Q73 / 160 · WBJEE 2012 · Mathematics

If $\frac{z-1}{z+1}$ is purely imaginary, then

  • A
    $|z|=\frac{1}{2}$
  • Correct answer
    B
    $|z|=1$
  • C
    $|z|=2$
  • D
    $|z|=3$

Solution

Let $\omega=\frac{z-1}{z+1}$ Then, using componendo and dividendo, we get $$ \begin{array}{l} \quad z=\frac{1+\omega}{1-\omega} \\ \Rightarrow \quad|z|=\left|\frac{\omega+1}{\omega-1}\right| \\ \Rightarrow \quad|z|=1 \\ \Rightarrow \quad|\omega+1|=|\omega-1| \end{array} $$ Let $6=u+i v$ Then, $|\omega+1|=|u+i v+1|$ $$ =|(u+1)+i v|=\sqrt{(u+1)^{2}+v^{2}} $$ and $|\omega-1|=\sqrt{(t u-1)^{2}+v^{2}}$ $\therefore$ From Eq. (i) $$ \begin{array}{rlr} & \sqrt{(u+1)^{2}+v^{2}}=\sqrt{(u-1)^{2}+v^{2}} \\ \Rightarrow & (u+1)^{2}+v^{2} & =(v-1)^{2}+v^{2} \\ \Rightarrow & u=0 \end{array} $$ $\therefore \omega=\frac{z-1}{z+1}$ is a pure imaginary number.

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

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