The time domain Poynting relationship was found to be

\label{eqn:poyntingTimeHarmonic:20}
0
=
\spacegrad \cdot \lr{ \BE \cross \BH }
+ \frac{\epsilon}{2} \BE \cdot \PD{t}{\BE}
+ \frac{\mu}{2} \BH \cdot \PD{t}{\BH}
+ \BH \cdot \BM_i
+ \BE \cdot \BJ_i
+ \sigma \BE \cdot \BE.

Let’s derive the equivalent relationship for the time averaged portion of the time-harmonic Poynting vector. The time domain representation of the Poynting vector in terms of the time-harmonic (phasor) vectors is

\label{eqn:poyntingTimeHarmonic:40}
\begin{aligned}
\boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}}
&= \inv{4}
\lr{
\BE e^{j\omega t}
+ \BE^\conj e^{-j\omega t}
}
\cross
\lr{
\BH e^{j\omega t}
+ \BH^\conj e^{-j\omega t}
} \\
&=
\inv{2} \textrm{Re} \lr{ \BE \cross \BH^\conj + \BE \cross \BH e^{2 j \omega t} },
\end{aligned}

so if we are looking for the relationships that effect only the time averaged Poynting vector, over integral multiples of the period, we are interested in evaluating the divergence of

\label{eqn:poyntingTimeHarmonic:60}
\inv{2} \BE \cross \BH^\conj.

The time-harmonic Maxwell’s equations are
\label{eqn:poyntingTimeHarmonic:80}
\begin{aligned}
\spacegrad \cross \BE &= – j \omega \mu \BH – \BM_i \\
\spacegrad \cross \BH &= j \omega \epsilon \BE + \BJ_i + \sigma \BE \\
\end{aligned}

The latter after conjugation is

\label{eqn:poyntingTimeHarmonic:100}
\spacegrad \cross \BH^\conj = -j \omega \epsilon^\conj \BE^\conj + \BJ_i^\conj + \sigma^\conj \BE^\conj.

For the divergence we have

\label{eqn:poyntingTimeHarmonic:120}
\begin{aligned}
\spacegrad \cdot \lr{ \BE \cross \BH^\conj }
&=
\BH^\conj \cdot \lr{ \spacegrad \cdot \BE }
-\BE \cdot \lr{ \spacegrad \cdot \BH^\conj } \\
&=
\BH^\conj \cdot \lr{ – j \omega \mu \BH – \BM_i }
– \BE \cdot \lr{ -j \omega \epsilon^\conj \BE^\conj + \BJ_i^\conj + \sigma^\conj \BE^\conj },
\end{aligned}

or

\label{eqn:poyntingTimeHarmonic:140}
0
=
\spacegrad \cdot \lr{ \BE \cross \BH^\conj }
+
\BH^\conj \cdot \lr{ j \omega \mu \BH + \BM_i }
+ \BE \cdot \lr{ -j \omega \epsilon^\conj \BE^\conj + \BJ_i^\conj + \sigma^\conj \BE^\conj },

so
\label{eqn:poyntingTimeHarmonic:160}
\boxed{
0
=
\spacegrad \cdot \inv{2} \lr{ \BE \cross \BH^\conj }
+ \inv{2} \lr{ \BH^\conj \cdot \BM_i
+ \BE \cdot \BJ_i^\conj }
+ \inv{2} j \omega \lr{ \mu \Abs{\BH}^2 – \epsilon^\conj \Abs{\BE}^2 }
+ \inv{2} \sigma^\conj \Abs{\BE}^2.
}