Question

Consider the complex number $z_1$ and $z_2$ satisfying the relation $|z_1+z_2{|}^2=|z_1{|}^2+|z_2{|}^2$, then one of the possible argument of complex number $i\frac{z_1}{z_2}$ is,

• A. $\frac{\pi }{2}$
• B. $-\frac{\pi }{2}$
• C. $0$
• D. none of these

Answer 1

The correct option is C $0$

$|z_1+z_2{|}^2=z_1^2+z_2^2$

$z_1^2+z_2^2+z_1\overline{z_2}+\overline{z_1}{z}_2=z_1^2+z_2^2$

$z_1\overline{z_2}+\overline{z_1}{z}_2=0$

let
$z_1=e^{i\theta }$

$z_2=e^{i\gamma }$

Substituting, in the above equation, we get

$e^{i(\theta -\gamma )}=-e^{-i(\theta -\gamma )}$

$e^{2i(\theta -\gamma )}=-1$

Hence
$2(\theta -\gamma )=(2n-1)\pi$

$\theta -\gamma =\frac{(2n-1)\pi }{2}$

Now $\theta -\gamma =arg\left(z_1\overline{z_2}\right)$

Therefore, $arg\left(z_1\overline{z_2}\right)=\frac{(2n-1)\pi }{2}$

Hence $z_1\overline{z_2}$ is purely imaginary.

Therefore
$i\left(z_1\overline{z_2}\right)$ will be purely real.

$arg\left(i\right(z_1\overline{z_2}\left)\right)$ will be 0 or $\pi$.

Hence, option ${}^{\prime }{C}^{\prime }$ is correct.