Q170 / 270 · NDA 2023 (Phase 1) · Mathematics

If $z$ is a complex number such that $\frac{z-1}{z+1}$ is purely imaginary, then what is $|z|$ equal to?

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

Solution

Let $z=x+i y$ $\frac{z-1}{z+1}=\frac{(x-1)+i y}{(x+1)+i y}$ $=\frac{[(x-1)+i y] \times[(x+1)-i y]}{[(x+1)+i y] \times[(x+1)-i y]}$ $=\frac{\left(x^2-1\right)+y^2+i[y(x+1)-y(x-1)]}{(x+1)^2+y^2}$ $=\frac{\left(x^2+y^2-1\right)}{(x+1)^2+y^2}+i \frac{(2 y)}{(x+1)^2+y^2}$ $\because$ It is purely imaginary, $\operatorname{Re}\left(\frac{z-1}{z+1}\right)=0$ $\frac{x^2+y^2-1}{(x+1)^2+y^2}=0$ $\therefore \quad x^2+y^2-1=0$ $x^2+y^2=1$ $|z|^2=1$ $|z|=1$

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

Relevant Questions