 Solution

Write ${\gamma }^{\pm }=-i{\beta }^{\pm }$, $\overline{\beta }=({\beta }^{+}+{\beta }^{-})/2$ and $\chi =({\beta }^{+}-{\beta }^{-})z/2$. Then $\gamma z=-i\overline{\beta }z$ and $\delta z=-i\chi$. With $\cosh (-i\chi )=\cos \chi$ and $\sinh (-i\chi )=-i\sin \chi$, the corrected matrix becomesThe scalar is a common propagation wave phase and the real rotation matrix rotates the input linear polarization through $\chi$. There is no attenuation and no generated ellipticity: the field map is a unitary matrix. The two circular eigenmodes accumulate different wave phases, giving circular birefringence and optical activity.

The correct unfactored matrix remains finite where $\cos \chi =0$. In contrast, the printed $\mathrm{sech}$ form would give $e^{i\overline{\beta }z}{\sec }^2\chi$ times the rotation matrix, including unphysical divergent gain in this stipulated lossless case. This provides a direct physical check on the correction in part (d).