mathbf{H}=\mathbf{J}+\epsilon_0\frac{\partial\mathbf{E}}{\partial t}$.
I would appreciate any help/hint. Thank you.

 A: If we linearize the equations, we make the assumption that all quantities can be written as:
$$
Q\left( \mathbf{x}, t \right) = Q_{o} + \delta Q
$$
where $Q_{o}$ is constant and $\delta Q \propto e^{i \left( \mathbf{k} \cdot \mathbf{x} - \omega \ t \right)}$.  This means that operators can be replaced in the following way:
$$
\begin{align}
  \nabla & = i \ \mathbf{k} \\
  \partial/\partial t & = -i \ \omega
\end{align}
$$
In the case of an electrostatic wave like the Langmuir oscillations to which you refer, the particles oscillate along the quasi-static magnetic field.  The wave itself is a longitudinal osciallation, i.e., the wavevector is aligned with electric field oscillations.
So if we look at Faraday's law, and we use the linear approximation above knowing that $\mathbf{k} \ \parallel \ \delta \mathbf{E}$ (i.e., the cross product of two parallel vectors is null) we find that:
$$
\begin{align}
  \nabla \times \mathbf{E} & = -\frac{\partial \mathbf{B}}{\partial t} \\
  \mathbf{k} \times \delta \mathbf{E} & = -i \ \omega \ \delta \mathbf{B} \\
  & = 0
\end{align}
$$
Thus, if we take the partial time derivative of Ampere's law (i.e., Equation 1.22 in your image), we find:
$$
\begin{align}
  \frac{\partial}{\partial t} \left( \nabla \times \mathbf{B} \right) & = \frac{\partial}{\partial t} \left( \mu_{o} \mathbf{j} + \frac{1}{c^{2}} \frac{\partial \mathbf{E}}{\partial t} \right) \\
  \nabla \times \frac{\partial \mathbf{B}}{\partial t} & = \mu_{o} \frac{\partial \mathbf{j}}{\partial t} + \frac{1}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}} \\
  0 & = \mu_{o} \frac{\partial \mathbf{j}}{\partial t} + \frac{1}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}
\end{align}
$$
which is precisely what is shown in your Equation 1.23.
A: Not sure if I understand your question...
Assuming that equation 1.23 is correct, and that the electric field is of the form $$E = E_0 e^{i\omega t}$$
(namely, "oscillating" with simple harmonic motion), then the second derivative with respect to time exactly cancels the $\omega^2 E$ term... and thus it's zero.
Is that what you were asking?
