Question

If $z_1$ and $z_2$ are two non zero complex numbers, satisfying the equation $|z_1|=|z_2|+|z_1-z_2|,$ then which of the following is/are true

• A. $\text{Im}\left(\frac{z_1}{z_2}\right)=0$
• B. $\text{Im}\left(z_1{z}_2\right)=0$
• C. $\arg \left(z_1{z}_2\right)=0$
• D. $\arg \left(\frac{z_1}{z_2}\right)=0$

Answer 1

Answer: (A), (D)

The correct options are
A $\text{Im}\left(\frac{z_1}{z_2}\right)=0$
D $\arg \left(\frac{z_1}{z_2}\right)=0$

We have,
$|z_1|=|z_2|+|z_1-z_2||z_1|-|z_2|=|z_1-z_2|$

Squaring both the sides, we get
$\Rightarrow (|z_1|-|z_2|{)}^2=|z_1-z_2{|}^2\Rightarrow |z_1{|}^2+|z_2{|}^2-2|z_1||z_2|=|z_1{|}^2+|z_2{|}^2-2|z_1||z_2|\cos ({\theta }_1-{\theta }_2)$
$({\theta }_1,{\theta }_2$ are the arguments of $z_1$ and $z_2$ respectively$)$
$\Rightarrow \cos ({\theta }_1-{\theta }_2)=1\Rightarrow {\theta }_1-{\theta }_2=0\Rightarrow \arg \left(z_1\right)-\arg \left(z_2\right)=0$
$\Rightarrow \arg \left(\frac{z_1}{z_2}\right)=0$
$\Rightarrow \frac{z_1}{z_2}$ is purely real
$\Rightarrow \text{Im}\left(\frac{z_1}{z_2}\right)=0$

$\text{Alternate Solution}$
$|z_1-z_2|=|z_1|-|z_2|$
So, $z_1,z_2,0$ are collinear and $z_1,z_2$ lies on same side of origin
$\Rightarrow \arg \left(z_1\right)=\arg \left(z_2\right)$
$\Rightarrow \arg z_1-\arg z_2=0$
$\Rightarrow \arg \left(\frac{z_1}{z_2}\right)=0$
So, $\frac{z_1}{z_2}=k(k>0)$
$\Rightarrow Im\left(\frac{z_1}{z_2}\right)=0$