Solution:
Given that
$\left|z_{1}\right|=\left|z_{2}\right|$ and $\arg \left(z_{1}\right)+\arg \left(z_{2}\right)=\pi$
Let’s assume $\arg \left(z_{1}\right)=\theta$
$\arg \left(z_{2}\right)=\Pi-\theta$
It is known that in the polar form, $z=|z|(\cos \theta+i \sin \theta)$
$\begin{array}{l}
z_{1}=\left|z_{1}\right|(\cos \theta+i \sin \theta) \ldots \ldots \ldots \text { (i) } \\
z_{2}=\left|z_{2}\right|(\cos (\pi-\theta)+i \sin (\pi-\theta)) \\
=\left|z_{2}\right|(-\cos \theta+i \sin \theta) \\
=-\left|z_{2}\right|(\cos \theta-i \sin \theta)
\end{array}$
Let us now find the conjugate of
$\bar{z}_{2}=-\left|z_{2}\right|(\cos \theta+i \sin \theta) \ldots \ldots \text { (ii) }\left(\text { since, }\left|\overline{Z_{2}}\right|=\left|Z_{2}\right|\right)$
So now,
$\begin{array}{l}
z_{1} / \bar{z}_{2}=\left[\left|z_{1}\right|(\cos \theta+i \sin \theta)\right] /\left[-\left|z_{2}\right|(\cos \theta+i \sin \theta)\right] \\
=-\left|z_{1}\right| /\left|z_{2}\right|\left[\text { since, }\left|z_{1}\right|=\left|z_{2}\right|\right] \\
=-1
\end{array}$
On cross multiplying we get,
$\begin{array}{l}
\mathrm{Z}_{1}=- \overline{\mathrm{Z}_{2}}
\end{array}$
As a result, hence proved.