f+2u\right)  }{3}}\theta$$ is a pseudo-Einstein contact form if and only if $$P_{1}f=iA_{11}\gamma_{\overline{1}}+R\gamma_{1}-\gamma_{1,1\overline{1}}.\label{27A}$$

By the fact that $\gamma_{1,\overline{1}}=0$, it’s easy to see $$\int_{M}\left(  P_{1}f\right)  \gamma_{\overline{1}}d\mu=\int_{M}\left(
f_{\overline{1}11}+iA_{11}f_{\overline{1}}\right)  \gamma_{\overline{1}}d\mu=i\int_{M}A_{11}f_{\overline{1}}\gamma_{\overline{1}}d\mu.$$

Then, substituting $\left(  \ref{27A}\right)  $ into the last equality and adding its conjugation, we have$$-\int_{M}Tor\left(  d_{b}f,\gamma \right)  d\mu=\int_{M}\left(  2R-Tor\right)
\left(  \gamma,\gamma \right)  d\mu+2\int_{M}\left \vert \gamma_{1,1}\right \vert
^{2}d\mu.\label{28A}$$

On the other hand, the equality $\left(  \ref{27A}\right)  $ and the commutation formulas enable us to get$$\begin{array}
[c]{ccl}\int_{M}\left(  P_{1}f\right)  f_{\overline{1}}d\mu & = & \int_{M}\left(
iA_{11}\gamma_{\overline{1}}+R\gamma_{1}\right)  f_{\overline{1}}d\mu-\int
_{M}\gamma_{1}f_{\overline{1}\overline{1}1}d\mu \\
& = & \int_{M}\left(  iA_{11}\gamma_{\overline{1}}+R\gamma_{1}\right)
f_{\overline{1}}d\mu \\
&  & +\int_{M}\gamma_{1}\left(  -f_{\overline{1}1\overline{1}}+if_{\overline
{1}0}-R\gamma_{1}f_{\overline{1}}\right)  d\mu \\
& = & \int_{M}iA_{11}\gamma_{\overline{1}}f_{\overline{1}}d\mu+\int_{M}\gamma_{1}\left(  -f_{\overline{1}1\overline{1}}+if_{\overline{1}0}\right)
d\mu \\
& = & \int_{M}iA_{11}\gamma_{\overline{1}}f_{\overline{1}}d\mu+\int_{M}\gamma_{1}\left(  -f_{\overline{1}1\overline{1}}+if_{0\overline{1}}-iA_{\overline{1}\overline{1}}f_{1}\right)  d\mu \\
& = & i\int_{M}\left(  A_{11}\gamma_{\overline{1}}f_{\overline{1}}-A_{\overline{1}\overline{1}}\gamma_{1}f_{1}\right)  d\mu \\
& = & -\int_{M}Tor\left(  d_{b}f,\gamma \right)  d\mu.
\end{array}$$

By the definition of  the CR Paneitz operator, we obtain $$\int_{M}\left(  P_{0}f\right)  fd\mu=-\int_{M}\left(  \left(  P_{1}f\right)
f_{\overline{1}}+\left(  P_{\overline{1}}f\right)  f_{1}\right)  d\mu
=2\int_{M}Tor\left(  d_{b}f,\gamma \right)  d\mu \label{29A}$$

Therefore, it follows from the equalities $\left(  \ref{28A}\right)  $ and $\left(  \ref{29A}\right)  $ that $$\int_{M}\left(  2R-Tor\right)  \left(  \gamma,\gamma \right)  d\mu+2\int
_{M}\left \vert \gamma_{1,1}\right \vert ^{2}d\mu+\frac{1}{2}\int_{M}\left(
P_{0}f\right)  fd\mu=0.$$

Then we are done.

Such equality enables us to prove Theorem \[t2\] as follows:

(Proof of Theorem \[t2\]) From the equality $\left(  \ref{26A}\right)  $ and 