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 $z+\bar{z}=0$