Q105 / 200 · Manipal (MET) 2012 · Mathematics

If $z_1$ and $z_2$ are two complex numbers such that $\left|z_1\right|=\left|z_2\right|+\left|z_1-z_2\right|$, then

  • Correct answer
    A
    $\operatorname{Im}\left(\frac{z_1}{z_2}\right)=0$
  • B
    $\operatorname{Re}\left(\frac{z_1}{z_2}\right)=0$
  • C
    $\operatorname{Re}\left(\frac{z_1}{z_2}\right)=\operatorname{Im}\left(\frac{z_1}{z_2}\right)$
  • D
    None of these

Solution

$\begin{array}{lc}\because & \left|z_1\right|=\left|z_2\right|+\left|z_1-z_2\right| \\ \Rightarrow & \left|z_1\right|-\left|z_2\right|=\left|z_1-z_2\right| \\ \Rightarrow & \left(\left|z_1\right|-\left|z_2\right|\right)^2=\left|z_1-z_2\right|^2 \\ \Rightarrow & \left|z_1\right|^2+\left|z_2\right|^2-2\left|z_1\right|\left|z_2\right|=\left|z_1\right|^2+\left|z_2\right|^2 \\ -2\left|z_1\right|\left|z_2\right| \cos \left(\theta_1-\theta_2\right) \\ \Rightarrow & \cos \left(\theta_1-\theta_2\right)=1 \\ \Rightarrow & \quad \theta_1-\theta_2=0 \\ \Rightarrow & \arg \left(z_1\right)-\arg \left(z_2\right)=0 \\ \Rightarrow & \frac{z_1}{z_2} \text { is purely real. } \\ \Rightarrow & \operatorname{lm}\left(\frac{z_1}{z_2}\right)=0\end{array}$

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

Relevant Questions