Question

If $z_1$ and $z_2$ are two complex numbers such that $|z_1|=|z_2|+|z_1-z_2|$, then

• A. $\text{Im}\left(\frac{z_1}{z_2}\right)=0$
• B. $\text{Re}\left(\frac{z_1}{z_2}\right)=0$
• C. $\text{Re}\left(\frac{z_1}{z_2}\right)=Im\left(\frac{z_1}{z_2}\right)$
• D. None of these

Answer 1

The correct option is A $\text{Im}\left(\frac{z_1}{z_2}\right)=0$

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