Question

If $z_{1}$ and $z_{2}$ are two complex numbers such that $\left| z_{1}\right|=\left| z_{2}\right|+\left| z_{1}-z_{2}\right|$, then

• A. $\text{Im}\left(\frac{z_{1}}{z_{2}}\right)=0$
• B. $\text{Re}\left(\frac{z_{1}}{z_{2}}\right)=0$
• C. $\text{Re}\left(\frac{z_{1}}{z_{2}}\right)=\text{Im}\left(\frac{z_{1}}{z_{2}}\right)$
• D. None of these

Answer 1

Answer: (A)

$\because \left| z_{1}\right|=\left| z_{2}\right|+\left| z_{1}-z_{2}\right|$
$\Rightarrow \left| z_{1}\right|-\left| z_{2}\right|=\left| z_{1}-z_{2}\right|$
$\Rightarrow \left(\left| z_{1}\right|-\left| z_{2}\right|\right)^{2}=\left| z_{1}-z_{2}\right|^{2}$
$\Rightarrow \left| z_{1}\right|^{2}+\left| z_{2}\right|^{2}-2\left| z_{1}\right|\left| z_{2}\right|=\left| z_{1}\right|^{2}+\left| z_{2}\right|^{2}$
$-2\left|z_{1} \| z_{2}\right| \cos \left(\theta_{1}-\theta_{2}\right)$
$\Rightarrow \cos \left(\theta_{1}-\theta_{2}\right)=1$
$\Rightarrow \theta_{1}-\theta_{2}=0$
$\Rightarrow \arg \left(z_{1}\right)-\arg \left(z_{2}\right)=0$
$\Rightarrow \frac{z_{1}}{z_{2}}$ is purely real.
$\Rightarrow \text{lm}\left(\frac{z_{1}}{z_{2}}\right)=0$