Q
$$
\begin{aligned}
\bar{z}=\frac{1}{z}, \bar{\omega}=\frac{1}{\omega}, \Rightarrow|z| & =1=|\omega \mid \\
\because \frac{z+\omega}{1+z \omega}-\left(\frac{\bar{z}+\bar{\omega}}{1+\bar{z} \bar{\omega}}\right) & =\frac{z+\omega}{1+z \omega}-\left(\frac{1 / z+1 / \omega}{1+1 / z \omega}\right) \\
& =\frac{z+\omega}{1+z \omega}-\frac{z \omega}{z \omega}\left(\frac{\frac{1}{z}+\frac{1}{\omega}}{1+\frac{1}{z \omega}}\right) \\
& =\frac{z+\omega}{1+z \omega}-\left(\frac{\omega+z}{z \omega+1}\right)=0 .
\end{aligned}
$$
Thus; $\Rightarrow \frac{z+\omega}{1+z \omega}=\frac{\bar{z}+\bar{\omega}}{1+\bar{z} \bar{\omega}}=\left(\overline{\frac{z+\omega}{1+z \omega}}\right), \Rightarrow \frac{z+\omega}{1+z \omega}$ is purely neal