Solve the following

Let |z1| = |z2| = r

Let arg z1 = θ1 and arg z2 = θ2

$\Rightarrow z_{1}=r e^{i \theta_{1}}$ and $z_{2}=r e^{i \theta_{2}}$

Since $\arg \left(\frac{z_{1}}{z_{2}}\right)=\pi$

i. e $\arg z_{1}-\arg z_{2}=\pi$

i. $e \theta_{1}-\theta_{2}=\pi$

i. e $\theta_{1}=\theta_{2}+\pi$

$e^{i \theta_{1}}=e^{i\left(\theta_{2}+\pi\right)}$

$=e^{i \pi} e^{i \theta_{2}}$

$=(\cos \pi+i \sin \pi)\left(\cos \theta_{2}+i \sin \theta_{2}\right)$

$=-1\left(\cos \theta_{2}+i \sin \theta_{2}\right)$

$e^{i \theta_{1}}=-e^{i \theta_{2}}$

$\Rightarrow z_{1}+z_{2}=r e^{i \theta_{1}}+r e^{i \theta_{2}}$

Hence, $z_{1}+z_{2}=r\left(e^{i \theta_{1}}+e^{i \theta_{2}}\right)=0$