Question

One of the possible argument of complex number $i\left(z_1/z_2\right)$

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

Answer 1

The correct option is C $0$

${\left|z_1{+z}_2\right|}^2={\left|z_1\right|}^2+{\left|z_2\right|}^2$
We know that ${\left|z_1{+z}_2\right|}^2=\left(z_1{+z}_2\right)(\overline{z_1}+\overline{z_2})$
${⟹\left|z_1\right|}^2+{\left|z_2\right|}^2=\left(z_1{+z}_2\right)(\overline{z_1}+\overline{z_2})$
$⟹{\left|z_1\right|}^2+{\left|z_2\right|}^2={\left|z_1\right|}^2+{\left|z_2\right|}^2+z_1\overline{z_2}+z_2\overline{z_1}$
$⟹z_1\overline{z_2}+z_2\overline{z_1}=0$ ...(1)
Let $z=\frac{z_1}{z_2}$
$z+\overline{z}=\frac{z_1}{z_2}+\frac{\overline{z_1}}{\overline{z_2}}=\frac{z_1\overline{z_2}+z_2\overline{z_1}}{{\left|z_2\right|}^2}=0$ ...{from (1)}
$⟹Re\left(z\right)=0$

$⟹Re\left(\frac{z_1}{z_2}\right)=0$
$\therefore \frac{z_1}{z_2}$ is purely imaginary.
$⟹i\frac{z_1}{z_2}$ is purely real.
$\therefore arg\left(\frac{{iz}_1}{z_2}\right)=0$

Ans: C