Question

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

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

Answer 1

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

$Since,$
$\left|z_1\right|=\left|z_2\right|+|z_1-z_2|$
$\therefore \left|z_1\right|-\left|z_2\right|=|z_1-z_2|$
$\Rightarrow {(\left|z_1\right|-\left|z_1\right|)}^2={|z_1-z_2|}^{2}squaringonbothsides$
$\Rightarrow {\left|z_1\right|}^2{+\left|z_2\right|}^2-2\left|z_1\right|\left|z_2\right|={\left|z_1\right|}^2+{\left|z_2\right|}^2-2\left|z_1\right|\left|z_2\right|\cos ({\theta }_1-{\theta }_2)$
$\Rightarrow \cos ({\theta }_1-{\theta }_2)=1\Rightarrow ({\theta }_1-{\theta }_2)=0$
$\Rightarrow arg\left(z_1\right)-arg\left(z_2\right)=0$
$\Rightarrow \frac{z_1}{z_2}ispurelyrear$
$\Rightarrow Im\left(\frac{z_1}{z_2}\right)=0$