Question

If $\left|\frac{z_1}{z_2}\right|=1$ and $\arg \left(z_1{z}_2\right)=0$, then

• A. $z_1=z_2$
• B. ${\left|z_2\right|}^2=z_1{z}_2$
• C. $z_1{z}_2=1$
• D. None of these

Answer 1

Answer: (B)

Let $z_1={\gamma }_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)$
Given that, $\left|\frac{z_1}{z_2}\right|=1$
$\Rightarrow \left|z_1\right|=\left|z_2\right|$
$\Rightarrow \left|z_1\right|=\left|z_2\right|={\gamma }_1$
Now, $\arg \left(z_1{z}_2\right)=0$
$\Rightarrow \arg \left(z_1\right)+\arg \left(z_2\right)=0$
$\Rightarrow \arg \left(z_2\right)=-{\theta }_1$
Therefore, $z_2={\gamma }_1(\cos \left(-{\theta }_1\right)+i\sin (-{\theta }_1)$
$={\gamma }_1\left(\cos {\theta }_1-i\sin {\theta }_1\right)={\bar{z}}_1$
$\Rightarrow {\bar{z}}_2=\left({\bar{z}}_1\right)=z_1$
$\Rightarrow z_2{\bar{z}}_2=z_1{z}_2$
$\Rightarrow {\left|z_2\right|}^2=z_1{z}_2$