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, $z_{2}=\gamma_{1}\left(\cos \left(-\theta_{1}\right)+i \sin (-\theta_{1})\right.$
$=\gamma_{1}\left(\cos \theta_{1}-i \sin \theta_{1}\right)=\bar{z}_{1}$
$\Rightarrow \bar{z}_{2}=\left(\bar{z}_{1}\right) =z_{1}$
$\Rightarrow z_{2} \bar{z}_{2}=z_{1} z_{2}$
$\Rightarrow \left |z_{2}\right|^{2} =z_{1} z_{2}$