Question

Assertion :Statement -1: If $z_1$ and $z_2$ are two complex numbers such that $|z_1|=|z_2|+|z_1-z_2|$, then $Im\left(\frac{z_1}{z_2}\right)=0$ Reason: Statement -2: $arg\left(z\right)=0\Rightarrow z$ is purely real.

• A. Statement -1 is true, Statement -2 is true ; Statement -2 is a correct explanation for Statement -1
• B. Statement-1 is true, Statement-2 is true ; Statement-2 is not a correct explanation for Statement-1
• C. Statement -1 is true, Statement -2 is false
• D. Statement -1 is false, Statement -2 is true

Answer 1

The correct option is A Statement -1 is true, Statement -2 is true ; Statement -2 is a correct explanation for Statement -1

Given, $\left|z_1\right|=\left|z_2\right|+|z_1-z_2|$
$\Rightarrow |z_1-z_2|=\left|z_1\right|-\left|z_2\right|$
$\Rightarrow {|z_1-z_2|}^2=(\left|z_1\right|-\left|z_2\right|{)}^2(if\left|z_1\right|\ge \left|z_2\right|)$
$\Rightarrow {\left|z_1\right|}^2-2\left|z_1\right|\left|z_2\right|\cos ({\theta }_1-{\theta }_2)+|z_2{|}^2={\left|z_1\right|}^2+{\left|z_2\right|}^2-2\left|z_1\right|\left|z_2\right|$
$\Rightarrow \cos ({\theta }_1-{\theta }_2)=1\Rightarrow {\theta }_1-{\theta }_2=2n\pi ,n\in I$
Now
$\frac{z_1}{z_2}=\left|\frac{z_1}{z_2}\right|\left[\cos \right({\theta }_1-{\theta }_2)+i\sin ({\theta }_1-{\theta }_2\left)\right]=\left|\frac{z_1}{z_2}\right|$
$\Rightarrow Im\left(\frac{z_1}{z_2}\right)=0$
Hence, Statement-1 is True.
Hence, option A is correct.