TM Waves

B_z & = & 0
E_z|_S & = & 0 The magnetic field is strictly transverse, but the electric field in the $z$ direction only has to vanish at the boundary - elsewhere it can have a $z$ component.

Thus: _&perp#perp;& = & i&mu#mu;&epsi#epsilon;&omega#omega;^2 - k^2( k_&perp#perp; E_z - &omega#omega;( ×_&perp#perp;B_z))
(&mu#mu;&epsi#epsilon;&omega#omega;^2 - k^2) _&perp#perp;& = & i k_&perp#perp; E_z
1ik(&mu#mu;&epsi#epsilon;&omega#omega;^2 - k^2) _&perp#perp;& = & _&perp#perp; E_z
which looks just perfect to substitute into: _&perp#perp;& = & i&mu#mu;&epsi#epsilon;&omega#omega;^2 - k^2( k_&perp#perp; B_z + &mu#mu;&epsi#epsilon;&omega#omega;( ×_&perp#perp;E_z))
(&mu#mu;&epsi#epsilon;&omega#omega;^2 - k^2) _&perp#perp;& = & i&mu#mu;&epsi#epsilon;&omega#omega;( ×_&perp#perp;E_z)
(&mu#mu;&epsi#epsilon;&omega#omega;^2 - k^2) _&perp#perp;& = & &mu#mu;&epsi#epsilon;&omega#omega;k(&mu#mu;&epsi#epsilon;&omega#omega;^2 - k^2)( × _&perp#perp;)
giving us: _&perp#perp;= ±&mu#mu;&epsi#epsilon;&omega#omega;k( × _&perp#perp;) or (as the book would have it): _&perp#perp;= ±&epsi#epsilon;&omega#omega;k( × _&perp#perp;) (where as usual the two signs indicate the direction of wave propagation).

Of course, we still have to find at least one of the two fields for this to do us any good. Or do we? Looking above we see: (&mu#mu;&epsi#epsilon;&omega#omega;^2 - k^2) _&perp#perp;& = & i k_&perp#perp; &psi#psi;
_&perp#perp;& = & ±ik(&mu#mu;&epsi#epsilon;&omega#omega;^2 - k^2)_&perp#perp; &psi#psi;
Where $\psi(x,y) e^{ikz} = E_z$ . This must satisfy the transverse wave function: (&nabla#nabla;_&perp#perp;^2 + (&mu#mu;&epsi#epsilon;&omega#omega;^2 - k^2))&psi#psi;= 0 and the boundary conditions for a TM wave: &psi#psi;|_S = 0