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)

$\because \left|z_1\right|=\left|z_2\right|+\left|z_1-z_2\right|$
$\Rightarrow \left|z_1\right|-\left|z_2\right|=\left|z_1-z_2\right|$
$\Rightarrow {\left(\left|z_1\right|-\left|z_2\right|\right)}^2={\left|z_1-z_2\right|}^2$
$\Rightarrow {\left|z_1\right|}^2+{\left|z_2\right|}^2-2\left|z_1\right|\left|z_2\right|={\left|z_1\right|}^2+{\left|z_2\right|}^2$
$-2\left|z_1\Vert z_2\right|\cos \left({\theta }_1-{\theta }_2\right)$
$\Rightarrow \cos \left({\theta }_1-{\theta }_2\right)=1$
$\Rightarrow {\theta }_1-{\theta }_2=0$
$\Rightarrow \arg \left(z_1\right)-\arg \left(z_2\right)=0$
$\Rightarrow \frac{z_1}{z_2}$ is purely real.
$\Rightarrow \text{lm}\left(\frac{z_1}{z_2}\right)=0$