Let $\omega=\frac{z-1}{z+1}$
Then, using componendo and dividendo, we get
$$
\begin{array}{l}
\quad z=\frac{1+\omega}{1-\omega} \\
\Rightarrow \quad|z|=\left|\frac{\omega+1}{\omega-1}\right| \\
\Rightarrow \quad|z|=1 \\
\Rightarrow \quad|\omega+1|=|\omega-1|
\end{array}
$$
Let $6=u+i v$
Then, $|\omega+1|=|u+i v+1|$
$$
=|(u+1)+i v|=\sqrt{(u+1)^{2}+v^{2}}
$$
and $|\omega-1|=\sqrt{(t u-1)^{2}+v^{2}}$
$\therefore$ From Eq. (i)
$$
\begin{array}{rlr}
& \sqrt{(u+1)^{2}+v^{2}}=\sqrt{(u-1)^{2}+v^{2}} \\
\Rightarrow & (u+1)^{2}+v^{2} & =(v-1)^{2}+v^{2} \\
\Rightarrow & u=0
\end{array}
$$
$\therefore \omega=\frac{z-1}{z+1}$ is a pure imaginary number.