Q63 / 200 · Manipal (MET) 2012 · Mathematics

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

  • A
    $z_1=z_2$
  • Correct answer
    B
    $\left|z_2\right|^2=z_1 z_2$
  • C
    $z_1 z_2=1$
  • D
    None of these

Solution

Let $z_1=\gamma_1\left(\cos \theta_1+i \sin \theta_1\right)$. Given that, $\left|\frac{z_1}{z_2}\right|=1$ $\Rightarrow\left|q_1\right|=\left|z_2\right| \Rightarrow\left|z_1\right|=\left|z_2\right|=\gamma_1$
Now, $\arg \left(z_1 z_2\right)=0 \Rightarrow \arg \left(z_1\right)+\arg \left(z_2\right)=0$ $\Rightarrow \quad \arg \left(z_2\right)=-\theta_1$
Therefore, $z_2=\gamma_1\left(\cos \left(-\theta_1\right)+i \sin \left(-\theta_1\right)\right.$ $\begin{array}{cc} & =\gamma_1\left(\cos \theta_1-i \sin \theta_1\right)=\bar{z}_1 \\ \Rightarrow & \bar{z}_2=\left(\bar{z}_1\right)=z_1 \Rightarrow z_2 \bar{z}_2=z_1 z_2 \\ \Rightarrow & \left|z_2\right|^2=z_1 z_2 \end{array}$

Hence, the correct answer is option: 'B'.

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