Show that $ E = E_o e^{-α' z} e^{j(ωt - kz)}$ is a possible solution to:
$\frac{d^2 E}{dz^2} - E_o E_r μ_o \frac{d^2 E}{dt^2} = μ_o σ \frac{∂E}{∂t} $
...
$ E = E_o e^{jωt - z(jk + α')}$
$\frac{d E}{dz} = -(jk + α') E_o e^{jωt - z(jk + α')}$
$\frac{d^2E}{dz^2} = (jk + α')^2 E_o e^{jωt - z(jk + α')} = (α'^2 + j2α'k - k^2) E_o e^{jωt - z(jk + α')}$
$\frac{∂E}{∂t} = jω E_o e^{jωt - z(jk + α')} $
How does jw replace the $(α'^2 + j2α'k - k^2)$ term?
Here's screenshot.
