Question

For any two complex numbers $z_1,z_2$ we have $|z_1+z_2{|}^2=|z_1{|}^2+|z_2{|}^2.$ Then

• A. $Re\left(\frac{z_1}{z_2}\right)=0$
• B. $Im\left(\frac{z_1}{z_2}\right)=0$
• C. $Re\left(z_1{z}_2\right)=0$
• D. $Im\left(z_1{z}_2\right)=0$

Answer 1

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

We know that,
$|z_1+z_2{|}^2=|z_1{|}^2+|z_2{|}^2+2Re\left(z_1{\overline{z}}_2\right)\dots \left(i\right)$
Given,
$|z_1+z_2{|}^2=|z_1{|}^2+|z_2{|}^2\dots \left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right),$ we get
$2Re\left(z_1\overline{z_2}\right)=0$
$\Rightarrow z_1\overline{z_2}+\overline{z_1\overline{z_2}}=0(\because 2Re(z)=z+\overline{z},\overline{\overline{z}}=z)$
$\Rightarrow z_1\overline{z_2}+\overline{z_1}{z}_2=0$
$\Rightarrow z_1\overline{z_2}=-\overline{z_1}{z}_2\Rightarrow \frac{z_1}{z_2}=-\frac{\overline{z_1}}{\overline{z_2}}\Rightarrow \frac{z_1}{z_2}+\overline{\left(\frac{z_1}{z_2}\right)}=0(\because \overline{\left(\frac{z_1}{z_2}\right)}=\frac{\overline{z_1}}{\overline{z_2}})$
$\Rightarrow 2Re\left(\frac{z_1}{z_2}\right)=0$
$\Rightarrow Re\left(\frac{z_1}{z_2}\right)=0$

$\text{Alternate Solution}$
We know that,
$|z_1+z_2{|}^2=|z_1{|}^2+|z_2{|}^2+2Re\left(z_1{\overline{z}}_2\right)\dots \left(i\right)$
Given,
$|z_1+z_2{|}^2=|z_1{|}^2+|z_2{|}^2\dots \left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right),$ we get
$2Re\left(z_1\overline{z_2}\right)=0$
$\Rightarrow Re\left(\frac{z_1{z}_2\overline{z_2}}{z_2}\right)=0$
$[\because z\overline{z}=|z{|}^2]$
$\Rightarrow Re\left(\frac{z_1}{z_2}|z_2{|}^2\right)=0$
$\Rightarrow \frac{z_1}{z_2}|z_2{|}^2$ is purely imaginary
So, $\frac{z_1}{z_2}$ will also be purely imaginary
$\therefore Re\left(\frac{z_1}{z_2}\right)=0$