Question

Assertion : If ${\left|z_1+z_2\right|}^2={\left|z_1\right|}^2+{\left|z_2\right|}^2$, then $\frac{z_1}{z_2}$ is purely imaginary. Reason : If $z$ is purely imaginary, then $z+\bar{z}=0$.

• A. Assertion is correct, Reason is correct; reason is a correct explanation for assertion.
• B. Assertion is correct, Reason is correct; reason is not a correct explanation for assertion
• C. Assertion is correct, Reason is incorrect
• D. Assertion is incorrect, Reason is correct.

Answer 1

Answer: (B)

We have, ${\left|z_1+z_2\right|}^2={\left|z_1\right|}^2+{\left|z_2\right|}^2$
$\Rightarrow {\left|z_1\right|}^2+{\left|z_2\right|}^2+2\left|z_1\right|\left|z_2\right|\cos \left({\theta }_1-{\theta }_2\right)={\left|z_1\right|}^2+{\left|z_2\right|}^2$
where ${\theta }_1=\arg \left(z_1\right),{\theta }_2=\arg \left(z_2\right)$
$\Rightarrow \cos \left({\theta }_1-{\theta }_2\right)=0$
$\Rightarrow {\theta }_1-{\theta }_2=\frac{\pi }{2}$
$\Rightarrow arg\left(\frac{z_1}{z_2}\right)=\frac{\pi }{2}$
$\Rightarrow Re\left(\frac{z_1}{z_2}\right)=0$
$\therefore \frac{z_1}{z_2}$ is purely imaginary.
If z is purely imaginary, then $z+\bar{z}=0$