If $z_{1}$ and $z_{2}$ are two complex numbers such that $\left|z_{1}\right|=\left|z_{2}\right|$ and $\arg \left(z_{1}\right)+\arg \left(z_{2}\right)=\pi$, then show that $z_{1}=-\overline{z_{2}}$.
Let $\theta_{1}$ be the $\arg \left(z_{1}\right)$ and $\theta_{2}$ be the $\arg \left(z_{2}\right)$.
It is given that $\left|z_{1}\right|=\left|z_{2}\right|$ and $\arg \left(z_{1}\right)+\arg \left(z_{2}\right)=\pi$.
Since, $z_{1}$ is a complex number.
$z_{1}=\left|z_{1}\right|\left(\cos \theta_{1}+i \sin \theta_{1}\right)$
$=\left|z_{2}\right|\left[\cos \left(\pi-\theta_{2}\right)+i \sin \left(\pi-\theta_{2}\right)\right]$
$=\left|z_{2}\right|\left[-\cos \left(\theta_{2}\right)+i \sin \left(\theta_{2}\right)\right]$
$=-\left|z_{2}\right|\left[\cos \left(\theta_{2}\right)-i \sin \left(\theta_{2}\right)\right]$
$=-\overline{z_{2}}$
Hence, $z_{1}=-\overline{z_{2}}$
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